Mensuration I (Area and Perimeter)

Transcription

1 Chapter-0 Mensuration I (Area and Perimeter) Mensuration Mensuration is the science of measurement of the length of lines, areas of surfaces and volumes of solids. Its knowledge is of immense use to the surveyor, architect and engineer. In this chapter, we will deal with the perimeter and area of the plane figures. Perimeter The perimeter of a figure is the sum of length of all its sides. This is measured in the units of length. For example, cm, m, etc. Area The area of any figure is the amount of surface enclosed within its bounding lines. This is measured by the number of square cm or square metres (or other units of square measures) it contains. Measures of Area Area, being the product of two linear measurements (see in the following pages), is expressed in square units of linear measure. The following will give you the measures of area in the Metric System: (i) Square Centimetre: A square centimetre is the amount of surface enclosed within a square of which each side is one centimetre in length. The term square metre, square hectometre, square kilometre etc are used in the same sense. (ii) Square Metre: The area of a region formed by a square of side metre is called a square metre and is written as m. The basic unit of length in SI system is a metre ( m) and the basic unit of area in SI system is a square metre ( m ). Since, m 00 cm m m m 00 cm 00 cm 0000 cm Thus, m 0000 cm. (iii) Square Decimetre: The area of a region formed by a square of side decimetre ( dm) is called a square decimetre and is written as dm. Since, m 0 dm m m m 0 dm 0 dm 00 dm Also, dm 0 cm dm dm dm 0 cm 0 cm 00 cm (iv) Square Millimetre: The area of a region formed by a square of side millimetre ( mm) is called a square millimetre and is written as mm. Since cm 0 mm cm cm cm 0 mm 0 mm 00 mm Also, m 0000 cm m 0000 mm 00 mm mm (v) Square Decametre or an Are: The area of a region formed by a square of side one decametre ( dam) is called a square decametre or an arc is denoted by dam or a. Since dam 0 m dam dam dam 0 m 0 m 00 m Thus, dam 00 m or are 00 m (vi) Square Hectometre or Hectare: The area of a region formed by a square of side hectometre ( hm) is called a square hectometre or a hectare and is denoted by hm or ha. Since, hm 00 m ha hm hm 00 m 00 m 0000 m Also, hm 0 dam ha 0 dam 0 dam 00 dam 00 a. Thus, ha 0000 m or ha 00 a. (vii) Square Kilometre: The area of a region formed by a square of side kilometre ( km) is called a square kilometre and is written as km. Since, km 000 m km 000 m 000 m m Since ha 0000 m. Therefore, km ha 00 ha 0000 Also, km 0000 are

2 50 Concept of Arithmetic We have introduced various standard units of area. Each can be converted into others as given below. Conversion of Units Units of Length Units of Area cm 0 mm cm (0 0) mm 00 mm m 0 dm m (0 0) dm 00 dm dm 0 cm dm (0 0) cm 00 cm m 00 cm m (00 00) cm 0000 cm dam 0 m dam (0 0) m 00 m are hm 00 m hm (00 00) m 0000 m hectare km 0 hm mm (0 0) hm 00 hm km 000 m km ( ) m m Rectangle and Square (iii) Diagonal of a rectangle (ii) Square (Length) (Breadth) ABC is a right-angled triangle. AC AB BC A square is a four-sided figure having all its angles right angles and all its sides are equal. In a square, the length is equal to the breadth. (i) Perimeter of a square length one of its sides (ii) Area of a square side side (side) From this formula we get, Side Area (iii) Diagonal of a square (Area) (i) Rectangle A rectangle is a four-sided figure having all its K angles right angles. KUNDAN The page of this book,the faces of a brick, the floor of a room are rectangles, for their opposite sides are equal and their angles are right angles. The sides of a rectangle are usually called its length and breadth. In the rectangle, the length and breadth are uneqaual. (i) Perimeter of a rectangle (Length + Breadth) (ii) Area of a rectangle (Length Breadth) From this formula we get, Area (a) Length Breadth If ABCD be a square, then AC AB BC AB AB AB [ AB BC] AC AB From the above, we have AB AC Area (AB) (AC) Area (b) Breadth Length Hence, the area of a square is half the square of its diagonal. Solved Examples Ex. : Find the area, in hectare, of a field whose length is 0 m and breadth 0 m. We have, Length of the field 0 m Breadth of the field 0 m. Area of the field (0 0) m m hectare 0000 [ 0000 m hectare].6 hectare Ex. : What will happen to the area of a rectangle if (i) its length is doubled and breadth is trebled. (ii) its length and breadth are doubled? Let l cm and b cm be the length and breadth of the rectangle. Further, let A be the area of the rectangle. Then, A l b... (i)

3 Mensuration I (Area and Perimeter) 50 (i) We have, New breadth l; New length b A New Area l b 6 (l b) 6A [Using (i)] Hence, the area of the new rectangle is 6 times the area of the old rectangle. (ii) We have, New length l, New breadth b A New Area l b (l b) A [Using (i)] Hence, the area of the new rectangle is times the area of the old rectangle. Ex. : What will happen to the area of a square if its side is (i) doubled (ii) halved? (i) Let the side of a square be x cm. Then, Area A x cm...(i) When the side is doubled, then, side of the new square x cm. A Area of the new square (x ) cm x cm A [Using (i)] Thus, if the side is doubled, then area becomes times. (ii) When the side is halved, then, Ex. 6: or, x y (i) And, Perimeter 6 cm or, (x + y) 6 cm or, x + y... (ii) Now, x + y or, (x + y) or, x + y + xy 59 or, 89 + xy 59 or, xy or, xy 0 or, xy 0 Hence, Area xy 0 cm. The length of a rectangle exceeds its width by 8 cm and the area of the rectangle is 0 sq cm. Find the dimensions of the rectangle. Let the breadth of the given rectangle be x cm. Then, length (x + 8) cm. Now, Area 0 cm or, length breadth 0 or, (x + 8) x 0 or, x + 8x 0 0 or, x + 0x x 0 0 or, x (x + 0) (x + 0) 0 or, (x + 0) (x ) 0 or, x or x 0 But x cannot be negative. So, x. Hence, breadth cm and length ( + 8 ) 0 cm. Ex. : Ex. 5: Side of the new square x A Area of the new square x cm x A cm. [Using (i)] Thus, if the side is halved, then area becomes one-fourth. The length of a rectangle is twice its breadth. Find the dimensions of the rectangle if its area is 88 sq cm. Let the breadth of the given rectangle be x cm. Then, length x cm [Given]. Area x x x or, x 88 [ Area 88 sq cm (given)] or, x or, x cm Hence, length cm, and breadth cm. If the diagonal of a rectangle is cm long and the perimeter of the rectangle is 6 cm. Find the area of the rectangle. Let the length and breadth of the given rectangle be x cm and y cm respectively. Th en, Diagonal cm or, x y Ex. : The cost of ploughing a rectangular field at 85 paise per square centimetre is Rs 6.5. Find the perimeter of the field if its sides are in the ratio 5 :. Let the length and the breadth of the rectangular field be 5x cm and x cm in length. Then, Area 5x x 5x cm... (i) It is given that the cost of ploughing the 85 rectangular field at the rate of Rs is 00 Rs 6.5. Area of the rectangular field Total cost Rate per sq m cm... (ii) From (i) and (ii), we get 5x 5 or x 9 or x Hence, the sides of the rectangular field are 5x (5 ) 5 cm and x ( ) cm respectively. Perimeter { (5 + ) } cm

4 50 Concept of Arithmetic Ex. 8: The length and breadth of a rectangular field are in the ratio :. If the area of the field is 56 cm, find the cost of fencing the field at Rs.50 per cm. Let the length and breadth of the rectangular field be x and x cm respectively. Then, Area of the rectangular field (x x) cm 6x cm Also, area of the rectangular field 56 cm (given) 6x 56 or, x 56 6 or, x 56 or, x 56 Length ( ) cm cm Breadth ( ) cm 8 cm Perimeter of the field (length + breadth) [ ( + 8)] cm 0 cm Rate of fencing Rs.50 per cm Cost of fencing Rs (0.50) Rs 80 Ex. 9: Find the area of a rectangular plot one side of which is 8 cm and its diagonal 50 cm. Let the other side be x cm. Since ABC is a right triangle, therefore, AC AB + BC or, x or, x (50) (8) or, x (50 + 8) (50 8) or, x 98 or, x or, x Thus, the other side of the rectangle is cm Area of the rectangle (8 ) cm 6 cm Ex. 0: The area of square ABCD is 6 cm. Find the area of the square joining the midpoints of the sides. We have, Area of square ABCD 6 cm Each side of square ABCD 6 cm cm In APS, we have AP AB cm and AS AD cm Also, PS AP + AS [Using Pythagoras theorem] PS Thus, each side of square PQRS is of length cm. Area of the square PQRS cm cm Ex. : The length of a rectangular field is increased by 50% and its breadth is decreased by 50% to form a new rectangular field. Find the percentage change in the area of the field. Let x and y denote the length and breadth of the given field. Then, its area is given by A xy Increase in length 50% of the original length of x 50 x x 00 x New length x x Decrease in breadth 50% of the original breadth 50% of y 50y y 00 y y New length y A Area of the new field y x xy Change in Area A A xy xy xy Percentage change in the area xy A A A xy 5% Hence, there is 5% decrease in the area. Ex. : A rectangular courtyard is m 8 cm long and 5 m 5 cm broad. It is desired to pave it with square tiles of the same size. What is the largest size of the tile that can be used? Also, find the number of such tiles.

5 Mensuration I (Area and Perimeter) 505 Clearly, the size of the tile should be a factor of both the length and breadth of the courtyard. Therefore, the size of the largest size tile should be the HCF of 8 cm and 55 cm Now, 8 and 55 5 Hence, HCF of 8 and 55 is Thus, the size of the largest size square tile cm Area of a tile (side) cm () cm cm Area of courtyard (8 55) cm 9850 cm Now, Number of tiles Area of courtyard Area of a tile Ex. : The floor of a rectangular hall is to be K covered with KUNDAN a carpet 50 cm wide. If the length and breadth of the hall are 0 m and 8 m respectively, find the cost of the carpet at the rate of Rs 0 per metre. We have, Area of the hall (0 8) m 60 m Width of the carpet 50 cm.5 m Length of the carpet Area of the hall 60 Width of the carpet.5 0 m Rate of the carpet Rs 0 per metre. Cost of the carpet Rs (0 0) Rs 00. Ex. : Find the height of the wall whose length is m and which can be covered by 00 tiles of size 5 cm by 0 cm. Area of a tile (5 0) cm 500 cm. Area of 00 tiles (00 500) cm cm m 0000 [ 0000 cm m ] 0 m. Let the height of the wall be h metres. Th en, area of the wall h m. Since 00 tiles completely cover the wall. Therefore, Area of the wall Area of 00 tiles or, h 0 h 0 or, [Dividing both sides by ] or, h 0 Hence, the height of the wall is 0 metres. Ex. 5: Find the perimeter of a rectangular field whose length is four times its width and which has an area equal to 096 cm. Let the width of the field be b cm. Then, Length of the field b cm. Area of the field (b b) cm b cm But, area of the field is given as 096 cm. b or, b or, b (88) or, b 88 Length of the field b cm ( 88) cm 5 cm Width of the field b cm 88 cm Hence, Perimeter of the field (length + breadth) (5 + 88) cm 880 cm Ex. 6: A 5 m wide lane was paved with bricks of size 0 cm by 5 cm. If the rate of bricks was Rs 50 per thousand and if bricks worth Rs 9500 were used for pavements, find the length of the lane. We have, Rate of bricks Rs 50 per thousand Total cost of bricks Rs Number of bricks thousand thousand Area of one brick (0 5) cm 00 cm Area covered by bricks cm cm m 980 m cm m Hence, area of the lane 980 m. It is given that the width of the lane is 5 m. Length of the lane Area width m 980 m 96 m 5

6 506 Concept of Arithmetic Ex. : The length and breadh of a playground are 5 m 0 cm and m 80 cm, respectively. Find the cost of levelling it at Rs.50 per square metre. How long will a boy take to go three times round the field, if he walks at the rate of.5 m/sec. We have, Length of the playground 5 m 0 cm 5.0 m Breadth of the playground m 80 m.80 Area of the playground m m Cost of levelling Rs Rs 95. Perimeter of the playground (length + breadth) ( ) m ( 0) m 0 m. Total distance to be covered by the boy (Perimeter of the playground) 0 m 660 m Speed of the boy.5 m/sec. Time taken by the boy 660 sec.5 Distance Time Speed 0 sec 0 minutes 60 minutes minutes 0 seconds Ex. 8: The carpet for a room 6.6 m by 5.6 m costs Rs 960 and it was made from a roll 0 cm wide. Find the cost of the carpet per metre. We have, Area of the carpet m 6.96 m. Width of the roll 0 cm 0. m Area Length of the roll Width 5.8 m Cost of the carpet Rs 960. Cost of the carpet per metre Rs Rs Hence, the carpet costs Rs 5 per metre. Ex. 9: The area of a rectangular field is calculated to be 00 m when its sides are measured with a faulty metre rod. If that metre rod is actually 0.90 metre long, find the correct area of the field. Let the actual length and breadth of the rectangular field be l and b metres respectively. The faulty metre-rod measures 0.90 metre as metre. It will measure metre as metre 0.90 It will measure l metres as l 0.90 metres. Thus, according to the faulty metre-rod the length of the field is l 0.90 metres. Similarly, breadth of the field measured by the faulty rod b 0.90 metres. l b lb Area of the field But, the faulty metre-rod measures the area of the field as 00 m. lb or, lb m or, lb 6 m Hence, the correct area of the field is 6 m. Areas of Paths Ex. 0: A rectangular grassy lawn measuring 0 m by 8 m is to be surrounded externally by a path which is m wide. Find the cost of levelling the path at the rate of Rs 5 per square metre. Let ABCD be the grassy lawn, and let PQRS be the external boundaries of the path. We have, length AB 0 m, breadth BC 8 m Area of lawn ABCD 0 8 m 80 m Length PQ (0 m + m + m) m Breadth QR 8 m + m + m m Area PQRS m 088 m

7 Mensuration I (Area and Perimeter) 50 Now, Area of the path Area PQRS - Area of the lawn ( ) 8 m Cost of levelling the path Rs (8 5) Rs 0 Ex. : A grassy plot is 80 m 60 m. Two cross paths each m wide are constructed at right angles through the centre of the field, such that each path is parallel to one of the sides of the rectangle. Find the total area used as path. Also, find the cost of gravelling them at Rs 5 per square metre. Let ABCD and EFGH be the cross paths. We have, AB 80 m and BC m Area of a square at one of the corners (8 8) m 6 m Area of the four squares 6 m 56 m Hence, required area Area of the rectangle ABCD Area of the four squares (880-6 ) 86 m Ex. : Calculate the area of the figure given below. Area of path ABCD (80 ) m 0 m Again, EF 60 m and FG m Area of path EFGH (60 ) m 0 m Clearly, area PQRS is common to both the paths. We have, Area PQRS ( ) 6 m Total area used as path Area of path ABCD + Area of path EFGH - Area PQRS ( ) 5 m We have, rate of gravelling the path Rs 5 per square metre Total cost of gravelling the path Rs (5 5) Rs 0 Ex. : Calculate the area of the shaded region shown in the figure given below: Complete the rectangles ABCQ and DEFR by drawing dotted lines NQ and PR. Now, Area of rectangle ABCQ ( ) m 6 m Area of rectangle DEFR ( ) m 6 m Area of rectangle PNQR (9 ) m 6 m [ PN PH + HM + MN ( + + ) m and, QN AQ AN ( 8) m] Area of rectangle LMHK (6 ) m 8 m Hence, required area 6 m + 6 m + 6 m + 8 m 6 m Ex. : A table cover, m m, is spread on a meeting table. If 5 cm of the table cover is hanging all around the table, find the cost of polishing the table top at Rs.5 per square metre. To find the cost of polishing the table top we have to find its area for which we require its length and breadth. We have, Area of the rectangle ABCD (60 8) m 880 m We have, Length of the cloth m Breadth of the cloth m

8 508 Concept of Arithmetic Since 5 cm width of cloth is outside the table on each side. Therefore, Length of the table ( 0.5) m.5 m Breadth of the table ( 0.5) m.5 m Area of the top of the table (.5.5) m It is given that the cost of polishing the table top is at the rate of Rs.5 per square metre. Therefore, cost of polishing the top Area Rate per sq metre Rs (.5.5.5) 9 Rs.8 Ex. 5: There is a square field whose side is m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at Rs.5 and Rs.50 per square metre respectively, is Rs 90. Find the width of the gravel path. Let the width of the gravel path be x metres. Then, each side of the square flower bed is ( - x) metres. It is given that the total cost of laying the flower bed and gravelling the path is Rs 90 ( x) + 6 (x x ) 90 or, (8 x + x ) + (6x 6x ) 90 or, 5x 0x or, 5x 0x or, x x or, x x x or, x (x ) (x ) 0 or, (x ) (x ) 0 or, x or x But x, as the side of the square is m. Therefore, x. Hence, the width of the gravel path is metres. Ex. 6: The length and breadth of a rectangular field are in the ratio of :. A path m wide running all around outside it has an area of 6 m. Find the length and breadth of the field. Let the length and breadth of the field be x and x metres respectively. Then, Area of the field (x x) m Now, area of the square field ( ) 96 m Area of the flower bed ( x) m Area of the gravel path Area of the field - Area of the flower bed 96 ( x) 96 (96 6x + x ) (6x x ) m Cost of laying the flower bed (Area of the flower bed) (Rate per sq m) ( x ) 5 00 ( x ) ( x) Cost of gravelling the path (Area of the path) (Rate per sq m) 50 ( 6x x) 6(x x ) 00 8x m Length of the field (including path) (x + 8) m Breadth of the field (including path) (x + 8) m So, Area of the field and path together (x + 8) (x + 8) m Area of the path [(x + 8) (x + 8) 8x ] m (88x + 6) m It is given that the area of the path is 6 m 88x or, 88x 6-6 or, 88x 5 or, x m Hence, length of the field x 8 m breadth of the field x 6 m Ex. : A chess board contains 6 equal squares and the area of each square is 6.5 cm. A border round the board is cm wide. Find the length of the side of the chess board. Let the length of the side of the chess board be x cm. Then,

9 Mensuration I (Area and Perimeter) 509 (ii) Hence the area of a triangle is equal to half the product of the base and the height. From the above, we have Base Area Height and Height Area Base Area of a Triangle when its sides are given (Hero s Formula) s( s a)( s b)( s c ) Where, is the area of the triangle; and a, Triangle area of 6 squares (x ) (x ) or, x 8x or, x 8x 8 0 or, x x + 6x 8 0 or, (x ) (x + 6) 0 or, x cm A figure bounded by the three sides is called a triangle. (i) Area of a triangle Base Height b, c are its sides and s (a + b + c) semi-perimeter of the triangle. Thus, from half of the sum of the three sides subtract each side separately. Multiply the half sum and the three remainders together. The square root of the product will be the area of triangle. Equilateral Triangle A triangle whose all the three sides are equal is called equilateral triangle. In an equilateral triangle ABC, a b c a a a s Hence, a ABC is the given triangle. Let BDEC be the rectangle on the same base BC and on the same height AF. Since AF is perpendicular to BC, each of the figures ADBF, AECF is a rectangle. Now triangle ABF Now triangle ACF By adding, we get ABF + ACF rectangle ADBF rectangle AECF rectangle ADBF + rectangle AECF ie ABC rectangle BCED a a a a (i) Area a a a (ii) Height a a a a (Side) Area Base a a a a (Side) Right-angled Triangle A triangle having one of its angles equal to 90 is called right-angled triangle. BC CE BC AF [As CE AF] Base Height The figure ABC is right-angled triangle, angle B being a right angle ie of 90. Here, BC is the base of the triangle, AB is the height of the triangle.

10 50 Concept of Arithmetic AC, the side opposite to the right angle, is called the hypotenuse. In case of a triangle ABC right-angled at B, AC AB + BC. This is known as the Pythagoras Theorem. It may be stated in words thus: In a right-angled triangle the square described on the hypotenuse is equal to the sum of the squares on the other two sides. Let b be the base, p be the perpendicular and h be the hypotenuse of a right-angled triangle. Then, (i) Perimeter (b + p + h) (ii) Area Base Height b p (iii) Hypotenuse b p Isosceles Triangle An isosceles triangle is one which has two of its sides equal. Its third side is usually called the base. Let ABC be an isosceles triangle such that AB AC b units and BC a units. Area of ABC Base (Equal side) (Base) Isosceles Right-angled Triangle For an isosceles right-angled triangle, each of whose equal side is a, we have Solved Examples Ex. 8: The area of a right-angled triangle is 50 m. If one of the legs is 0 m, find the length of the other leg. In a right-angled triangle, if one side is the base, then the other side is its altitude or height. Let the given leg be the base. Then, the other leg is the altitude. Here, Area of the triangle 50 m One leg of the triangle 0 m The other leg of the triangle Height of the triangle Area 50 m 5 m Base 0 Ex. 9: Find the area of an isosceles rightangled triangle, if one of the equal sides is 0 cm long. We know that in an isosceles right angled triangle, any one of the two sides which are at right angle can be taken as the base and the other perpendicular side is the altitude. Therefore, base 0 cm and altitude 0 cm So, area of the given triangle 0 0cm 00 cm Ex. 0: The area of a triangle is equal to that of a square whose each side measures 60 metres. Find the side of the triangle whose corresponding altitude is 90 metres. We have, Area of the square (60 60) m 600 m Area of the square 600 m Altitude of the triangle 90 m Side of the triangle Area Corresponding Altitude (i) Hypotenuse a a a (ii) Perimeter a a a (iii) Area Base Height a a a 600 m 80 m 90 Ex. : The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 60 per hectare is Rs.0, find its base and height. Let the altitude of the triangular field be x metres. Then, base x metres (given). Area (Base Height)

11 Mensuration I (Area and Perimeter) 5 (x x) x sq m... (i) It is given that the cost of cultivating the field at the rate of Rs.60 per hectare is Rs.0. Total cost Rate.5 hectares ( ) sq m. Area.0.60 [ hectare 0000 sq m] 5000 sq m... (ii) From (i) and (ii), we have Th en, Area of ABC (Base Height) x 5000 or, x 5000 or, x or, x 00 Hence, height 00 m and base x 900 m. Ex. : Find the area of a right-angled triangle with hypotenuse 5 cm and base cm. Let ABC be the right-angled triangle with base BC cm and hypotenuse AC 5 cm. Using Pythagoras Theorem, we have AC AB + BC or, (5) AB + or, AB or, AB. Hence, area of ABC (Base height) ( ) 8 cm. Ex. : The length of the sides forming rightangle of a right-angled triangle are 5x cm and (x ) cm. If the area of the triangle is 60 cm, find its hypotenuse. Let ABC be a right-angled triangle with right-angle at B. Let AB 5x and BC x. or, 60 (AB BC) or, 60 5x (x ) or, 0 5x (x ) or, x (x ) or, x x 0 or, x 9x + 8x 0 or, x (x ) + 8 (x ) 0 or, (x ) (x + 8) 0 or, x 0 or, x or, x or, x 8 or, x [ x 8 ] AB 5x (5 ) 5 cm and BC (x ) ( ) 8 cm. Now, AC AB + BC [By Pythagoras Theorem] or, AC (5) + (8) or, AC 89 or, AC cm. Hence, hypotenuse cm. Ex. : The perimeter of a right-angled triangle is 60 cm. Its hypotenues is 5 cm. Find the area of the triangle. Let ABC be the given right-angled triangle such that base BC x cm and hypotenuse AC 5 cm.

12 5 Concept of Arithmetic Now, perimeter 60 cm or, AB + BC + AC 60 or, AB + x cm or, AB 5 x. By Pythagoras Theorem, we have AB + BC AC or, ( 5 x) x 5 or, x 0x or, x 5x 00 0 or, x 0x 5x 00 0 or, ( x 0)( x 5) 0 or, x 0 or x 5 If x 0, then AB 5 x 5 and BC x 0 Area (BC AB) (0 5) 50cm If x 5, then AB 5 x 0 and BC x 5 Area (BC AB) or, Hypotenuse 50 cm Hence, the dimensions of the given triangle are Base 0 cm, Altitude 0 cm and Hypotenuse 50 cm. Ex. 6: Find the perimeter of an equilateral triangle whose area is cm. Let each side of the triangle be a cm. Then, its area is a. a or, a 6 or, a. Hence, perimeter of the given triangle a cm ( ) cm cm. Ex. : If each side of an equilateral triangle is increased by cm, then its area increases by cm. Find the length of each side and its area. Let ABC be an equilateral triangle of side a cm. Th en,. ( 5 0) 50 cm Hence, area 50 cm Ex. 5: The area of a right-angled triangle is 600 sq cm. If the base of the triangle exceeds the altitude by 0 cm, find the dimensions of the triangle. Let the altitude of the given triangle be x cm long. Then, base (x + 0) cm. Now, Area 600 cm or, (Base Height) 600 A Area of ABC a cm...(i) Let DEF be the new equilateral triangle of side (a + b) cm. Then, A Area of DEF It is given that A A or, ( a ) a [Using (i) and (ii)] ( a b) cm... (ii) or, (x + 0)x 600 or, x + 0x 00 or, x + 0x 00 0 or, x + 0x 0x 00 0 or, x (x + 0) 0 (x + 0) 0 or, (x + 0) (x 0) 0 or, x 0 or x 0 But x cannot be negative. So, x 0. So, base x + 0 (0 + 0 ) 0 cm. Since the triangle is right-angled. Therefore, (Hypotenuse) (Base) + (Perpendicular) or, (Hypotenuse) or, (Hypotenuse) 500 or, ( a ) ( a ) or, ( a ) a or, a a a or, a 8 or, a So, length of each side of ABC cm. And, area of ABC () cm. Ex. 8: Find the area of an isosceles triangle having the base 6 cm and the length of each equal side 5 cm.

13 Mensuration I (Area and Perimeter) 5 We know that, Area of an isosceles triangle Base (Equal side) (Base) Here, base 6 cm, equal side 5 cm. Area of the given triangle cm (5) (6) cm cm 8 6 cm (6 8) cm 8 cm Ex. 0: Find the perimeter of an isosceles rightangled triangle having an area of 00 cm. Let ABC be an isosceles right-angled triangle with right angle at B such that AB BC a cm. Then, 6 cm cm Ex. 9: The base of an isosceles triangle is cm and its perimeter is cm. Find its area. We have, base cm and perimeter cm. Let the length of each of the two equal sides be b cm. Then, Perimeter cm or, b + or, b or, b 0 or, b 0 Thus, we have Base cm and equal side 0 cm. Area of the given triangle Base (Equal side) (Base) (0) () cm cm 6 6 cm 6 8 cm 8 cm Alternative Method: Let the length of the two equal sides be x cm. Then, Perimeter cm or, x + or, x or, x 0 or, x 0 Thus, the sides of the given triangle are a 0 cm, b 0 cm and c cm, and s cm. s 6 cm, s a (6 0) cm 6 cm, s b (6 0) 6 cm and s c (6 ) cm cm. Area of the triangle s( s a)( s b)( s c ) cm Area of ABC (Base Height) or, 00 a a (a a) [ Area 00 cm (given)] or, a 00 or, a 0 Now, AC AB + BC [By Pythagoras Theorem] or, AC a + a or, AC a or, AC a cm. Hence, perimeter AB + BC + AC a + a [ a 0] cm. Ex. : The area of an isosceles triangle is 60 cm and the length of each one of its equal sides is cm. Find its base. Let ABC be the given isosceles triangle in which AB AC cm. Draw AD perpendicular from A on BC. Let BC x cm.

14 5 Concept of Arithmetic Then, BD DC x cm. In ABC, we have AB AD BD or, AD x [ By Pythagoras Theorem] Ex. : Find the percentage increase in the area of a triangle if its each side is doubled. Let a, b, c be the sides of the old triangle and s be its semi-perimeter. Then, s (a + b + c). or, AD x 69 x. Now, area 60 cm or, (BC AD) 60 or, 69 x x 60 or, x 69 x 60 or, x (69 x ) 600 or, x 69x or, ( x )( x 5) 0 or, x or x 5 x or x 5 Hence, Base x cm or 0 cm Ex. : The perimeter of a triangular field is m and the ratio of the sides is : : 5. Find the area of the field. Let a, b, c be the lengths of the sides of the triangular field. Then, a : b : c : : 5 or a x, b x and c 5x. Now, perimeter m or (x + x + 5x) m or x x Thus, the sides of the triangle are a ( ) m 6 m, b ( ) m 8 m and c (5 ) m 60 m Now, s ( a b c) or The sides of the new triangle are a, b and c. Let s be its semi-perimeter. Then, s (a + b + c) a + b + c s. Let and be the areas of the old and new triangles respectively. Then, s( s a)( s b)( s c ) and s( s a )( s b )( s c ) s(s a )(s b )(s c ) s( s a)( s b)( s c ) [ s s] Increase in the area of the triangle. Hence, percentage increase in area 00 00%. Ex. : The lengths of the sides of a triangle are 5 cm, cm and cm. Find the length of perpendicular from the opposite vertex to the side whose length is cm. Here a 5, b and c. s (a + b + c) s ( ) s a ( 6) 6, s b ( 8) and s c ( 60) Area of field s ( s a)( s b)( s c ) 6 m m 6 6 m 6 86 m (5 + + ) 5 Let A be the area of the given triangle. Th en, A s ( s a)( s b)( s c ) or, A 5 (5 5)(5 )(5 ) or, A 5 0 cm...(i) Let p be the length of the perpendicular from vertex A on the side BC. Then,

15 Mensuration I (Area and Perimeter) 55 Area Base Height BC x cm or, A () p... (ii) From (i) and (ii), we get p 0 60 p cm. Hence, the length of the perpendicular from the opposite vertex to the side whose 60 length is cm is cm. Ex. 5: A field in the form of a parallelogram has one of its diagonals m long and the perpendicular distance of this diagonal from either of the outlying vertices is 0 m 8 dm (see Fig). Find the area of the field. Now, Area of FBE 08 cm (given) or, (FB BE) 08 or, x x 08 or, x 96 or x 9 or, x 6 cm. In ABC, we have AC AB + BC AC 6 6 or, AC 6 or, AC 6 Hence, AC 6 Parallelogram cm. A parallelogram is a four-sided figure whose oppostie sides are parallel. Thus ABCD is a parallelogram in which AB DC and AD BC. We have, AC m and DL BM 0 m 8 dm 0 m 80 cm 0.8 m. Area of the field Area of ACD 0.8 m 5.6 m Ex. 6: ABCD is a square. F is the mid-point of AB and BE is one third of BC. If the area of the FBE is 08 sq cm, find the length of AC. Let the side of the square be x cm. Since F is the mid-point of AB. So, BF AF x cm. Area of parallelogram ABCD Base Height Let CDEF be a rectangle on the same base DC and of the same height FC. Then since parallelograms on the same base and of the same height are equal in area. Parallelogram ABCD Rectangle CDEF CD FC Base Height Hence area of a parallelogram is equal to the product of its base and height. Area Base Height From the above we have, (i) Area Base of a parallelogram Height Also, BE is one third of BC. or, Side of a parallelogram Area Corresponding altitude (ii) Area Height of a parallelogram Base or, Altitude of parallelogram Area Corresponding side

16 56 Concept of Arithmetic Solved Examples Ex. : The base of a parallelogram is thrice its height. If the area is 86 cm, find the base and height of the parallelogram. Let the height of the parallelogram be x cm. Then, base x cm. Area of the parallelogram (x x) cm x cm But, area of the parallelogram is given as 86 cm. x 86 or, x 89 x 89 Thus, height cm and base ( ) cm 5 cm. Ex. 8: In the figure given below, ABCD is a parallelogram, CM AB and BL AD. (i) If AB 6 cm, AD cm and CM 0 cm, find BL. (ii) If AD 0 cm, CM 8 cm and BL cm, find AB. Ex. 9: A field in the form of a parallelogram has base 5 dm and altitude 8 dm. Find the cost of watering the field at the rate of 50 paise per square metre. We have, Base 5 dm (5 0) m [ dm 0 m] 50 m Altitude 8 dm (8 0) m 80 m Area of the field (50 80) m 000 m Rate of watering the field 50 paise per square metre Re per square metre Cost of watering the field Rs 000 Rs Ex. 50: In the figure given below, ABCD is a parallelogram. DL AB, AB 8 cm and AD 5 cm. If the area of the parallelogram is cm, find AL. (i) We have, base AB 6 cm and altitude CM 0 cm. Area of parallelogram ABCD Base Altitude (6 0) cm 60 cm...(i) Now, taking AD as the base, we have Area of parallelogram ABCD Base Altitude ( BL) cm...(ii) From (i) and (ii), we have BL BL. cm (ii) We have, AD 0 cm, BL cm Area of parallelogram ABCD Base Height 0 cm cm 0 cm...(iii) Now, taking AB as the base, we have Area of parallelogram ABCD AB CM (AB 8) cm...(iv) From (iii) and (iv), we get AB 8 0 Rhombus We have, base AB 8 cm Area cm Area Base Altitude or, 8 DL or, DL cm cm. 8 Now, in ALD, we have AD 5 cm, DL cm. By Pythagoras theorem, we have AD AL + DL or, 5 AL + or, AL or, AL or, AL A rhombus is a parallelogram all of whose sides are equal. In a rhombus the diagonals bisect each other at right angles. Thus in the rhombus ABCD, 0 or, AB cm 8 or, AB 5 cm. AB BC CD DA and

17 Mensuration I (Area and Perimeter) 5 AO OC, BO OD BOC 90 Area of the rhombus ABCD BCD BOC BO OC Ex. 5: Find the area of a rhombus whose each side is of length 5 m and one of the diagonals is of length 8 m. Let ABCD be a rhombus whose each side is of length 5 m. Let the diagonals AC and BD intersect at O. Let AC be 8 m. BD AC BD AC (Product of diagonals) Hence, the area of rhombus is equal to the half of the product of its diagonals. Solved Examples Ex. 5: Find the area of a rhombus whose diagonals are of lengths 0 cm and 8 cm. Area of the given rhombus Product of diagonals Since the diagonals of a rhombus bisect each other at right angles. Therefore, AOB is a right-triangle or, AB OA + OB [Using: Pythagoras theorem] or, 5 + OB or, OB 5 6 or, OB 9 or, OB or, OB m BD OB ( )m 6m. Hence, area of rhombus ABCD 0 8 cm 80 cm. Ex. 5: The area of a rhombus is cm. If its perimeter is cm, find its altitude. We have, perimeter of the rhombus cm (side) cm [ Perimeter (side)] side cm 8 cm Now, area of the rhombus cm or, (Side Altitude) or, 8 Altitude Product of diagonals 8 6 m m Ex. 55: If the area of a rhombus be 8 cm and one of its diagonal is cm, find its altitude. Let ABCD be a rhombus of area 8 cm and diagonal BD cm. Altitude cm 9 cm 8 Ex. 5: Find the altitude of a rhombus whose area is 6 m and perimeter is 6 m. We have, perimeter of the rhombus 6 m and, area of the rhombus 6 m Now, side of the rhombus Perimeter 6 m 9 m Altitude of the rhombus Area 6 m m. Side 9 Now, Area 8 cm or, AC BD 8 or, AC 8 or, 6 AC 8 AC 8 cm 8 cm 6

18 58 Concept of Arithmetic Since the diagonals of a rhombus bisect each other at right angles. OA AC cm OB BD cm Also, AB OA + OB [Using Pythagoras Theorem] or, AB + or, AB or, AB 5 cm Since a rhombus is a parallelogram also. Therefore, Area of rhobmus AB Altitude or, 8 5 Altitude or, AB 0 cm 0 cm Hence, perimeter of rhombus 0 cm 8 0 cm ABCD. Ex. 5: If the side of a square is m and it is converted into a rhombus whose major diagonal is 6 m, find the other diagonal and the area of the rhombus. Let AB m be the side of a square ABPQ which is converted into a rhombus ABCD such that diagonal AC 6 m or, Altitude cm cm 9. cm 5 5 Ex. 56: If the area of a rhombus be cm and one of its diagonals be cm, find the perimeter of the rhombus. Let ABCD be a rhombus such that its one diagonal AC cm. Suppose the diagonals AC and BD intersect at O. or, OB Now, Area of rhombus ABCD cm or, AC BD or, BD or, BD or, BD Thus, we have, AC cm and BD cm OA AC cm and OB BD 6 cm Since the diagonals of a rhombus bisect each other at right angle. Therefore, OAB is a right triangle, right angled at O. By Pythagoras theorem, we have AB OA + OB or, AB Since the diagonals of a rhombus bisect each other at right angle, therefore OA AC m and AOB 90. In OAB, we have AB OA + OB or, + OB or, OB 6 9 m BC OB m. Hence, area of rhombus ABCD AC BD Trapezium 6 m 6 m. A trapezium is a four-sided figure having a pair of opposite sides parallel. Thus ABCD is a trapezium in which AB DC. Area of trapezium Height Sum of the parallel sides (Distance between parallel sides) (Sum of parallel sides)

19 Mensuration I (Area and Perimeter) 59 Draw AE and BF perpendicular from A and B to DC. Trapezium ABCD ADE + Rectangle ABFE + BFC DE AE + AE EF + BF FC AE (DE + EF + FC) [ BF AE] AE (DE + EF + FC + AB) [ EF AB] AE (DC + AB) Height Sum of the parallel sides Hence, the area of a trapezium is equal to the product of half the sum of parallel sides and height. 50 cm 5 cm 6 0 cm 9 cm 80 cm + cm cm Ex. 60: The area of a trapezium is 80 cm and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides. Let one of the parallel sides be of length x cm. Then, the length of the other parallel side is (x + 6) cm. Area of the trapezium x 6 9 x cm Solved Examples Ex. 58: Find the area of a trapezium whose parallel sides are of lengths 0 cm and cm and the distance between them is cm. We have, Area of the trapezium (Sum of the parallel sides) (Distance between the parallel sides) 0 cm cm cm Ex. 59: Find the area of the figure given below: 6 9 x cm (9x + ) cm But the area of the trapezium is given as 80 cm. 9x + 80 or, 9x 80 5 or, x 5 9 Thus, the two parallel sides are of lengths cm and ( + 6) cm cm. Ex. 6: If the perimeter of a trapezium be 5 cm, its non-parallel sides are equal to 0 cm each and its altitude is 8 cm, find the area of the trapezium. We have, Perimeter of the trapezium 5 cm or, Sum of the parallel sides + Sum of the non-parallel sides 5 cm or, Sum of the parallel sides or, Sum of the parallel sides (5 0)cm cm Altitude of the trapezium 8 cm. Area of the trapezium (Sum of the parallel sides) Altitude We have, Area of the given figure Area of trapezium ABCD + Area of trapezium CDEF 8 cm 8 cm. Ex. 6: The parallel sides of a trapezium are 0 cm and 0 cm. Its non-parallel sides are both equal, each being cm. Find the area of the trapezium. Let ABCD be a trapezium such that AB 0 cm, CD 0 cm and AD BC cm.

20 50 Concept of Arithmetic Draw CL AD and CM AB. Now, CL AD and CD AB. ALCD is a parallelogram. AL CD 0 cm and CL AD cm. In CLB, we have CL CB cm CLB is an isosceles triangle. LM MB BL 0 cm 5 cm BL AB AL (0 0)cm 0 cm Applying Pythagoras theorem in CLM, we have CL CM + LM or, CM + 5 or, CM 69 5 or, CM Area of CLB BL CM 0 cm 60 cm. Area of parallelogram ALCD AL CM (0 )cm 0 cm. Hence, area of trapezium ABCD Area of parallelogram ALCD + Area of CLB (0 + 60)cm 80 cm. Ex. 6: The parallel sides of a trapezium are 5 cm and cm, while its non-parallel sides are 5 cm and cm. Find the area of the trapezium. Clearly, AECD is a parallelogram. Now, EB AB AE AB DC [ AE DC] (5 ) cm cm Also, EC AD 5 cm. Thus, in ECB, we have EB cm, EC 5 cm and BC cm Let s be the semi-perimeter of the ECB. Th en, s 5 cm cm. Area of ECB s( s a)( s b)( s c ) ( )( 5)( ) cm 6 8 cm cm cm cm 8 cm...(i) Also, Area of ECB (Base Height) (EB CF) ( CF) cm ( CF) cm...(ii) From (i) and (ii), we get CF 8 or, CF 8 cm cm. Area of parallelogram AECB Base Height AE EF ( ) cm cm Now, Area of trapezium ABCD (Area of parallelogram AECB) + (Area of ECB) ( + 8) cm 6 cm Note : We can find the area of trapezium directly as follows: We have, lengths of parallel sides cm and 5 cm Height of the trapezium cm Area of trapezium Height (Sum of the parallel sides) Let ABCD be a trapezium such that AB DC, AB 5 cm, DC cm, AD 5 cm and BC cm. Draw CE DA and CF AB. (5 ) sq cm 6 [6 6] 6 sq cm

21 Mensuration I (Area and Perimeter) 5 Quadrilateral A quadrilateral is a plane figure bounded by four sides. Thus ABCD is a quadrilateral. 0 (8.5 ) (0 9.5) m 95 m. Ex. 65: In quadrilateral ABCD shown in figure given below, AB DC and AD AB. Also, AB 8 m, DC BC 5 m. Find the area of the quadrilateral. m Area of quadrilateral ABCD (Length of a diagonal) (Sum of the lengths of perpendiculars from the remaining two vertices on the diagonal). Join DB. Draw AE and CF perpendiculars to DB Quadrilateral ABCD ABD + BDC BD AE + BD CF BD (AE + CF) Hence, the area of quadrilateral is equal to the product of one diagonals and half the sum of perpendiculars drawn on it from the other two vertices. Note: (i) The area of the quadrilateral can also be found if the lengths of all its sides and one diagonal is known, then the area of each of the two triangles into which the diagonal divides the quadrilateral can be found. (ii) This formula is also applicable to a rectangle, a square, a parallelogram, a rhombus or a trapezium for which we have special formulae. Solved Examples Ex. 6: The diagonal of a quadrilateral is 0 m in length and the perpendiculars to it from the opposite vertices are 8.5 m and m. Find the area of the quadrilateral. In quadrilateral ABCD, we have AC 0 m. Let BL AC and DM AC such that BL 8.5 m and DM m. Clearly, ABCD is a trapezium, and AD CE is its height. We have, BE AB AE AB DC (8 5) m m. In BCE, we have BC BE + CE or, 5 + CE or, CE 5 9 or, CE 6 or, CE 6 m m. Area of quadrilateral ABCD (AB + DC) CE (8 5) m 6 m. Ex. 66: Find the area of the quadrilateral ABCD, in which AB cm, BC 6 cm, CD cm, DA 5 cm and AC 9 cm. The diagonal AC divides the quadrilateral ABCD into two triangles ABC and ACD. Area of quadrilateral ABCD AC (BL + DM) Area of quadrilateral ABCD Area of ABC + Area of ACD For ABC, we have 6 9 s cm

22 5 Concept of Arithmetic Area of ABC s ( s a)( s b)( s c ) ( 6)( )( 9) Area of ACD s( s a)( s b)( s c ) ( )( 8)( 5) 5 0 sq cm 0.98 cm For ACD, we have 9 5 s 8 cm. Area of ACD 8 (8 9)(8 )(8 5) cm 5 cm Hence, area of quadrilateral ABCD ( ) cm.98 cm. Ex. 6: Find the area of a quadrilateral ABCD whose sides are 9 m, 0 m, 8 m and 5 m respectively and the angle between the first two sides is a right angle. Let ABCD be the given quadrilateral such that ABC 90 and AB 9 m, BC 0 m, CD 8 m, AD 5 m. 9 6 m Hence, Area of quadrilateral ABCD (Area of ABC) + (Area of ACD) (80 + 6) m 06 m Regular Polygons A polygon is figure bounded by more than four straight lines. A polygon is said to be regular when all its sides and angles are equal. Polygon can be either convex or concave as mentioned below. A polygon in which none of its interior (internal) angles is more than 80, is known as a convex polygon. On the other hand, if at least one angle of a polygon is more than 80º then it is said to be concave polygon. We use the following terminology depending upon the number of sides of a polygon. Number of sides Polygon 5 Pentagon 6 Hexagon Septagon 8 Octagon 9 Nonagon 0 Decagon Undecagon Dodecagon (i) Area of a regular polygon of n sides having given the length of a side and the radius of the inscribed circle In ABC, we have AC AB + BC [Using Pythagoras Theorem] or, AC or, AC m Now, Area of ABC (Base Height) number of sides (n) length of a side radius of the inscribed circle (AB BC) (9 0) m 80 m In ACD, we have AC m, CD 8 m and DA 5 m. Let a AC m, b CD 8 m and c DA 5 m. Then, s (a + b + c) ( ) Let ABCDE be a regular polygon in which AB a and OG (the radius of the inscribed circle) r Join OA, OB, OC, OD, OE. Thus the polygon is divided into as many triangles as its number of sides. Area of polygon Area of AOB Number of sides of the polygon n AB OG n a r

23 Mensuration I (Area and Perimeter) 5 (a) Hexagon It can be easily seen that triangle AOB is equilateral. Area of equilateral triangle AOB a Let ABCDE be a regular polygon of n sides with AB a and OB (the radius of the circumscribed circle) R Draw OG perpendicular to AB. Then OG is the radius of the inscribed circle. Area of hexagon ABCDEF (b) Octagon 6 a a OG OB GB Area of a regular polygon n n ar a a R a R In the case of a hexagon, R a and n 6. area of a regular hexagon Here it will be seen that the radius of the inscribed circle OG OH + HG a + HG HG KB OG a a a a Area of octagon ABCDEFLM 8 a a a n a r (ii) The area of a regular polygon of n sides having given the length of one side and the radius of the circumscribed circle number of sides side Radius Side 6 a a a a (iii) Some Important Results (A) Interior Angle of a Regular Polygon: Each interior angle of a regular polygon of n sides is º n equal to 80 or 80 (Exterior Angle). n (B) Exterior Angle of a Regular Polygon: Each exterior angle of a regular polygon of n sides is º 60 equal to. n (C) In a convex polygon of n sides, we have: (i) Sum of all interior angles (n ) right angles (n ) 80 (ii) Sum of all exterior angles right angles (iii) Number of diagonals of a polygon of n sides n( n ) n Some Particular cases: Regular Polygon Internal Angle Triangle 60 Quadrilateral 90 Pentagon 08 Hexagon 0 Octagon 5 Nonagon 0 Decagon

24 5 Concept of Arithmetic (D) Circum-circle of a Regular Polygon: A regular polygon can be inscribed in a circle which is known as the circum-circle or circumscribing circle. The centre of this circle is also the centre of the polygon and the radius is known as the circum-radius which is generally denoted by R. Particular cases: l Radius of in-circle of a regular hexagon a cot 0º a l Area of the in-circle of a regular hexagon a a If a is the length of each side of a regular polygon and R is the circum-radius, then we have the following results: a 80º (a) R cosec n 80º (b) Area of the polygon na cot n or, Area of the polygon 80º 80º nr sin cos n n (c) Area of the circum-circle of an n-sided regular polygon 80º a cosec n Particular cases: l Area of a regular hexagon 6 a cot 0º a l Area of circum-circle of a regular hexagon a l Perimeter 6a. Each angle 0º. (E) In-circle of a Regular Polygon: A regular polygon can also circumscribe a circle. A circle having centre at the centre of a regular polygon and touching all sides of it is called the in-circle. If a is the length of a side of a regular polygon and r is the radius of the in-circle, then we have the following results: a 80º (a) r cot n 80º (b) Area of polygon nr tan n (c) Area of the in-circle of an n-sided regular polygon 80º a cot n Solved Examples Ex. 68: Find the area of a regular hexagon whose side is 0 cm long. Area of a regular hexagon side Here, side 0 cm. Hence, Area of the given regular hexagon 0 50 cm Ex. 69: The area of a regular hexagon is 600 cm. Determine its perimeter. We know that the area of a regular hexagon is equal to side Area 600 cm or, (side) 600 or, 00 (side) or, (side) 00 or, side 0 cm. So, perimeter 6 (side) (6 0) cm 0 cm. Ex. 0: Find to the nearest metre the side of a regular octagonal enclosure whose area is hectare. Area of a regular octagon a a hectare ( ) sq cm. a or, a 0 sq m approx. a 6 metres approx. Ex. : A square and a regular hexagon have equal perimeters. Compare their areas. Let P be the perimeter of both a square and a regular hexagon.

25 Mensuration I (Area and Perimeter) 55 Th en, side of the square P and side of the regular hexagon 6 P. P A Area of the square (side) 6 A Area of the regular hexagon P 6 sdie P A P 6 A. P Hence, areas of the square and the hexagon are in the ratio :. Ex. : The side of a regular pentagon is 0 cm. Find its area. [Take Cot 6.6] The area of an n-sided regular polygon is n 80º cot n sides. Here, n 5 and side 0 cm Area of the pentagon 5 (cot 6 ) (0) 5 cot 6 cm cm Ex. : Find the difference between the area of a regular hexagon each of whose side is cm and the area of the circle inscribed in it. (Take ). We know that the area of an n-sided regular polygon is area of the incircle is 80º na cot n and 80º a cot ; n where a is the side of the polygon. Here, a and n 6. Required area 80º 80º na cot a cot n n 6 cot 0º cot 0º 6 9. cm ( ) m 9 cm. Ex. : A regular hexagon is inscribed in a circle of radius 5 cm. Find the area of the circle which is outside the hexagon. Circle [Use. and.] Area of the circumcircle (radius) 5 5 cm 5. cm 8.5 cm We know that the area of a regular polygon of n sides is given by nr sin cos ; n n where R is the radius of the circumcircle. Here, n 6, R 5 cm. So, Area of the regular hexagon 6 5 sin 0 cos 0 cm 50 cm 6.85 cm Hence, required area Area of the circumcircle Area of the hexagon cm.65 cm A circle is a geometrical figure consisting of all those points in a plane which are at a given distance from a fixed point in the same plane. The fixed point is called the centre of the circle and the constant distance is known as its radius. (plural radii) In the given figure, O is the centre and r is the radius of the circle. A circle with centre O and radius r is generally denoted by C (O, r). The word circle is often used for the circumference. (i) Some Important Terms (a) Circular Region: The part of the circle that consists of the circle and its interior is called the circular region. A circular region is also called a circular disc as shown in the figure given below.

26 56 Concept of Arithmetic (b) Chord of Circle: A line segment joining any two points on a circle is called a chord of the circle. It should be noted that a chord is not a part of the circle. In the figure given below, PQ is a chord of the circle. (c) Diameter: A chord passing through the centre of a circle is known as its diameter. Note that a circle has many diameters and a diameter of a given circle is one of the largest chords of the circle. Also, all diameters are of the same length. (d) Semi-Circle: Clearly, if d is diameter of a circle of radius r, then d r. A diameter of a circle divides the circumference K of a circle into two KUNDAN equal parts each of which is called a semi-circle. (e) Quadrant: Two perpendicular diameters of a circle divide its circumference into four equal parts each of which is known as a quadrant. (f) Concentric Circles: Circles having the same centre but with different radii are said to be concentric circles. The figure given below shows two concentric circles. Thus, we have. (approximately) (approximately) Circumference Now, Diameter C or, C r r Thus, circumference C of a circle of radius r is given by C r If d denotes the diameter of the circle. Then, d r C d Note:The number is not a rational number, but its value upto two decimal places coincides with. So, we take the value of as. In the remaining part of this chapter, unless stated otherwise, the value of will be taken as. (iii) Area of a Circle In this section, we shall first obtain the formula for the area of a circle and then the same will be used to solve some simple problems. To obtain the formula for the area of a circle, let us consider the following. Draw a circle of any radius (say, cm) on a thin card-board. Cut it out, and by folding divide it into four equal sectors. Cut these four sectors out, and bisect each of them by folding. Now you have got eight sectors, arrange these as shown in the figure below. (g) Congruent Circle: Two circles are said to be congruent if and only if either of them can be superpassed on the other so as to cover it exactly. It follows from the above definition that two circles are congruent if and only if their radii are equal. (ii) Circumference of a Circle The perimeter of a circle is called its circumference. The ratio of the circumference of a circle and its diameter is always constant. Circumference The ratio. (approximately) Diameter This ratio is denoted by (Pi). Next bisect each sector, as before, and so get 6 equal sectors. Rearrange these as shown in the figure below. Now notice that as the number of sectors is increased, each arc is decreased; so that

27 Mensuration I (Area and Perimeter) 5 (i) the outlines AB, DC tend to become straight lines, and (ii) the angles at D and B tend to become right angles. Thus, when the number of sectors is indefinitely increased the figure ABCD ultimately becomes a rectangle whose length is the semicircumference of the circle, and whose breadth is its radius. Hence area of the circle (circumference) (radius) Hence, (a) Arc of a sector of D Arc of a sector of D 60 Hence, Arc of the sector circumference 60 Sector angle Circumference 60 of the circle circumference (b) Area of a sector of area of circle 60 r r r Hence, area A of a circle of radius r cm is given by A r A Also, r Note:Area of a semi-circle (Area of the circle) r Area of a quadrant of a circle r. K KUNDAN (Area of the circle) (iv) Area Encl osed By Two Concentric Circles If R and r are radii of two concentric circles, then (c) D Area of a sector of D 60 Hence, Area of the sector area of circle Sector angle Area of the circle 60 To show that area of sector radius length of arc. Proof. Area of a sector of D D 60 area of circle D radius circumference 60 D circumference radius 60 arc radius radius length of arc Area enclosed by the two circles R r (R r ) (R r )(R r ) (v) Sector Its arc and Area The angle at the centre of a circle contains four right angles or 60. Hence, if through the centre of a circle we draw 60 radii making equal angles with one another, 60 angles of degree each would be formed at the centre. Since equal angles at the centre are subtended by equal arcs, the whole circumference would be divided into 60 equal arcs and the area of the circle would be divided into 60 equal sectors. Note: A quadrant is a part of the circle contained between two perpendicular radii. Hence a quadrant is a sector of 90. (vi) Area of Segment Any chord of a circle, which is not a diameter, such as AB, divides the circle into two segments, one greater and one lesser than a semi-circle. Greater segment is called major segment and lesser segment is called minor segment.

28 58 Concept of Arithmetic It will be seen from the figure that area of segment ACB sector OACB OAB. The area of the segment ADB will be found by subtracting the area of the segment ACB from the area of the circle. Note: Area of a minor segment of angle in a circle of radius r is given by A θ r 60 sin θ (vii) Some Particular Cases (a) (b) (Always Remember) Area of a semi-circle : The sector of a semicircle is 80. Area of a semi-circle 80 r r 60 Area of a quadrant : The sector angle of a quadrant of a circle is 90. Area of a sector of circle 90 r r 60 (c) Angle described by minute-hand in 60 minutes 60. Angle described by minute-hand in one Thus, minute-hand rotates through an angle of 6 in one minute. (d) Angle described by hour-hand in hours 60 Angle described by hour-hand in one minute minute 0 60 Thus, hour-hand rotates through 60 in one minute. 60 Solved Examples Ex. 5: The ratio of the radii of two circles is : 5. What is the ratio of their circumferences? We have, ratio of radii : 5. So, let the radii of two circles be r and 5r respectively. Let C and C be the circumference of two circles of radii r and 5r respectively. Th en, C r r and C 5r 0r C r C 0r 5 or, C : C : 5. Ex. 6: A piece of wire in the form of a rectangle 8.9 cm long and 5. cm broad is reshaped and bent into the form of a circle. Find the radius of the circle. We have, Length of the wire Perimeter of the rectangle (l + b) ( ) cm 8.6 cm Let the wire be bent into the form of a circle of radius r cm. Then, Circumference 8.6 cm or, r 8. 6 or, r 8. 6 or, r 8.6 cm 86 or, r cm.55 cm. 0 Ex. : A copper wire, when bent in the form of a square, encloses an area of 8 cm. If the same wire is bent in the form of a circle, find the area enclosed by it. (Use Area of the square 8 cm. Side of the square 8 cm cm [ Area (side) side Area ) ). So, perimeter of the square (side) ( ) cm 88 cm. Let r be the radius of the circle. Then, Circumference of the circle Perimeter of the square or, r 88 or, r 88 or, r cm r cm Area of the circle 66 cm Ex. 8: The diameter of the wheel of a car is cm. How many revolutions will it make to travel km. We have, diameter of the wheel of the car cm Circumference of the wheel of the car d cm cm Note that in one revolution of the wheel,

29 Mensuration I (Area and Perimeter) 59 the car travels a distance equal to the circumference of the wheel. Distance travelled by the car in one revolution of the wheel cm. Total distance travelled by the car km 000 m cm Number of revolutions Ex. 9: The circumference of a circle exceeds the diameter by 0 cm. Find the radius of the circle. Let the radius of the circle be r cm. Th en, Circumference of the circle r cm Diameter of the circle r cm It is given that the circumference of the circle exceeds its diameter by 0 cm. r r + 0 or, r 66 [ Area r ] or, r or, r or, r 8 or, r 96 r 96 cm cm Hence, radius of the circle cm. Ex. 8: A race track is in the form of a ring whose inner circumference is 5 m, and the outer circumference is 96 m. Find the width of the track. Let the outer and inner radii of the ring be R metres and r metres respectively. Th en, or, r r + 0 r or, r 0 r r or, 0 r r or, 0 0r or, 0 0 r cm cm 0 Ex. 80: The circumference of a circle is cm. Find its area. Let the radius of the circle be r cm. Then, Circumference cm or, r or, r or, r cm cm Area of the circle r cm cm 5 cm Ex. 8: The area of a circle is 66 cm. Find the radius of the circle. Let the radius of the circle be r cm. We have, Area of the circle 66 cm R 96 and r 5 or, R 96 and or, R r 5 96 and r 5 or, R 6 m and r 56 m. Hence, width of the track (R r) metres (6 56) metres metres. Ex. 8: The inner circumference of a circular track is 0 m. The track is m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs per metre. (Use ). Let the inner and outer radii of the circular track be r metres and R metres respectively. Then, Inner circumference 0 metres or, r 0

30 50 Concept of Arithmetic or, r 0 r 5 m Since the track is metres wide everywhere. Therefore, R Outer radius r + (5 + ) m m. Outer circumference R m 6 m. Rate of fencing Rs per metre Total cost of fencing (Circumference Rate) Rs (6 ) Rs 58. Ex. 8: A circular grassy plot of land, m in diameter, has a path.5 m wide running round it on the outside. Find the cost of gravelling the path at Rs per square metre. K Radius of the KUNDAN plot m m. Radius of the plot including the path ( +.5) m.5 m. Area of the path (.5) () (.5) () m m.5.5 m m m Hence, cost of gravelling the path Rs (500.5 ) Rs 00. Ex. 85: A paper is in the form of a rectangle ABCD in which AB 0 cm and BC cm. A semicircular portion with BC as m diameter is cut off. Find the area of the remaining part. Length of the rectangle ABCD AB 0 cm Breadth of the rectangle ABCD BC cm Area of rectangle ABCD (0 ) cm 80 cm. Diameter of the semi-circle BC cm Radius of the semi-circle cm. Area of the semi-circular portion cut off from the rectangle ABCD r cm cm Area of the remaining part Area of rectangle ABCD - Area of semi-circle (80 - ) cm 0 cm Ex. 86: The circumferences of two circles are in the ratio :. Find the ratio of their areas. Let r and r be the radii of two given circles and C and C be their circumference. Then, C r and C r Now, C : C : or, C C or r r or r r Let A and A be the areas of two circles. Th en, A r and A r A r r A r r r r r r 9 A : A : 9 Hence, the areas of two given circles are in the ratio : 9. Ex. 8: The areas of two circles are in the ratio 6 : 5. Find the ratio of their circumferences. Let r and r be the radii of two circles and let their areas be A and A respectively. Th en, A 9 r and A r

31 Mensuration I (Area and Perimeter) 5 Now, A : A 6 : 5 or, r : r 6 : 5 [Given] Let ABCD be the given square each side of which is cm long. Clearly, the radius of each circle is cm. or, r r 6 5 r or r 5 r r 5... (i) [Taking square root of both sides] Let C and C be the circumferences of two circles. Then, C r and C r. or, C r r C r r C C 5... [Using (i)] C : C : 5 Hence, the circumferences of the two circles are in the ratio : 5. Ex. 88: A square park has each side of 00 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius m as shown in the figure giv en below. Find the area of the remaining part of the park. (Take ). Area of each quadrant of radius m r [ r ] 5 m Area of quadrants ( 5) m 66 m. Area of square park having side 00 m long (00 00) m 0,000 m Hence, area of the remaining part of the park ( ) 98 m. Ex. 89: Four equal circles are described about the four corners of a square so that each touches two of the others as shown in figure. Find the area of the shaded region, each side of the square measuring cm. We have: Area of the square of side cm long ( ) cm 96 cm Area of each quadrant of a circle of radius cm r cm 8.5 cm Area of quadrants 8.5 cm 5 cm. Hence, area of the shaded region Area of the square ABCD Area of quadrants (96 5) cm cm. Ex. 90: An arc subtends an angle of 6º at the centre of a circle of radius.6 cm, find the length of the arc. We know that the length of an arc of a circle of radius r is given by D D circumference r Here, D 6 and r.6 cm Length of the arc 6. 6 cm.6 cm. 60 Ex. 9: A sector is cut from a circle of radius cm. The angle of the sector is 50. Find the length of its arc and area. The arc length l and area A of a sector of angle D in a circle of radius r are given D by l r and 60 A D 60 r respectively. Here, r cm and D l cm 55 cm And, A cm 55 cm 5.5 cm

32 5 Concept of Arithmetic Ex. 9: The area of a sector of a circle is 0 th of the area of the circle. Find the angle of the sector. Let the radius of the circle be r cm and the sector angle be of x. Then, Area of the sector x r cm 60º and, Area of the circle r cm It is given that: Area of the sector 0 Area of the circle or, xº r 60º 0 r 60 or, x r 0 r or, x 6 Hence, the sector angle is of 6. Ex. 9: A 6 sector of a circle has area.85 cm. What is the length of the arc of the sector? Let r cm be the radius of the circle. We have, sector angle 6 and area of the sector.85 cm. We have, Area of the sector Sector angle Area of circle 60º the or,.85 6º r 60º or, r 6 or, r.5 or, r. 5 cm.5 cm Now, Area of the sector length of the arc radius Let A and A be the areas of sectors OAB and OCD respectively. Then, A Area of a sector of angle 0 in a circle of radius cm cm Using A D 60 r cm 6 Area of a sector of angle 0 in a circle of radius.5 cm cm cm cm Area of the shaded region A A 6 cm cm cm 9.65 cm 8 Ex. 95: The perimeter of a sector of a circle of radius 5. cm is 6. cm. Find the area of the sector. Let OAB be the given sector. Then Perimeter of sector OAB 6. cm or,.85 length of the arc.5 or, length of the arc.85 cm. cm.5 Ex. 9: In the figure given below are shown sectors of two concentric circles of radii cm and.5 cm. Find the area of the shaded region. (Use ) or, OA + OB + arc AB 6. cm or, arc AB 6. or, arc AB 6 cm Area of sector OAB length of the arc radius 6 5. cm 5.6 cm

33 Mensuration I (Area and Perimeter) 5 Ex. 96: The area of an equilateral triangle is 9 cm. Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in figure. Find the area of the triangle not included in the circle. Let each side of the triangle be a cm. Then, Area 9 cm or a 9 Area side or, a 9 a cm Thus, radius of each circle is cm. Now, required area Area of ABC (Area of a sector of angle 60 in a circle of radius cm) 60 9 cm 60 [ 9 ] cm [9. ] cm. cm Ex. 9: The length of minute-hand of a clock is cm. Find the area swept by the minute-hand in one minute. (Use ). Clearly, minute-hand of a clock describes a circle of radius equal to its length ie cm. Since the minute-hand rotates through 6 in one minute, therefore, area swept by the minute hand in one minute is the area of a sector of angle 6 in a circle of radius cm. sector angle Hence, required area r cm cm 60 5 cm 0.6 cm 5 Ex. 98: The minute-hand of a clock is 0 cm long. Find the area on the face of the clock described by the minute-hand between 9 am and 9:5 am. Angle described by the minute-hand in one minute 6º. So, angle described by the minute-hand in 5 minutes (6 5) 0 Area swept by the minute-hand in 5 minutes Area of a sector of angle 0 in a circle of radius 0 cm cm 8. cm Ex. 99: The short and long hands of a clock are cm and 6 cm long respectively. Find the sum of distances travelled by their tips in days. (Take ). In days, the short hand will complete rounds. Distance moved by its tip (circumference of a circle of radius cm) cm 0 cm In days, the long hand will complete 8 rounds. Distance moved by its tip 8 (circumference of a circle of radius 6 cm) 8 6 cm 6 cm Hence, the sum of the distances moved by the tips of two hands of the clock 0 6 cm 90.5 cm Ex.00: Find the area of a segment of a circle of radius cm if the arc of the segment has measure 60. Let O be the centre of the circle and PXQ the arc of the segment such that m(pxq) 60º POQ 60º Now, Area of sector POQ 60º cm 60º cm cm

34 5 Concept of Arithmetic Since OP OQ and POQ 60º. Therefore, OPQ is an equilateral triangle. Therefore, Area of OPQ side cm. cm 90. cm Area of segment PXQ Area of sector OPQ Area of OPQ ( 90.) cm 0. cm. Ex.0: If the arc of a segment of a circle has measure 0. If the radius of the circle is 6 cm, find the area of the segment. Let O be the centre of the circle and PXQ the arc of the segment such that m (PXQ) 0. POQ 0 In PNO, ON sin 0 OP ON sin 0 OP ON (Hypotenuse) [since, sin 0 ] OP 6 cm cm In OPN, we have OP PN ON PN OP ON or, PN OP PN or, PN 6 9 cm cm PQ PN cm 6 cm Area of OPQ PQ ON 6 cm 9 cm Hence, Area of segment PXQ Area of sector OPQ Area of OPQ 6 9 cm ( ) cm.55 cm [.] Ex.0: A horse is placed for grazing inside a rectangular field 0 m by 5 m and is tethered to one corner by a rope m long. On how much area can it graze? Shaded portion indicates the area which the horse can graze. Clearly, shaded area is the area of a quadrant of a circle of radius r m. Now, Area of sector POQ 0º 6 6 cm 60º 6 6 cm 0º 60º r 6 cm Let ONPQ. Then OPN is right angled triangle, right angled at N and OPN 0. Hence, required area r cm 69 cm 6.5 cm

35 Mensuration I (Area and Perimeter) 55 Ex.0: PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn on PQ and QS as diameters as shown in the figure given below. Find the perimeter of the shaded region. Ex.05: Find the area of the shaded region in the figure given below. We have PS diameter of a circle of radius 6 cm cm PQ QR RS cm QS QR + RS ( ) cm 8 cm Hence, required perimeter Arc of semi-circle of radius 6 cm + Arc of semi-circle of radius cm + Arc of semicircle of radius cm ( 6 ) cm cm Ex.0: In the figure giv en below, AOBCA represents a quadrant of a circle of radius.5 cm with centre O. Calculate the area of the shaded portion. (Take ). Clearly, radius of the bigger semi-circle cm Area of the bigger semi-circles r 08 cm cm Radius of each of the smaller circle cm Area of smaller semi-circles cm 5 cm Hence, required area (08 + 5) cm 6 cm Ex.06: ABCD is a flower bed. If OA m and OC m, find the area of the bed. (Take ). Area of quadrant AOBCA.5 r cm cm Area of AOD OA OB.5 cm Base Height.5 cm Hence, area of the shaded portion Area of quadrant Area of AOD (9.65.5) cm 6.5 cm We have OA R m and OC r m. Area of the flower bed area of a quadrant of a circle of radius R Area of a quadrant of a circle of radius r R r R cm r R m and r m m 5 m 9.5 m

36 56 Concept of Arithmetic Ex.0: ABCP is a quadrant of a circle of radius cm. With AC as diameter, a semicircle is drawn. Find the area of the shaded portion. BD CD cm The centre of the inscribed circle will coincide with the centroid of ABC Therefore, OD AD In the right-angled triangle ABC, We have AC AB + BC or, AC + or, AC cm AC or, cm Now, required area Area APCQA area ACQA Area ACPA Area ACQA (Area ABCPA Area of ABC) (Area of semi-circle with AC as diameter) [Area of a quadrant of a circle with AB as radius Area of ABC] cm 9 cm cm 98 cm Ex.08: In an equilateral triangle of side cm, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle. (Take. ). Let ABC be an equilateral triangle of side cm, and let AD be perpendicular from A on BC. Since the triangle is equilateral, so D bisects BC. In ABD, we have AB AD BD [Using Pythagoras Theorem] or, AD or, AD OD 6 cm AD cm Area of the incircle OD cm 8 cm cm Area of the triangle ABC side 9. cm Area of the remaining portion of the triangle cm cm Ex.09: Two circles touch externally. The sum of their areas is 0 sq cm and the distance between their centres is cm. Find the radii of the circles. Note that if two circles touch externally, the distance between their centres is equal to the sum of their radii. Let the radii of the two circles be r cm and r cm respectively. Let C and C be the centres of the given circles. Then, CC r r or, r r C cm (given) C or, r r... (i) It is given that the sum of the areas of two circles is equal to 0 cm.

37 Mensuration I (Area and Perimeter) 5 r r 0 or, r r 0... (ii) Now, r r r r r r or, 0 r r [Using (i) and (ii)] or, 96 0 r r r r... (iii) Now, r r r r r r or, r r 0 [Using (ii) and (iii)] or, r r 6 r r 8... (iv) Solving (i) and (iv), we get r K cm and r KUNDAN cm Hence, radii of the two circles are cm and cm. Ex.0: In the given figure, a semicircle is drawn with segment PR as diameter. Q is the mid-point of segment PR, two semicircles with segment PQ and QR as diameters are drawn. A circle is drawn which touches the three semicircles. If PR 8 cm, find the area of the shaded region. [.] Let A be the centre of the circle touching the three semicircles at points D, E and F respectively. Let r be the radius of the circle Then r PR [From Geometry] 6 r 8 cm 6 cm Now area of shaded region Area of semi-circle PDR Area of semicircle PEQ Area of QFR Area of circle with centre A cm. Ex.: Find the area of a right-angled triangle, if the radius of its circum-cirlce is 5 cm and the altitude drawn to the hypotenues is cm. We know that the circumcentre of a rightangled triangle is the mid-point of its hypotenues and the circum-radius is half of the hypotenues. Let ABC be the given triangle with rightangle at B. Let O be the mid-point of hypotenues AC. Let BD be the perpendicular from B on AC. Then, AC (OA) 5 0 cm [ OAradius of the circumcirlce5 cm] and, BD cm (given) Area of ABC (Base Height) AC BD (0 ) cm 0 cm.

38 58 Concept of Arithmetic. Calculate the area of a rectangle metres decimetres long and metres decimetres 8 centimetres wide.. Find the diagonal of a rectangle whose sides are metres and 5 metres.. How many metres of carpet 5 cm wide will be required to cover the floor of a room which is 0 metres long and metres broad?. How many paving stones, each measuring metres by metres, are required to pave a rectangular courtyard 0 metres long and 6 metres broad? 5. A hall room, 9 m 0 cm long and 5 m 0 cm broad, is to be paved with equal square tiles. Find the largest tile which will exactly fit and the number required. 6. A wire is in the shape of a square of side 0 cm. If the wire is rebent into a rectangle of length cm, find the breadth. Which encloses more area, the square or the rectangle?. The area of a square and a rectangle are equal. If the side of the square is 0 cm and the breadth of the rectangle is 5 cm, find the length of the rectangle. Also find the perimeter of the rectangle. 8. A map is drawn to a scale of 0 cm to the km. How many square cm on the map will represent a hectare of ground? 9. Find the width of a roller which traverses 8 sq km while cutting 6. hectares of grass. 0. The diagonal of a rectangular field is 5 m and its area is 08 m. What will be the cost of fencing this field if the cost of fencing for one metre is Rs 5.. A strip of paper. km long and.05 mm thick is rolled up into a solid cylinder. Assuming the area of a circle to be times the square of its radius, find approximately the radius of the circular ends of the cylinder.. A square field containing 68 square metres is to be enclosed with wire placed at heights,, and metres above the ground. What length of the wire will be needed, if the length required for each circuit is 5% greater than the perimeter of the field?. The area of a rectangular field is 000 square metres and the ratio between its length and Practice Exercise Exercise (Rectangle, Square and Area of Path) breadth is 6 : 5. Find the cost of the wire required to go four times round the field at Rs 0 per kilometre of length of the wire.. A rectangular park is 00 metres long and 80 metres wide. There are two paths, each 5 metres wide, in the middle of the park running one parallel to the length and the other parallel to the width of the park. Find, (i) the area of the paths, (ii) the expenditure involved in constructing the paths at 5 paise per square metre, and (iii) the expenditure involved in laying grass in the remaining portion of the park at 5 paise per square metre. 5. A rectangular field 50 metres long and 00 metres wide, has within it a 0 metres wide uniform path running round it. Find, (i) the area of the path, and (ii) the cost of cultivating the remaining part of the field at Rs.50 per square metre. 6. A school hall 0 m long and 5 m broad is surrounded by a verandah of uniform width of m. Find the cost of flooring the verandah at Rs.50 per square metre.. A room is 8 m long and 6 m wide. It is surrounded by a verandah. Find the width of the verandah if it occupies sq metres. 8. A path m wide, running all around outside a square garden occupies 0 sq metres. Find, (i) the length of the square garden. (ii) the area of the part of the garden enclosed by the path. 9. A square carpet is spread in the centre of a hall 9 m square leaving some margin of equal width all around. The total cost of carpeting at Rs.50 per sq m and decorating the margin at 0 paise per sq m is Rs 6.0. Find the width of the margin. 0. A rectangular field is 00 metres long and metres broad. It is planted with trees in rows perpendicular to the length, one metre from row-to-row, and one metre from tree-totree in the same row. If the width of a metre all-round the field remains unplanted, find the number of trees.. A path metres wide running all-round a square garden has an area of 9680 sq metres. Find the area of the part of the garden enclosed by the path.. A marginal walk all-round the inside of a

39 Mensuration I (Area and Perimeter) 59 rectangular space m by 0 m occupies 50 sq metres. Find the width of the walk.. A garden, whose length is metres, has a path.5 metres wide on the two sides and at one end. If it costs Rs 60 to turf the remainder at Rs 5 a sq metre, what is the width of the garden?. In the centre of a room 0 metres square, there is a square of turkey carpet, and the rest of the floor is covered with oilcloth. The carpet and the oilcloth cost respectively Rs 50 and Rs 65 per square metre, and the total cost of the carpet and the oilcloth is Rs 85. Find the width of the oilcloth border.. The base of triangular field is 880 metres and its height 550 metres. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of Rs.50 per sq hectometre.. The base of a triangular field is three times its height. If the cost of cultivating the field at Rs per hectare is Rs 0.5, find its base and height.. Find the area of triangular field whose sides are 50 metres, 8 metres, metres respectively and also find the perpendicular from the opposite angle on the side metres. If it is lent at Rs 0000 per hectare, find the rent of the field.. X is a point on side CD of a square ABCD such that CX 5 cm. If area of the triangle ADX is cm, find the length of the side of the square. 5. Find the area of a triangle, one of whose angles is 90, hypotenuse is.5 cm and the base is.5 cm. 6. The area of a triangle equals the area of a Exercise (Triangle). Find the area of an isosceles right triangle, the length of whose each side containing the right angle is 5 cm. 8. The sides of an equilateral triangle is 8 cm. Find its area and the height. 9. From a point in the interior of an equilateral triangle, perpendiculars drawn to the three sides, are 6 cm, 0 cm and cm respectively. Find the area of the triangle. 0. A triangular park ABC has sides 0 m, 80 m and 50 m (see figure). A gardener Dhania has to put a fence all round it and also plant grass inside. How much area does he need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 0 per metre leaving a space m wide for a gate on one side. square whose side is 5 m. Find the length of the side of the triangle which is 5 m from the opposite vertex. Exercise (Quadrilateral, Parallelogram, Rhombus, Trapezium and Regular Polygon). Find the area of the quadrilateral ABCD in which the diagonal DB 0 m and the perpendiculars AL and CM drawn on it from A and C are respectively m and 6 m. (These perpendiculars are called offsets).. Find the area of a quadrilateral piece of ground ABCD in which AB 85 metres, BC metres, CD 65 metres, DA 85 metres and DB 5 metres.. The sides of a parallelogram are 0 m and 0 m respectively and its diagonal is 50 m. Find its area.

40 50 Concept of Arithmetic. The perimeter of a rhombus is 6 cm and one of its diagonals is 55 cm. Find the other diagonal and the area of the rhombus. 5. The parallel sides of a field, which is in the shape of a trapezium, are 0 m and m and the remaining two sides are 0 m and m. Find the cost of levelling the field at the rate of Rs 0 per sq metre. 6. Ratio between the parallel sides of the trapezium is :, while ratio between unparallel sides of the trapezium is :. Ratio between bigger parallel and unparallel sides is :. If height of the trapezium is 5 5, then find the area of the trapezium?. A regular hexagon of side 6 cm is inscribed in a circle. Find the area of the region in the circle which is outside the hexagon. [Use.,. ] 8. The diagonals of a rhombus are 8 cm and 6 cm, find the sides and the area.. The radius of a circular wheel is m. How many revolutions will it make in travelling km?. The circumference of a circular garden is 0 metres. Find the area. Outside the garden a road of.5 metres width runs round it. Calculate the area of this road and find the cost of gravelling the road at Rs per 00 sq metres.. A bicycle wheel makes 5000 revolutions in moving km. Find the diameter of the wheel.. A boy is cycling such that the wheels of the cycle are making 0 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling. Exercise (Circle) 0. Find the ratio of area of a square inscribed in a semi-circle of radius r to the area of another square inscribed in the entire circle of radius r.. Find to the three places of decimals the radius of the circle whose area is the sum of the areas of two triangles whose sides are 5, 5, 66 and, 56, 65 measured in centimetres. (Take /).. PQRS is a diameter of a circle of radius 6 cm. The length PQ, QR and RS are equal. Semicircles are drawn on PQ and QS as diameters as shown in the figure below. Find the area of the shaded region. 5. The diameter of the wheel of a bus is 0 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour? 6. A copper wire, when bent in the form of a square, encloses an area of 8 cm. If the same wire is bent in the form of a circle, find the area enclosed by it. (Use /). A wire is looped in the form of a circle of radius 8 cm. It is re-bent into a square form. Determine the length of the side of the square. 8. A bucket is raised from a well by means of a rope which revolves round a wheel of diametre 5 cm. If the bucket ascends in minute seconds with a uniform speed of. m per second, find the number of complete revolutions made by the wheel in raising the bucket. 9. The radius of circle is 0 cm. Three more concentric circles are drawn inside it in such a manner that it is divided into equal parts. Find the radius of the smallest circles?. A square water tank has its side equal to 0 m. There are four semi-circular grassy plots all round it. Find the cost of turfing the plots at Rs.5 per sq cm. (Use.). A rectangular park is 00 m by 50 m. It is surrounded by semi-circular flower beds all around. Find the cost of levelling the semicircular flower beds at 60 per sq metre. (Use.) 5. A park is in the form of a rectangle 0 00 m. At the centre of the park there is a circular lawn. The area of the park excluding the lawn is 800 m. Find the radius of the circular lawn. (Use /)

41 Mensuration I (Area and Perimeter) 5 6. In the figure given below, ABCD is a rectangle. The radius of the semicircles drawn on AD and BC as diameters and radius of circle drawn in between is the same. If BC cm, find the area of the shaded region. in the figure. Find the area of the shaded region. [ /,.]. An athletic track m wide consists of two straight sections 0 m long joining semicircular ends whose inner radius is 5 m. Calculate the area of the shaded region.. In a circle of radius cm an arc subtends an angle of 60º at the centre. Find (i) the length of the arc, (ii) the area of the sector bounded by the arc and (iii)the area of the segment made by this arc.. A circular swimming pool is surrounded by a circular path which is m wide. If the area of the path is th part of the area of the 5 swimming pool, then find the radius of the swimming pool (in metres).. If the circumference of a circle is 80 cm, then find the side of a square inscribed in the circle. 8. In the given figure, the centre of the circle is K A and ABCDEF KUNDAN is a regular hexagon of side 6 cm. Find the following: (i) Area of segment BPF (ii) Area of the shaded portion. 9. Quadrilateral ABCD is a rectangle. Sectors with centre C and D are drawn as shown in the figure. If AB cm, CB cm, find the area of the shaded portion.. In the figure given below, square OABC is inscribed in a sector OPBQ. If OC 0 cm, find the area of the shaded region. (use.) 0. An equilateral ΔABC has each of its sides cm with each of its vertices as centres, and radius as cm, arcs are drawn as shown

42 5 Concept of Arithmetic 5. In the given figure, cresent is formed by two circles which touch at the point A. O is the centre of the bigger circle. If CB 9 cm and DE 5 cm, find the area of the shaded portion. 0. In a right ABC, A 90, AB cm, AC cm. On its three sides as diameters, three semi-circles are drawn as shown in the figure given below. Find area of shaded parts.. The area of the shaded circular ring is 0 sq cm and the difference between the radii of the two circles is cm. Find the area of the unshaded region. 6. Two circles touch internally. The sum of their areas is 6 sq cm and the distance between their centres is 6 cm. Find the radii of the circle.. Find the area of the shaded portion of the given diagram. Give your answer correct to three significant figures. 8. Find the area of the shaded portion in the given figure, where the arcs are quadrants of a circle.. In the figure given below, ABC is an equilateral triangle of side cm and segment BC is diameter of the semicircle. Find the area of the shaded region. (.). In the given diagram AC is a diameter of a circle with radius 5 cm. If AB BC and CD 8 cm, calculate area of the shaded region. 9. Find the area of the shaded portion in the figure given below: [use /]. In the figure as mentioned below, POQ and ROS are diameters of a circle with centre O

43 Mensuration I (Area and Perimeter) 5 and radius cm. Find the area of shaded region. (use /).. Find the area of the shaded region in the given figure, where ABCD is a square of side cm. 5. A rectangular field is surrounded by four semicircular flower-beds. If the length and the breadth of the field are 6 m and m respectively, find the cost of raising the flower-beds at the rate of Rs 8 per m. (Take.) 6. The length of the side of a square is cm. Taking vertices of the square as centres, circles are drawn each with a radius of cm. Find the area of the region of the square that remains outside the region of any of the circles.. A brick, 5 cm thick (high), is placed against a wheel to act for a stop. The horizontal distance of the face of the brick stopping the wheel from the point where the wheel touches the ground is 5 cm. What is the radius of the wheel? 8. The length of the minute-hand of a clock is 0 cm. What is the area swept by the minutehand in one minute? (Use.) 9. A chord AB of a circle of radius 0 cm makes. Find the area of the shaded region in the given figure, where ABCD is a square of side 0 cm and semi-circle are drawn with each side of the square as diameter. (use.). The circumference of a circle is 6.8 cm more than its diameter. What will be the radius of the circle? (use.) 5. A square has been inscribed in a circle. What will be the ratio of the areas of circle and the square? 6. The length of sides AB, BC and CA of a ABC a right angle at the centre of the circle. Find the area of the major and minor segments. (Take.). 0. A chord AB of a circle of radius 5 cm makes an angle of 60 at the centre of the circle. Find the area of the major and minor segment. (Take.,.).. In the given figure, two circular flower beds have been shown in two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds. are cm, 6 cm and 8 cm respectively. Three sectors of circles drawn with centres A, B, C and each with one centimetre of radius, starting and terminating with the sides of the triangle are cut off. Find the area of the remaining position of the triangle.. In a circle of radius 8 cm, an arc subtends an angle of at the centre. Find the length of an arc and the area of the sector so formed. 8. The radii of three concentric circles are in the ratio : :. Find the ratio of the area between the two inner circles to that between the two outer circles.

44 5 Concept of Arithmetic. Length.0 metres Breadth.8 metres Area (.0.8) square metres.6 square metres. Length of the diagonal 5 metres 69 metres metres. Area of carpet Area of the floor (0 ) sq metres Width of carpet 5 cm metre Length of carpet 0. Area of courtyard 0 6 sq m 95 sq metres 0 metres sq m Area of each paving stone Number of stones required m 0 cm 90 cm 5 m 0 cm 50 cm The side of the largest square tile HCF of 90 and 50 0 cm m 0 cm Number of tiles Side of the square 0 cm Length of the wire Perimeter of the square side ( 0 ) 0 cm Length of the rectangle l cm. Let b be the breadth of the rectangle. Perimeter of the rectangle Length of wire 0 cm Perimeter of rectangle ( + b) Thus, 0 ( + b) or, 0 + b b (0 ) 8 cm The breadth of the rectangle 8 cm. Area of the square (Side) 0 cm 0 cm 00 cm Area of the rectangle l b cm 8 cm 96 cm So, the square encloses more area even though its perimeter is the same as that of the rectangle. Answers and explanations Exercise. Area of the square (Side) 0 cm 0 cm 600 cm It is given that, the area of the rectangle The area of the square Area of the rectangle 600 cm Breadth of the rectangle 5 cm Area of the the rectangle l b or, 600 l l 6 cm 5 Hence, the length of rectangle is 6 cm Perimeter of the rectangle (l + b) (6 + 5) cm 8 cm Hence, the perimeter of the rectangle is 8 cm. 8. hectare 0000 sq m 000 metres are represented by 0 cm. ( ) sq m are represented by (0 0) sq m. 0 0 sq m is represented by sq cm sq m are represented by sq cm sq cm km 8000 m 6. hectare ( ) sq m Imagining the grass area to be 8000 m long, and as wide as the roller, we have Width required m m 50 cm Let the length of the rectangle be x metres and breadth be y metres. Area of rectangular field x y 08 m... (i) Area of rectangle length breadth And 5 x + y or, 5 x + y... (ii) In a right-angled triangle Hypotenuse Base + Height From equations (i) and (ii) (x + y) x + y + xy () or, x + y

45 Mensuration I (Area and Perimeter) 55 Now, the perimeter of the field (length + breadth) Perimeter of field (x + y) metres Cost of fencing the field ( 5 ) Rs 0. By rolling up a rectangular piece of paper it will be seen that the area of the longer edge of the paper rolled into solid cylinder is equal to that of the circular end of the cylinder.. km cm.05 mm.005 cm (radius) sq cm (radius) radius.9 cm (Approx.). Area of the field 68 sq m perimeter 68 metres 8 metres sq cm common area from the sum of the areas of the two paths. Area of the paths ( ) 85 sq units. (ii) The expenditure involved in constructing the paths 5 85 Rs 85 Rs Rs (iii) Area of the park sq m Area of the remaining portion of the park ( ) 5 sq m The expenditure of laying grass in the remaining portion of the park 5 Rs 5 Rs (i) Area of the field sq m Length of the inner rectangle {50 (0 + 0) } 0 m 05 length of each circuit 8 m 00 Since the wire goes round times. total length of wire required K 8 KUNDAN m 990. m. Let the length be 6x m and the breadth be 5x m. area ( 6x 5x) sq m 000 sq m or, 0x 000 or, x 900 x 0 Hence length 80 m and breadth 50 m Length of wire required to go round the field four times [ ( )] m.6 km required cost Rs (.6 0) Rs (i) The area of the path HG (00 5 ) 500 sq m Width of the inner rectangle {00 (0 + 0) } 80 m Area of the inner rectangle (0 80 ) 000 sq m Area of the path ( ) 600 sq m (ii) The area of the remaining part of the field to be cultivated 000 sq m Cost of cultivating the remaining part of the field Rs 000 Rs The area of the school hall ABCD (5 0 ) 00 sq m The length of the rectangular region PQRS 0 5 m The area of the path EF (80 5 ) 00 sq m Area of the shaded portion (5 5 ) 5 sq m The shaded portion is common to both the roads, so while finding the actual area of the paths, we should subtract this

46 56 Concept of Arithmetic The width of the rectangular region PQRS 5 m 0 m Area of the rectangular region PQRS (5 0 ) 500 sq m Area of the verandah (500 00) sq m 00 sq m Cost of flooring the verandah Rs 00 5 Rs 500. Let the width of the verandah be x metres. Area of the room ABCD (8 6 ) 8 sq m Area of square PQRS (x + ) sq m (x + ) x + 0 or, x + 8x + 6 x + 0 or, 8x or, x m 8 The length of the square garden ABCD m (ii) Area of the part of the garden enclosed by the path, ie, of ABCD sq m 9. Let the width of the margin be x m. Area of the square hall ABCD (9 9 ) 8 sq m Area of the square hall PQRS (9 x) sq m 8 6x + x sq m Length of PQRS (8 + x + x) m (8 + x) m Width of PQRS (6 + x + x) m (6 + x) m Area of PQRS (8 + x) (6 + x) sq m (8 + 8x + x ) sq m Area of the room + area of the verandah (8 + 8x + x ) sq m 8 8 8x x or, x 8x or, x x 8 or, x x 8 0 or, x 9x x 8 0 or, x( x 9) ( x 9) 0 or, ( x 9)( x ) 0 Either x 9 or x x cannot be 9 because width of the verandah cannot be negative x Width of the verandah m 8. (i) Let the length of the square garden ABCD be x metres. Area of the square garden ABCD (x x ) x sq m Length of square PQRS ( x )m ( x ) m Area of the margin 8 (8 6x + x ) sq m x x sq m 6x x sq m 5 Cost of the carpet Rs 8 6x x Rs Cost of decorating the margin Rs 6 x x Rs x 0x 6x x 5 It is given that total cost of carpeting at Rs.50 per sq m and decorating the margin at 0 paise per sq m is Rs x 0x 6x x 6. 5 or, x 00x x 8x or, x 00x x 8x 6 or, 9x 88x 9 0 or, 9x 6x 8x 9 0 or, 6x(x ) 9(x ) 0 or, (6x 9) (x ) 0

47 Mensuration I (Area and Perimeter) 5 0. Either 6x 9 0 or, x 0 If 6x 9 0 then x x m If x 0 then m 8 8 m x m But the margin cannot be length 9 m is less than x m Width of the margin m 8 m because the 8 8 m m 50 cm Area of the rectangle LTBQ sq metres The width LT metres 0 LQ 0 metres LM (0 ) 08 metres Area of the square LMNP (08) sq metres 596 sq m.596 sq km. Area of inner rectangle ( 0 50) sq m 50 sq m A little reflection will show that the difference between the sides of the inner rectangle must be equal to the difference between the sides of the outer rectangle. Hence, we must find two numbers whose product is 50 and difference ( 0 ). Now, if two numbers be x and y, then (a + b) (a b) + ab a + b. But a b a, b 0. the required width 5 m Total length of the rectangular field in which. trees are planted ( ) 9 m Hence, the total number of trees in a line (9 + ) 0 trees. Distance up to which trees are planted (00 ) 98 m total number of lines in which trees are to be planted (98 + ) 99 lines total number of trees Area of ABCD (Rs 60 Rs 5) sq m 6 sq metres Now, length AB (.5 ) 0.5 metres 0.5 metres BC 6 sq metres BC 6 8 metres The required width ( ) metres metres Let ABCD be the garden and LMNP be the part enclosed by the path. Produce LM, MN, NP, PL to meet the sides of the outer square in Q, R, S, T. The rectangles TQ, QR, RS, ST are all equal. Area of the square room 0 m 0 m 00 sq m Let the areas of the carpet and oilcloth be x sq metres and y sq metres respectively. Hence, according to the question, x + y 00...(i) and 50x + 65y 85 or, 0x + y 6...(ii)

48 58 Concept of Arithmetic Solving equations (i) and (ii), we have x 8 The area of the square carpet is 8 sq metres. Therefore, the carpet is 9 metres in length and breadth. But the room is 0 metres in length and breadth. Hence double the width of the border is (0 9 ) metre the width of the border metre 5 dm Alternative Method: The area of the square room 00 sq metres 85 The mean cost per sq metre Rs 00 Rs.85 8 : 9 By the Alligation Rule, the area of the square carpet is 8 sq metres. Therefore, the carpet is 9 metres in length and breadth. But room is 0 metres in length and breadth. Hence double the width of the border is (0 9 ) metre the width of the border metre 5 dm. Area of the field Base Height sq metres Exercise (s a) (0 50 ) 0 metres (s b) (0 8 ) metres (s c) (0 ) 8 metres Area sq m Area Base 0 metres 680 metres Perpendicular sq hectometres sq hectometres Cost of supplying water to sq hectometre Rs.50 Cost of supplying water to the whole field Rs.0.5 Rs Area of the field Also, area of the field 0.5 hectares Base Height Rent per hectare Rs 0000 required rent Rs Rs 680. Let ABCD be the given square and X is point on side CD. CX 5 cm Height Height (Height) (Height) (Height) Height hectares 9 hectares sq metres m 00 m Also, Base Height 900 m. Here, a 50 metres, b 8 metres, c metres s (50 8 ) metres 0 metres 0 metres Let the length of the side of square be x cm. XD (x 5) cm Now, ADX is a right-angled triangle (DX) (AD) or, ( x 5) ( x) or, x 5x 8 0 or, x x + x 8 0

49 Mensuration I (Area and Perimeter) or, x (x ) + (x ) 0 or, (x ) (x + ) 0 or, x because x The side of square cm Area of AOC x 6 8x sq cm Let ABC be a right-angled triangle, whose ABC 90 Hypotenuse, AC.5 cm Base, BC.5 cm Perpendicular (AB) AC BC (.5) (.5) (.5.5) (.5.5) cm required area Base Height sq cm 6. Area of the square (5 5 ) 05 sq m Area of the triangle 05 sq m Height of the triangle 5 m Required side of the triangle 05 5 m 5. Area of an isosceles right triangle (length of one of its two equal sides) 5 5 sq cm.5 sq cm side 8 8 sq cm. 6. sq cm 8. Area of an equilateral triangle Height of an equilateral triangle side cm 9. Let each side of ABC be x cm. Area of BOC x x sq cm Area of AOB x 0 0x sq cm Area of ABC area of BOC + area of AOC + area of AOB ( x 8x 0x ) sq cm 9x sq cm But area of equilateral ABC x 9x or, x or, x cm Area of ABC x x sq cm sq cm 0. For finding area of the park, we have s 50 m + 80 m + 0 m 50 m 50 ie s 5 m Now, (s a) (5 0 ) 5 m (s b) (5 80 ) 5 m (s c) (5 50 ) 5 m Therefore, area of the park s ( s a) ( s b) ( s c ) m 5 5 m Also perimeter of the park AB + BC + CA 50 m Therefore, length of the wire needed for fencing 50 m m (to be left for gate) m And so the cost of fencing Rs (0 ) Rs 90

50 550 Concept of Arithmetic Exercise. Here the area of the quadrilateral ABCD the area of the triangle ABD + the area of the. Area of quadrilateral ABCD Area of triangle ADB + Area of triangle DBC. triangle BCD BD AL + BD CM BD (AL + CM) ie, The area of a quadrilateral diagonal sum of offsets. Therefore, the required area of the quadrilateral sq m sq m. Area of parallelogram ABCD area of ABC + area of ACD area of ABC ( Each diagonal of a parallelogram bisects it). In triangle ADB, perimeter 6 m Area of the triangle ADB 6 (6 85) (6 85) (6 5) sq m 6 8 sq m In triangle DBC, 65 5 perimeter m Area of the triangle DBC ( 65)( )( 5) sq m sq m Area of the quadrilateral ABCD ( + 06) sq m 96 sq m Semi-perimeter (s) of ABC m Area of ABC s ( s a) ( s b) ( s c ) 60 (60 0) (60 0) (60 50) sq m Area of parallelogram ABCD 600 sq m 00 sq m Let ABCD be the rhombus in which AC 55 cm Perimeter of the rhombus 6 cm AB and AO cm 55.5 cm

51 Mensuration I (Area and Perimeter) BO ( 6.5) (.5) cm cm Hence the other diagonal BD 8 cm Area of the rhombus AC BC sq cm 6. Let x be present in the parallel sides of the trapezium. parallel sides are x and x. Let k be present in the non-parallel sides of the trapezium. Non-parallel sides are k and k. According to the question, or, x k x k Let ABCD be the trapezium such that AB CD, AB m, DC 0 m, AD 0 m and BC m Draw CE DA and CF AB. Clearly, AECD is a parallelogram Now, EB AB AE AB DC [ AE DC] ( 0 ) m Also, EC AD 0m K Thus, in ECB, KUNDAN we have EB m, EC 0 m and BC m Let s be the semi-perimeter of the ECB. Then 0 s m Area of ECB s ( s a) ( s b) ( s c ) ( )( 0)( ) sq m 8 sq m...(i) Also, area of ECB Base Height Combining triangle AEC and BFD, we get a triangle of base 6k k k and two other sides are k and k. Semi-perimetre of combined triangle k k k x Area of the triangle 9k s ( s k )( s k )( s k ) 9k 9k 9k 9k k k k 9k 5k k k k 9 5 k 5 Also area of the triangle 5 5 k k 5 5 (CF)...(ii) From equations (i) and (ii), we get CF 8 8 CF 8 m Area of parallelogram AECD Base Height AE CF (0 8 ) 60 sq m Now, area of trapezium ABCD (Area of parallelogram AECD) + (Area of ECB) (60 + 8) sq m sq m Cost of levelling the field at the rate of Rs 0 per sq metre Rs (0 ) Rs 0 k or, 5 k 5 5 k 0 units Parallel sides are 0 units and 60 units. Area of trapezium units 5 5 units.. Each side of the hexagon inscribed in the circle is 6 cm, the radius of the circle is 6 m. Area of the circle r. (6). 6.0 cm

52 55 Concept of Arithmetic Area of the hexagon (side) (6) 5 cm cm Hence the area of the region of the circle which is outside the hexagon.0 cm 9.5 cm 9.5 cm. 8. We know that the diagonals of a rhombus bisect one another at right angles. Therefore from the given figure the area of the rhombus ABCD area of the triangle ABC + area of the triangle ADC area of the triangle ABC. AC OB AC OB AC BD OB BD ie, the area of a rhombus product of its two diagonals. Here, AC 8 cm and BD 6 cm AO cm and BO cm AB 5 cm Each side of the rhombus is 5 cm. The required area of the rhombus 8 6 sq cm sq cm. Distance to be travelled km 000 m m Radius of the wheel Circumference of the wheel m m In travelling m the wheel makes revolution. In travelling 000 m the wheel makes revolutions.. Area r, circumference r r 0 metres. r 0 m 6 m area of the garden r 66 sq m 866 sq m Exercise Area of the road area of bigger circle area of the garden. Radius of the bigger circle 6 m area of bigger circle 9 m 9 9 sq m 8506 sq m area of road sq m 6 required cost Rs Rs Distance covered by the wheel in one revolution Distance moved Number of revolutions km cm cm Circumference of the wheel 0 cm Let the radius of the wheel be r cm. Then, Circumference 0 cm or, r 0 cm

53 Mensuration I (Area and Perimeter) 55 or, r 0 or, r 5 cm diameter r cm ( 5) cm 0 cm Hence, the diameter of the wheel is 0 cm. 60. Radius of the wheel r cm 0 cm Circumference of the wheel r 0 cm 0 cm Distance covered in one revolution 0 circumference cm Distance covered in 0 revolutions 0 0 cm (0 0) cm cm m m km 000 It is given that the wheels are making 0 revolutions per minute. So, distance covered in one minute Distance covered in 0 revolutions or, r 88 or, r cm Area of the circle r cm 66 cm Total time taken to pull up bucket is minute seconds (60 + ) seconds Distance travelled in one second. m Distance travelled in seconds (. ) 9.6 m or 960 cm Let number of revolutions be N. According to the question, Distance covered by bucket perimeter of 6 km 000 Distance covered in one hour 6 60 km 5.8 km 000 Hence, the speed with which the boy is cycling 5.8 km/hr 5. Distance covered by the wheel in one minute cm 60 Circumference of the wheel 0 0 cm Number of revolutions in one minute Area of the square 8 cm side of the square 8 cm cm [ Area (side) side So, Perimeter of the square (side) ( ) cm 88 cm Let r be the radius of the circle. Then, circumference of the circle Perimeter of the square r 88 Area ]. Length of the wire circumference of the circle 8 cm [Using C r ] 6 cm... (i) Let the side of the square be x cm. Then, perimeter of the square length of the wire or, x 6 [Using (i)] or, x cm Hence, the length of the side of the square is cm. 8. Radius of the wheel 5 cm Now, perimeter of the wheel r 5 5 cm wheel number of revolutions made by the wh eel N (perimeter of the wheel) 960 cm 5 or, N or, N Number of complete revolutions Area of circle r, where r is the radius. Area of the biggest circle 0 00 cm Area of the smallest circle 00 cm 00 cm

54 55 Concept of Arithmetic Now, if the radius of the smallest circle be r. Now, according to the question, r 00 or, r 00 or, r 0 radius of the smallest circle 0 cm 0. Each side of the square inscribed in semicircle BC OA a In right angle triangle OAB OA + AB OB or, or, a a r 5 a r r a 5 Area of the square inscribed in semicircle r a 5 Diagonal of the square inscribed in a circle r Area of this square (r ) r Required ratio r : r : 5 5. For the first triangle, we have a 5, b 5 and c 66 or : 5 a b c s cm Area of the first triangle s ( s a) ( s b) ( s c ) ( 5)( 5)( 66) cm...(i) For the second triangle, we have a, b 56, c 65 s a b c cm Area of the second triangle s ( s a) ( s b) ( s c ) ( )( 56)( 65) 9 cm...(ii) Let r be the radius of the circle. Then, Area of the circle Sum of the areas of two triangles or, r or, r 9 9 or, r 88 or, r or, r cm. For semi-circle with diameter PS, Radius 6 cm Area of such semi-circle r 96 (6) cm Diameter PS cm PQ QR RS cm QS QR + RS 8 cm For semi-circle with diameter QS, radius m Its area 6 r () cm For semi-circle with diameter PQ, radius cm Its area cm Area of shaded region cm.. Diameter of each of the semi-circle 0 m. 0 Radius, r

55 Mensuration I (Area and Perimeter) 555 Area of each semi-circle r Area of semi-circle grassy plots r r sq m Cost of turfing the plots at the rate of Rs.5 per sq m Rs (.5 5 ) Rs Radius of flower beds I and II 50 m. Let r be the radius of circular lawn r 00 or r 00 or, r 00 r / 050 (050) Radius of the circular lawn.0 m. 6. Area of rectangle ABCD Length Breadth AB BC ( ) 98 cm Area of unshaded region r r r r Area of each of flower beds I and II r sq m 95 sq m Total area of flower beds I and II sq m Radius of each of flower beds III and IV K KUNDAN 50 5 m Area of each of flower beds III and IV r. 5 5 sq m 98.5 sq m Total area of flower beds III and IV 98.5 sq m 96.6 sq m Total area of semi-circular beds ( ) sq m sq m Cost of levelling the flower beds at the rate of 60 p per sq m Rs Rs Area of rectangular park 0 m 00 m 000 m (.5).5. 5 cm Area of the shaded region Area of rectangle ABCD Area of unshaded region 98 cm cm cm.. Area of the shaded region (Area of rectangle with sides 0m and m) + [Area of the semi-circle with radius (5+) cm ie 9 cm - Area of the semi-circle with radius 5 m] (0 ) + (9 5 ) (680) (9 5)(9 5) 60 8 m ( ) m ( ) m 056 m 8. (i) Area of segment BPF Area of sector ABPF Area of ABF Area of the park excluding the circular lawn 800 m Area of circular lawn ( ) m 00 m Draw AK BF BAF is an angle of regular hexagon BAF 0...() The perpendicular from the centre of a circle to a chord disects the chord BK KF...() AB AF (Side of a regular hexagon) AK AK (common side)

56 556 Concept of Arithmetic ABK AFK (SSS congruence rule) BAK FAK Also BAK FAK BAF 0º BAK 60º ABK is a 0º 60º 90º triangle AK hypotenuse (AB) 6 0. Area of equilateral (side) and BK (AB) 6 cm BF BK 6 cm. Now Area of sector ABPF sector angle r cm 60 Area of ABF BF AK cm Area of segment BPF Area of sector ABPF Area of ABF.68 m 5.5 cm. cm. (ii) Area of shaded portion Area of hexagon ABCDEF area of ABF) cm 5.5 cm.85 cm. 9. Radius of sector CXB cm cm Area of three sectors sector angle r cm Area of shaded region Area of equilateral Area of three sectors (8.868 ).868 cm.. The radius of circle cm An arc ABC subtends an angle of 60º at the centre OA OB cm OAB is equilateral. OAB OBA A B cm 60 O (i) Length of the arc cm cm 60 circumference 60 (ii) Area of the sector area of the circle 60 Radius of sector DXY cm cm cm 90 5 cm 60 Area of sector CXB cm 60 Area of sector DXY Area of rectangle ABCD AB BC cm cm 9 cm Area of the shaded portion Area of rectangle Area of two sectors 9 ( ) cm. 60 sq cm sq cm 60 (iii) Area of segment Area of sector Area of equilateral AOB of side cm sq cm. sq cm ( 90.95) sq cm 0.0 sq cm.

57 Mensuration I (Area and Perimeter) 55. Let the radius of the swimming pool be r m. Width of the path m Hence side of square Then area of the swimming pool r sq m and area of the swimming pool including the path ( r ) sq m 0 a. OC CB 0 cm In right-angled OCB, OB cm 0 cm ( r ) r r 5 It is given that the area of the path is th 5 part of the area of the swimming pool. 5 or, r r r or, 5( r 8r 6) 5r r or, 5r 00r 00 5r r or, r 00r r r , 0 0 0, 0 Since radius of the swimming pool cannot be negative. Hence radius 0 m.. Circumference of circle 80 cm Radius of sector OPBQ 0 cm. Area of sector OPBQ sector angle 60 90º 60 r cm Area of square OABC {(0) } 00 cm. Hence the area of the shaded region Area of sector OPBQ Area of square OABC 68 cm 00 cm 8 cm. 5. Let r be the radius of inner circle and R be the radius of outer circle or, r 80 cm or r 0 cm Let the side of square be a cm. Since from the centre of a circle to a chord bisects the chord, OM a and AM a Now in right-angled triangle OAM, we have OA or,, or, OM AM or r a a 0 a 0 a 0 or a Then, R r 9 ΔAOD ~ ΔDOC...(i) OD OC OA OD or, OD OA OC or R 5 RR 9 or, R 0R 5 R 9R or R 5 (5 r) 9 [using (i)] or, r 50 9 or, r

58 558 Concept of Arithmetic r 0.5 Area of shaded portion R r [(5) (0.5) ] cm. 6. Let the radii of the given circles be R and r respectively. Sum of their areas or, R r 6 (R r ) 6 or, R r 6 6 cm...(i) If O and O be the centres of the given circles, then OO R r Total area of unshaded corners 8.5 m Area of shaded portion Area of rectangle Area of corners (8 ) m. 8. Required area Area of square Area of one sector 96 9 (96 5 ) cm. 9. Area of shaded region Area of a square (Area of a semi-circle of radius cm) (96 5 ) cm. 0. In ABC, A 90 By pythagoras theorem, we get BC AB AC () () 5 BC 5 cm 5 cm 6 R r (Given)...(ii) Now, R r R r R r r 6 6 [Using (i) and (ii)] or, R or, R r 6 96 R r 96 Solving (ii) and (iii), we get R 0...(iii) 0 R 0 From (iii), 0 + r or r 0 Hence radii of the given circles are 0 cm and cm respectively.. Area of rectangle (8 6 ) 8 m Area of shaded portion cm.. Let r cm be the radius of the inner circle, then The radius of the outer circle (r + ) cm. Area of the shaded region 0 cm or, ( r ) r 0 or, ( r r 9 r ) 0 Area of one corner (unshaded) (0) cm or, r 9 0 or, r 5 or, r 8 r Radius of outer circle ( + ) cm cm.

59 Mensuration I (Area and Perimeter) 559. Side of the square cm cm The area of the unshaded region of the square () 0 (6 0) cm 99 cm. Area of the shaded region Area of the equilateral + Area of the semicircle side r [ BC cm is the diameter of semicircle r cm ] cm.. AC radius 5 cm 0 cm. As angle in a semi-circle is 90º. So ADC 90 and ABC 90 AD 5, BC 5 ] cm.. From the given figure it can be seen that Radius OP OR OQ OS cm Taking the semi-circle POQS first Area of shaded small circle OS 8 5. (OS) 5 cm Now taking the semi-circle OPRQ Area of the semi-circle OPRQ (OR) 08 cm Area of triangle PRQ base height PQ OR 96 cm Area of shaded region in semi-circle OPRQ (08 96) cm cm The area of the shaded region Area of shaded region in semi-circle OPQR + Area of shaded region in POQS (5 + ) cm 66 cm By pythagoras theorem; AD AC CD AD 6 6 cm and AB + BC AC or, AB + AB 00 or, AB 00 or, AB 50 or, AB 50 5 cm The area of the shaded region area of the circle area of ADC area of ABC r AD CD AB BC [ AD 5 cm, CD 8 cm In the given figure PQRS is a rectangular field in which SR 6 m and PS m. We have to find the cost of shaded portion in the given figure. The semicircle PAS and QBR makes a complete circle of radius QR PS m area of circle r. m Similarly, PCQ and SDR also makes a complete circle of radius PQ SR 6 m area of circle r. 9 m

60 560 Concept of Arithmetic 6. Total area in which flower is raised m total cost (. 8 ) Rs 6.56 Area of the square ( cm) 96 cm. Area of circular part at one vertex 90º 60º 5 cm 5 Total area of circular parts 5 cm. Area of the region of the square that remains outside the region of any of the circle (96 5) cm cm.. In the following figure, BDEF is the brick put to stop the wheel. Here, r 0 cm and D A. 0 0 sq cm sq cm 0 9. Area of the minor segment Area of sector OAB Area of the rightangled triangle OAB cm (8.5 50) cm 8.5 cm Area of the major segment Area of the circle Area of the minor segment [ ] cm [ 8.5] cm 85.5 cm Note: We know that the area of a minor segment of angle in a circle of radius r is given by A r sin [Always Remember] 60 Here BD AC EF 5 cm AB CD 5 cm OA OD r (say) [Radius of the wheel] In the OCD; OC OA CA (r 5) cm CD 5 cm Here, OD OC + CD or, r ( r 5) 5 Here, r 0, 90 A. 90 (0) sin 90 cm 60. (0) cm [. 5 50] cm (8.5 50)cm 8.5 cm 0. We know that the area of a minor segment of angle in a circle of radius r is given by or, r r 5 0r 5 or, 0r 50 r 5 cm Therefore, the required radius of the wheel 5 cm. 8. Angle described by minute-hand in 60 minutes 60 Angle described by minute-hand in one 60 minute 6 60 We know that the area A of a sector of angle D, in a circle of radius r is given. A D 60 r A r 60 sin. 60 (5) sin 60 cm cm 5 [ ] cm cm 0. cm

61 Mensuration I (Area and Perimeter) 56 Area of the major segment Area of the circle Area of the minor segment [. (5) 0.] cm [ ] cm cm Note: We can have another method for finding area of the minor segment. Clearly, from the figure, triangle OAB is an equilateral triangle with the side 5 cm. Now, area of the minor segment Area of the sector OAB Area of the triangle OAB 60 (5) (5) 60 cm. 5 6 cm 0. cm. Area of the square lawn ABCD (56 56) sq m Let OA OB x metres so, x + x 56 or, x or, x 8 56 Now, area of sector OAB x x Also, area of OAB 8 56 sq m [Putting the value of x 8 56] So, area of flower bed AB sq m (AOB 90 ) sq m 8 56 sq m sq m Similarly, area of the other flower bed sq m Therefore, total area sq m 8 56 sq m sq m 0 sq m Alternative Method: Total area Area of sector OAB + Area of sector ODC + Area of OAD + Area of OBC sq m sq m (56 ) 0 sq m. Area of square ABCD ( ) sq cm 96 sq cm. Diameter of each circle m Radius of each circle cm Area of one circle r sq cm Therefore, area of the four circles 5 sq cm Hence, area of the shaded region (96 5 ) sq cm sq cm Let us mark the four unshaded regions as I, II, III and IV as in the figure. Area of I + Area of III Area of ABCD Areas of two semicircles of each of radius 5 cm sq cm (00. 5) sq cm.5 sq cm Similarly, Area of II + Area of IV.5 sq cm So, area of the shaded region Area of ABCD Area of (I + II + III + IV) (00.5) sq cm (00 ) 5 sq cm