Fdaytalk.com. Acute angle The angle which is less than Right angle The angle which is equal to 90 0

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1 Acute angle The angle which is less than 90 0 Right angle The angle which is equal to 90 0 Obtuse angle The angle which is greater than 90 0 Straight angle The angle which is Reflexive angle The angle which is between to Complete angle The angle which is Supplementary Angles When sum of two angles is equal to 180 ( A + B = Complementary Angles When sum of two angles is equal to 90 A + B = 90 0 LINES When two lines crossed each other 1 = 3 and = 4 [vertically opposite angles] 1 + = = = = Transversal

2 1,, 7 and 8 are exterior angles 3, 4, 5 and 6 are interior angle 1 and 7 & and 8 are alternate exterior angles 3 and 5 & 4 and 6 are alternate interior angles Corresponding angles are equal 1 = 5, = 6, 4 = 8, 3 = 7 Alternate interior angles are equal 4 = 6, 3 = 5 Alternate exterior angles are equal = 8, 1 = 7 And = 180, = l, m and n are three parallel lines and two transverse lines are passed across Then, P Q = X Y EXAMPLE

3 ABC is a Triangle and D, E are the mid points of AB and AC sides Then, AD = AE DB EC ABC is a Triangle A, B and C are interior angles A + B + C = And D is a exterior Exterior angle is sum of the opposite interior angles D = A + B TRIANGLES Area of triangle (Hero s formula) = s(s a)(s b)(s c) S = a+b+c Based on sides Triangles are divided into 1 Scaline Triangle: No two sides are equal Isosceless Triangle: Any two sides are equal 3 Equilateral Triangle: All sides are equal Based on angles 1 Acute angle Triangle: The angle which is less than 90 0 Right angle Triangle: One angle should be Obtuse angle Triangle: The angle which is greater than 90 0 In Triangle ABC, if a is the largest side a < b + c => Acute angle Triangle a = b + c => Right angle Triangle a > b + c => Obtuse angle Triangle Congruency of Triangle Congruency means same (same in size and shape)

4 Three sides and three angles must be equal The above two Triangles ABC & DEF have equal sides and equal angles Therefore, ABC DEF 1 SIDE SIDE SIDE (SSS) In both two Triangles all corresponding sides are equal The above two Triangles, ABC DEF SIDE ANGLE SIDE (SAS) In this case, angle must be between two sides of Triangles The above two Triangles, BCA FED 3 ANGLE SIDE ANGLE (ASA) In this case, two angles and one side consider in both Triangles and side must be on angles side So, CAB EFD 4 RIGHT ANGLE HYPOTENUS SIDE (RHS) Here, in both the two Triangles, one angle must be 90 0 and Hypotenuse is same and one side is same

5 From above two Triangles, BCA EFD Similar Triangles In case of Similar Triangles, shape is the same From above two Triangles, angles are equal in both the Triangles and having same shape So, ABC ~ DEF AND a d = b e = c f = h 1 = 1 = P 1 = S 1 h P S Here, h 1 and h are the heights of both the Triangles ABC & DEF 1 and are the Areas of both the Triangles ABC & DEF P 1 and P are the Perimeters of both the Triangles ABC & DEF S 1 and S are the Semi Perimeters of both the Triangles ABC & DEF Mid Point Theorem In Triangle ABC, D & E are the mid points of AB & AC and DE is parallel to BC Then, DE = 1 BC MEDIANS & ITS PROPERTIES: A line segment joining from the mid point of the side to the opposite vertex

6 In Triangle ABC, DC is the median Here, AB, BC & CA are sides AD, BE & CF are Medians G is Centroid Centroid: The point of intersection of all the three medians of a Triangle is called Centroid 1 Median divides the Triangle into two equal parts Centroid divides the Median in : 1 3 Centroid always lies in inside the Triangle 4 Sum of the sides is greater than sum of the Medians AB + BC + CA > AD + BE + CF 5 Formula for Median, AD = 1 AB + AC BC BE = 1 AB + BC AC CF = 1 AC + BC AB ALTITUDE & ITS PROPERTIES The Altitude of a Triangle is a line segment perpendicularly drawn from vertex to the side opposite to it. The side on which the perpendicular is being drawn is called its base In Triangle ABC, AB, BC & CA are sides AD = altitude

7 In Triangle ABC AB, BC & CA are sides AD, BE & CF are altitudes O = Ortho centre ORTHO CENTRE The point of intersection of all the three Altitudes of a Triangle is called its Ortho centre. 1 Sum of the sides of Triangle is greater than sum of the Altitudes AB + BC + CA > AD + BE + CF In Acute angletriangle, Orho centre lies in inside the Triangle 3 In Right angle Triangle, Ortho centre lies on the right angle 4 In Obtuse angle Triangle, Ortho centre lies on the outside the Triangle Here, O = Ortho centre Then, AOC = B AND, BOC = A AOB = C ANGLE BI SECTOR & ITS PROPERTIES The Angle Bi sector is, the line which is bi sects the angle equally (divided into two equal parts) Here AB, BC & CA are sides of Triangle BD = angle bi sector line

8 In Triangle ABC AB, BC & CA are sides AD, BE & CF are Angle bi sector lines I = In centre of a Trianlge IN CENTRE: BIC = A CIA = B AIB = C EXAMPLE: ABC is a Triangle BI and CI are angle bisectors of ABC and ACB is BAC = 50 0, find BIC.. SHORT METHOD: BIC = A BIC = (answer) [Theorem] EX CENTRE Ex centre is formed by two external angles bi sector and one internal angle bi sector.

9 1 There is only one in centre And, there are three Ex centre formed 3 In centre is equidistance from the sides IN CIRCLE BD = Angle of bi sector BC = CD BA DA PERPENDICULAR BI SECTOR & ITS PROPERTIES Perpendicular bi sector is, a line passes through the mid point of opposite vertex perpendicularly

10 In Triangle ABC, D is the mid point of BC CIRCUM CENTRE In Triangle ABC, S = Circum centre The circle is formed is Circum circle 1 Circum centre is equidistance from all the vertices In Acute angle Triangle, circum centre lies in inside the Triangle 3 In Right angle Triangle, circum centre lies on mid point of Hypotenuse 4 In Obtuse angle Triangle, circum centre lies on outside the Triangle RIGHT ANGLE TRIANGLE In Right angle Triangle ABC I = In centre S = Circum centre In radius (r) = b+c a Circum radius (R) = a Area of Right angle Triangle = 1 base height

11 PYTHOGOROS TRIPLEX 3, 4 & 5 => = 5 5, 1 & 13 => = 13 7, 4 & 5 => = 5 8, 15 & 17 => = 17 9, 40 & 41 => = 41 11, 60 & 61 => = 61 ISOSCELES TRIANGLE The Triangle which have two equal sides and other is different Here, AB = AC B = C ABD ACD h = D is the mid point of BC then, BD = DC EQUILATERAL TRIANGLE In Equilateral Triangle, all sides and all angles are equal Here, AB = BC = CA = a A = B = C = 60 0 h = height From diagram, a = h + ( a )

12 By solving, we get h = 3 4 a Area = 1 a h => 1 = 3 4 a In radius (r) = a 3 Circum radius (R) = a 3 Fdaytalk.com a 3 4 a ***** In Equilateral Triangle Centroid, Ortho centre, In centre and Circum centre are coincide at the same point QUADRILATERALS Sum of the angles in a Quadrilateral is Quadrilateral means contain four sides PARALLELOGRAM In Parallelogram, opposite sides are equal and opposite angles are equal Here, AB //DC BC//AD AB = DC & BC = AD Opposite angles are equal, A = C & B = D Adjacent angles sum = A + B = B + C = C + D = D + A = Diagonals bi sect each other, OA = OC & OB = OD Here, AC & BD are diagonals Diagonals divide the Parallelogram into four equal areas AOB COD AOD BOC

13 Sum of the squares of the sides = Sum of the squares of the Diagonals AB + BC + CD + DA = AC + BD If AB//DC, then CDE is Transversal RECTANGLE Rectangle has two equal opposite sides and all angles are same with 90 0 Here, AB = DC AD = BC l = length b = breadth d 1 & d = diagonals Area = l b Perimeter = (l +b) Diagonal = l + b RHOMBUS All sides are equal Diagonals are perpendicular to each other Diagonals bi sect the interior angles AB = BC = CD = DA d 1 & d = diagonals

14 Side = 1 d1 + d Area = 1 (d 1 d ) Perimeter = d1 + d The four Triangles formed are congruent to each other AOB AOD BOC DOC SQUARE AB = BC = CD = DA = a = side d = diagonal Area = a Perimeter = 4a Diagonal = a TRAPEZIUM Here, AB// CD a, b are the lengths of the parallel sides h = height between the parallel lines Area = 1 h (a + b) Median = 1 (a + b) Area of AOC = Area of BOD Perimeter = AB + BC + CD + DA EXAMPLE: The parallel sides of a Trapezium are in a ratio : 3 and their shortest distance is 1 cm. If the area of the Trapezium is 480 sq. cm., the longer of the parallel sides is of length.. Sides of the Trapezium (l 1 & l ) = x and 3x cm Height (shortest distance), h = 1 cm Area = 1 ( l 1 + l ) h [Formula]

15 Here, l 1 & l are parallel sides And h = distance between the, l 1 & l 480 = 1 (x + 3x) 1 x = 16 Therefore, longer side = 3x => 3 16 = 48 cm (answer) POLYGONS Fdaytalk.com Polygon is nothing but contains more than four sides In Regular Polygon = All interior angles Sum of interior angles of any Polygon = (n ) No. of Diagonals of Polygon = n = no. of sides REGULAR POLYGON All interior angles are equal Each exterior angle = 3600 n Each interior angle = (n ) 1800 n (n 3) Exterior angle = interior angle CUBE n Volume = a 3 ( a = side of cube ) Lateral surface area = 4 a Total surface area = 6 a Diagonal = 3 a

16 CUBOID Volume = length (l) breadth (b) height (h) Total surface area = (lb + bh + lh) Area of four walls of a room = h (l + b) Diagonal = l + b + h CYLINDER Here, r = radius, h = height Volume (v) = πr h Curved surface area = πrh Total surface area = πrh + πr Cylinder is nothing but no. of circles placed one by one vertically CONE Here, r = radius, h = height And, l = r + h Volume (v) = 1/3 πr h Curved surface area = πrl Total surface area = πrl + πr Cone is formed by, when Cylinder is divided into three parts

17 Frustum of Cone Here, r 1 & r are radii of frustum And, h= height, l = length Then, = 1 3 π (r 1 + r + r 1 r ) h Surface area = π l (r 1 + r ) + πr 1 + πr SPHERE Here, r = radius Volume (v) = 4 3 πr3 Total surface area = 4 πr HEMISPHERE Here, r = radius Spherical Cell Volume = 4 π 3 (R3 r 3 ) Total surface area = 4 π (R r ) CIRCLES

18 Here, r = radius O = Centre Area = πr Perimeter = πr FOR SEMICIRCLE Area = πr / Perimeter = πr + r or r 36 Area of the ring or circular path = π (R r ) 7 Length of the arc = πr θ 360 Area of sector = πr θ 360 Perimeter of sector = πr θ r CHORD:

19 Here, O = Centre AB = Chord **** largest chord is diameter Here, AB = Chord O = Centre From Chord to any point on circle joins the lines, angle occurred at different places is same x = x Here, AB & CD are two equal chords (AB = CD), then 1 = The angle subtended by an arc of circle at the centre is double the angle subtended by it at any point on the remaining part of the circle i. e AOB = ACB

20 Angle in the same segment of a circle is equal AOB = ACB AOB = ADB ACB = ADB The opposite pairs of angles of cyclic quadrilateral is supplementary each other A + C = 180 B + D = 180 From above diagram, AB = Chord O = Centre P = Mid point of A & B r = radius Then, r = y + ( x ) TANGENTS & SECANTS Tangent touch the circle at one point while secant touch the circle at two points Here, AP = Tangent

21 BCP = Secant and A = 90 always If two chords AB and CD intersect internally at a point P PA PB = PC PD If two chords AB & CD intersect externally at a point P. Then PA PB = PC PD If ABP is secant to a circle intersecting the circle at A & B, and PT is tangent, then PA PB = PT (Tangent Secant theorem) The side of BC of ABC is produced to D The bisector of LA meets BC at L, then ABC + ACD = ALC

22 Here, Y = 75 & Z = 45 (Alternate angles are equal) X = 60 [180 (75+45)] Here, external angle = 135 And, 135 = X + 65 ( A + B = 135) X = 70 Here, X + 35 = X = 7 Here, R = circum radius X = side of a equilateral triangle O = centre of circle When an equilateral triangle is inscribed in a circle Radius (R) = x 3

23 Here, r = in-radius X = side of a equilateral triangle O = centre of circle When a circle is inscribed in an equilateral triangle Radius ( r ) = x 3 Here, r = radius O = centre of circle T = tangent P = external point from center And, d = distance between the center of circle and external point Then, length of the Tangent = d r Direct Common Tangent If the two circles are on the same side of a line, the common tangent is said to be direct common tangent Here, r = radius of one circle R = radius of another circle O = centre of circle And, d = distance between two centers of circles Then, the length of the direct common tangent = d (R r)

24 Transverse Common Tangent If the two circles are on the opposite side of a line, the common tangent is said to be transverse common tangent Here, r = radius one circle R = radius of another circle O = center of circles And, d = distance between the centers Then, the length of the Transverse common tangent = d (R + r) EXAMPLE: The radii of two circles are 5 cm and 3 cm respectively and the distance between their centres is 4 cm. Then the length of the Transverse common tangent is Transverse common tangent = d (R + r) Here, d = 4 cm, r = 3 cm and R = 5 cm = 4 (5 + 3) = 16 cm (answer)

25 Here, O = Centre 1 & are Direct Common Tangents 3 & 4 are Transverse Common Tangents Here, O = Centre 1 & are Direct Common Tangent and 3 is a Transverse Common Tangent Here, O = Centre 1 & are Direct Common Tangent Transverse Common Tangent = 0 Here, 1 is a Direct Common Tangent Transverse Common Tangent = 0

26 Here, O = Centre Direct Common Tangent = 0 Transverse Common Tangent = 0 MENSURATION Regular Polygon A polygon is a regular polygon if its all sides and all angles are equal Each internal angle of a regular polygon of n sides = (n 4) 90 n Exterior angle of a Regular Polygon of n sides = 3600 EXAMPLE: Each interior angle of Regular Polygon is 18 0 more than eight times an exterior angle. The no. of sides of the Polygon is Each internal angle of a regular polygon of n sides = (n 4) 90 n Exterior angle of a Regular Polygon of n sides = 3600 According to question n n n = 0 (answer) (n 4) 90 n = 360 n Area of regular Hexagon = (side) Area of circum circle of regular hexagon = π (side) Radius of in- circle of a regular hexagon = 3 side Area of in- circle = 3 π (side) 4

27 No. of diagonals in n sides polygon = n (n 3) EXAMPLE: A Polygon has 54 diagonals. The number of sides in the Polygon is. No. of diagonals in n sides polygon = n (n 3) 54 = n(n 3) n = 1 (answer) [Formula] Total surface area of Prism = Curved surface area + area of base Regular Tetrahedron Volume = 1 a3 Surface area = 3 a Here, a = side of tetrahedron Right Pyramid Volume = 1 (area of the base) vertical height 3 Lateral surface area = 1 (perimeter of the base) slant height Total surface area = Area of the base + Lateral surface area Volume of the Prism = Area of the base height 1 Each edge of a regular tetrahedron is 3 cm, then its volume is Tetrahedron is a figure composed of four triangular faces. For a regular tetrahedron, Surface area of each face = 3 4 Total surface area = a = 3 a Height of pyramid = 3 a a [Area of equilateral triangle] Volume = = 9 cm3 (answer) a a = a3 33, In this case volume = 3 6 6

28 If the radii of the circular ends of a truncated conical bucket which is 45cm high be 8 cm and 7 cm, then the capacity of the bucket in cubic centimeter is (use π = 7 ) r h l R This is the case of frustum of a cone Volume of frustum of a cone = 1 3 πh [r + Rr + R ] Lateral surface area = πl (R + r) In this case, volume = [ ] = cm 3 (answer) 3 A solid cone of height 9 cm with diameter of its base 18 cm is cut out from a wooden solid sphere of radius 9 cm. The percentage of wood wasted is % of wood wasted = [ = 75% (answer) 4 3 π π ] π 93 4 If the four equal circles of radius 3 cm touch each other externally, then the area of the region bounded by the four circles is D C Area of shaded region = 6 π 3 A B = 36-9π = 9 (4 π) (answer) 5 A path of uniform width runs all around the inside of rectangular field 116 m by 68 m and occupies 70 sq. m. Find the width of the path: x 116 m 68 m 116x + (68 x) x = x + 68x x = 360 x 9x = 0 x = m (answer)

29 6 ABCD is a square, 4 equal circles are just touching each other whose centres are the vertices A, B, C, D of the square. What is the ratio of the shaded to the unshaded area within square? If radius is r, Unshaded area = πr Area of square = 4r D C Shaded area = 4r πr = (4 π) r A B Shaded : Unshaded = 4 π : π 6 : 7 7 i.e. 3 : 11 (answer) 7 A goat is tied on the corner of a rectangular field of size 30 m 0 m by a 14 m long rope. The area of the region that she graze is.. Required area = = 154 m (answer) If each side of tetrahedron is 1 cm long then the volume of tetrahedron will be Volume of tetrahedron = 1 (side)3 [Formula] = 1 (1)3 = 144 cm 3 (answer) 9 In an Equilateral Triangle of side 4 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle is Given that, Side of Equilateral Triangle (a) = 4 cm Radius of inner circle (r) = a [Formula] 3 r = 4 => r = Remaining area of Triangle = Area of Equilateral Triangle Area of Circle = 3 4 a πr

30 = = sq. cm (answer) Fdaytalk.com 10 A Conical cup is filled with ice cream. The ice cream forms a Hemi Spherical shape on its open top. The height of the Hemi Spherical part is 7 cm. the radius of Hemi Spherical part equals the height of the cone. Then the volume of the ice cream Given that, radius of the Hemi Sphere (r) = 7 cm and Height of the Cone (h) = 7 cm Volume of the Hemi Sphere = 3 πr3 [Formula] = = cm 3 Volume of the Cone = 1 3 πr h [Formula] = = cm 3 Therefore, Volume of the Ice cream = Volume of the Hemi Sphere + Volume of the Cone = => 1078 cm 3 (answer) 11 If the radius of the Cylinder is decreased by 8% while its height is increased by 4%. What will be the effect on volume.. PERCENTAGE METHOD: Given that, Radius (r)= 8% decrease Height (h) = 4% increase Volume of the Cylinder = πr h r h r r h Effect on Volume = = % (answer) 1 If the radius of the Cylinder is increased by 5% and its height remains unchanged, then find the percentage increase in Volume..

31 PERCENTAGE METHOD: Given that, Radius (r) = 5% increase Height (h) = Same (100)% Volume of the Cylinder = πr h r h r r h Percentage increase in Volume = = 56. 5% (answer) 13 The radius of the base of a right circular Cone is increased by 15% keeping the height fixed. The volume of the cone will be increased by PERCENTAGE METHOD: Given that, radius (r) = 15% increase height (h) = Same (100%) Volume of the Cone = 1 3 πr h r h r r h = 3. 5% (answer) 13 On decreasing each side of and Equilateral Triangle by cm, there is a decrease of 4 3 cm in its area. The length of the each side of the Triangle is. Let the side of the Equilateral Triangle = a cm Area of Equilateral Triangle = 3 4 a [Formula] After decreasing each side, Area of Equilateral Triangle = 3 According to question 3 4 a 3 4 (a ) = 4 3 Side (a) = 5 cm (answer) 4 (a ) 14 With vertices of Triangle ABC as centers, three circles are drawn, each touching the other two externally. If the sides of the Triangle are 9 cm, 7 cm and 6 cm, then the radii of the circles in centimeters are

32 Given that, Sides of Triangle AB = 9cm, BC = 7cm & CA = 6 cm Let the radius of the three circles are = R 1 R and R 3 From above diagram, R 1 + R = 9. (1) R + R 3 = 7.. () R 3 + R 1 = 6.. (3) From equations 1, & 3, we get R 1 = 4cm, R = 5 cm & R 3 = cm (answer) 15 Three Spherical balls of radius of 1 cm, cm and 3 cm are melted to form a single Spherical ball. In the process, the loss of material is 5%. The radius of new ball is.. Given that, radius of three Spherical balls = R 1 = 1 cm, R = cm and R 3 = 3 cm Volume of the Sphere = 4 3 πr3 [Formula] Volume of the three Spherical balls = 4 3 π ( ) = 4 3 π 36 Volume of the new ball = 4 75 π πr3 = 4 75 π Radius (R) = 3 cm (answer) [5% material loss, given) 16 The ratio of the area of Sector of a Circle to the area of the Circles is 1 : 4. If the area of the circle is 154 cm, the perimeter of the sector is.. Area of Sector of Circle = πr Area of Circle = πr θ 360 [Formula] Given that, Area of Sector Area of Circle = 1 4 πr θ 360 πr = 1 4 θ = 90 0 Given that, πr = 154 => r = 7 cm Length of the Sector (l) = rθ => 7 90 = l = 11 cm π 180

33 Perimeter of the Sector = l + r = = 5 cm (answer) [Formula] 17 The lengths of three medians of a Triangle are 9 cm, 1 cm and 15 cm. The area of the Triangle is.. Let lengths of the medians a = 9 cm, b = 15 cm & c = 1 cm Area of the Triangle (When Medians given) = 1 3 (a b + b c + c a ) (a 4 + b 4 + c 4 ) = 1 3 ( ) ( ) By solving, we get = 7 sq. cm (answer) 18 In a Triangle, distances from Centroid to vertices are respectively 4 cm, 6 cm and 8 cm. Find Medians.. The point of intersection of Medians is called the Centroid of a Triangle Given that, AG = 4 cm, BG = 6 cm & CG = 8 cm So, Median AD => AG = 3 BG = 3 CG = 3 AD = 4 => AD = 6 cm (answer) BE = 6 => BE = 9 cm (answer) CF = 8 => CF = 1 cm (answer) 19 The length of each edge of a Rectangular Tetrahedron is 1 cm. The area of the total surface area of the Tetrahedron is. Given that, side of Rectangular Tetrahedron (a) = 1 cm

34 Total surface area of the Tetrahedron = a = = sq cm (answer) 0 The sides of a Triangle are 50 cm, 78 cm and 11 cm. The smallest Altitude is. Semi Perimeter of Triangle (s) = Area of the Triangle = s (s a)(s b)(s c) = 10(10 50)(10 78)(10 11) = 1680 sq cm The Altitude will be smallest when base is largest 1 11 h = 1680 h = 30 cm (answer) = 10 cm 1 The ratio of the areas of circum circle and the in circle of a Square ism Let side of a Square is x Radius of in circle = r Radius of ex circle = R Radius of a in circle (r) = side of a square [from above fig.) = x Radius of ex circle (R) = = x Required ratio = = 1 (answer) x x Diagonal of a Square

35 35 ml paint is required to paint a circular plate of 5 cm radius. How much paint is required to paint a similar plate of radius 75 cm. Ratio of radius of two circular plates = 5 : 75 = 1 : 3 Ratio of Areas of two plates = π(1) π(3) = 1 : 9 i. e area of the second plate is 9 times the area of the first plate Therefore, paint required = 35 9 = 315 ml (answer) 3 An Iron Cube of edge 3 cm weighs 15 grams. What is the weight of a similar Iron Cube whose edge is 1 cm Ratio of edges of Iron Cubes = 3 : 1 = 1 : 4 Ratio of their Volumes = = 1 : 64 As Volume is 64 times, weight will also be 64 times Therefore, weight of new Cube = = 960 grams (answer) 5 The base of a right Prism is an Equilateral Triangle of area 173 cm and the volume of the Prism is cm 3. The area of the lateral surface of the Prism is. Given that, area of Equilateral Triangle (base) = 173 cm Volume of the Prism = cm 3 Volume of Prism = Area of Base Height (h) = 173 h h = 60 cm [Formula] Now, Area of an Equilateral Triangle = 3 Side (a) = 0 cm 4 a = 173 [from question] Perimeter of Equilateral Triangle = 3 side=> 3 0 = 60 cm Therefore, Area of the Lateral surface of the Prism = Perimeter of the Base Height (h)

36 = = 3600 sq. cm (answer) 6 The radius of a Cylinder is 10 cm and height is 4 cm. The no. of centimeters that may be added either to the radius or to the height to get the same increase in the volume of the Cylinder Given that, Radius of Cylinder = 10 cm Height of Cylinder = 4 cm Let radius be increased by x cm Volume of the Cylinder = π (10 + x) 4 [V = πr h] If height of Cylinder be increased by x cm Volume of Cylinder = π 10 (4 + x) According to question π (10 + x) 4 = π 10 (4 + x) x = 5 cm (answer) 7 The ratio between the no. of sides of two regular polygons is 1 : and the ratio between their interior angles is : 3. The no. of sides of these polygons is respectively. Let the no. of sides of Polygon be = a & a Interior angle of Polygon = (n ) 180 n [Formula, n = side] According to question a = 4 (a ) 180 a (a ) 180 a = 3 So, sides = 4, 8 (answer) 8 The angles of a Triangle are in Arithmetic Progression. The ratio of the greatest angle in degrees to the no. of radians in the greatest angle is 60 : π. The angles in degrees are. Let the angles of Triangle in A. P are = (a d) 0, a 0 and (a + d) 0 a d + a + a + d = a = 60 0 According to question a d a + d = 60 π

37 60 d = d 180 d = 30 0 [ π = ] Therefore, angles = a d, a & a + d = 60 30, 60, = 30, 60, 90 (answer) 9 If the length and the perimeter of a Rectangle are in the ratio 5 : 16, then its length and breadth will be in the ratio.. Let the length of the Rectangle be x cm and breadth be y cm Therefore, Perimeter of the Rectangle = (x + y) According to question x By solving, we get x : y = 5 : 3 (answer) x + y = The circumference of a Circle is 100 cm. The measure of a side of the square inscribed in this circle is.. Given that Circumference of a Circle = πd = 100 cm d = 100 cm [d = diameter of Circle] π Therefore, Diagonal of inscribed Square = = 100 cm Therefore, Side of the Square = 1 Diagonal π = cm (answer) 31 Find the ratio of the diameter of the Circles inscribed in an Equilateral Triangle, the diameter circumscribing that Equilateral Triangle and the height of the same Equilateral Triangle. Let the side of Equilateral Triangle = a Then height = 3 a Diameter of the inner circle = ( 1 3 Diameter of the outer circle = ( 3 3 a ) = a 3 3 a ) = a 3

38 Required ratio = a 3 a = : 4 : 3 (answer) 3 3 a Fdaytalk.com 3 Two Goats are tethered to diagonally opposites vertices of a field formed by joining the mid points of the adjacent sides of another square field of side 0 m. What is the total grazing area of the two Goats if the length of the rope by which the Goats are tethered is 10 m.? The length of rope of Goat = 10 m Then the two Goats will graze an area = Area of a Semi circle with radius 10 m So, total area grazed = πr = 100π m (answer) 33 Find the area of the shaded region if the radius of each of the circles is 1 cm Length of the AB = = cm Area of Triangle ABC (Equilateral Triangle) = 3 4 As we can see Triangle ABC is a Equilateral Tringle CAB = ABC = ACB = 60 0 Required answer = 3 π = 3 π (answer) 34 What is the in r5adius of the in circle shown in the figure? 360

39 In radius of the Triangle = In radius = ( ) = 4 cm (answer) Area of Triangle Semi perimeter of Triangle 35 ABC and PQR both are similar and perimeter of ABC and PQR are 4 cm and 36 cm respectively if AB = 10 cm. Find PQ.. If two Triangles are similar, then Ratio of Side Ratio of Perimeter 4 36 = AB PQ PQ = PQ = 15 cm (answer) = Perimiter of ABC Perimeter of PQR 36 A Cylindrical Iron rod whose height is eight times of its radius is melted and cast into smaller Spheres whose radius is half of the radius of the Cylinder. Thus the no. of Spheres so formed is. Let the radius of the base of the Cylinder = r unit Height = 8r units Volume of Cylinder =πr h [Formula] = π r 8r = 8πr 3 Cubic unit Radius of Sphere = r unit Volume of Sphere = 4 3 = 4 3 π (r )3 π r3 [Formula] = πr3 6 Cubic unit Therefore, no. of Spheres = 8πr3 πr 3 = 48 (answer) 6 37 Each of the height and the base of a right Circular Cone is increased by 100%. What will be the increase in the volume of the Cone.

40 Let radius (base) = r Height = h Volume of the Right Circular Cone = = 1 3 πr h [Formula] PERCENTAGE METHOD: Volume of the Right Circular Cone = 1 3 πr h = 1 3 π [r r h] = r (100% increase) r (100% increase) height (100% increase) Required answer = = 700% (answer) 38 Each interior angle of Regular Polygon is The number of sides of the Polygon is Let the no. of sides of Polygon = n Then, each exterior angle = = 36 0 Therefore, no. of sides (n) = 360 n = 10 (answer) A wire in the form of a circle of radius 4 m is cut and again bent in the form of a square. What is the diagonal of the square Given that, radius (r) = 4 m Let side of the square = a m According to question πr = 4 a 4 = 4 a 7 a = 66 m So, the diagonal of square = a = 66 (answer) 40 The circumference of the front wheel of a cart is 30 feet long and that of the back wheel is 36 feet long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel The circumference of the wheel is 30 feet and that of the rear wheel is 36 feet Let the rear wheel make n revolutions. At this time, the front wheel should have made n + 5 revolutions

41 As both the wheels would have covered the same distance n 36 = (n + 5) 30 n = 5 Therefore, distance covered = 5 36 = 900 feet (answer) 41 The area of an Equilateral Triangle inscribed in Circle is 16 3 m. Find the area of the Circle.. Area of Equilateral Triangle = 3 4 a [Formula] 3 4 a = 16 3 a = 8 cm Height of the Equilateral Triangle = 3 8 = 4 3 cm Radius of the Circumcircle = = 8 3 cm Therefore, area of the Circle = π = 64 3 π cm (answer) If ABCD is a Cyclic Quadrilateral in which A = 4x 0, B = 7x 0, C = 5y 0 and D = y 0, then x : y is The sum of opposite angles of Cyclic Quadrilateral is A + C = => 4x + 5y = (1) B + D = x + y = () From equations (1) and () => () 5 (1) x = 70 31

42 y = = Therefore, x : y = = 4 : 3 (answer) 43 A is the centre of Circle whose radius is 8 and B is the centre of a Circle whose diameter is 8. If these two Circles touch externally, then the area of the circle with diameter AB is. From above figure, Diameter AB = => 1 units Radius = 1 = 6 units Therefore, Area of the Circle = πr = π 6 6 = 36π sq. units (answer ) 44 By decreasing 15 0 each angle of a Triangle the ratios of their angles are : 3 : 5, the radian measure for greatest angle is.. Ratio of angle after decreasing 15 0 each angle = : 3 : 5 According to question x + 3x + 5x = x = Greatest angle = = or 8. 5 π 180 = 11π 4 (answer) 45 The in radius of an Equilateral Triangle is of length 3 cm. Then the length of each of its medians is. In radius of Equilateral Triangle = side = Media 3 3 Therefore, Median = 3 In radius Median = 9 cm (answer) 46 The Perimeters of two Squares are 40 cm and 3 cm. The Perimeter of third Square whose area is the difference of the area of the two Squares is

43 Let sides of Squares are a, b and c Given that, Perimeters 4a = 40 => a = 10 cm And 4b = 3 => b = 8 cm According to question c = (10) (8) = 36 c = 6 cm So, Perimeter of a third Square = 4 6 = 4 cm (answer) 47 An angle is 30 0 more than the one half of its complement. Find the angle in degrees Let the angle be x, then its Complementary angle be (90 x) According to question x (90 x) = 30 x = 50 0 (answer) 48 If an angle of a Parallelogram is two third of its adjacent angle, then the largest angle of Parallelogram. Since, adjacent angles of a Parallelogram are Supplementary. Therefore, x + x = x = x => = 7 0 Therefore, angles of Parallelogram = 108 0, 7 0, and 7 0 Largest angle = (answer) 49 The ratio between the length and the breadth of a Rectangular park is 3 :. If a man cycling along the boundary of the park at the speed of 1 kmph completes one round in 8 minutes, then the area of the park is.. Given that, Ratio between length & breadth = 3 : So, distance covered by Man in 8 minutes = = 1600 meters Then, distance = (3x + x) = 1600 = 160 meter So, Area = 3x x => = m (answer)

44 50 The length of the floor of a room is 0 m more than its breadth. The area of the floor remains unchanged even if its length is decreased by 10 m and breadth is increased by 5 m. The area of floor is Let width of floor = x meter length = (x + 0) meter Area of floor = (x + 0) x m In second condition (x + 10) (x +5) = (x + 0) x By solving, we get x = 10 meter Length = x + 0 => = 30 meter Area of floor = = 300 m (answer) 51 A Cube of 384 cm surface area is melted to make x number of small cubes each of 96 cm surface area. The value of x is. Surface area of big Cube = 6A A = side or edge of the big Cube Given that, 6A = 384 A = 8 cm Surface area of small cube = 6a 6a = 96 [From question] a =4 cm 8 3 = x (4) 3 x = 8 (answer) 5 A spherical metal of radius 10 cm is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by: Volume of the Sphere = 4 3 πr3 From question, 4 3 πr3 = πr3 => R = 10r R = Radius of big sphere, r = radius of small sphere 10 = 10 r [R = 10 cm] r = 1 cm Initial surface area of Sphere = 4πR => 400π Final Surface area of 1000 smaller spheres = πr = π 1 = 4000π

45 Increase in Surface area = 4000π 400π Increase in Surface area = 3600π i. e 9 times (answer) 53 The area of paper can be divided into 144 Squares, but if the dimensions of each Square were reduced by cm each, then the no. of Squares so formed are 400. The area of the paper initially was.. Let edge of Square = x cm According to question 114x = 400 (x ) By solving, we get x = 5 and 5 4 edge of Square = 5 cm initial area = 144 x = = 3, 600 cm (answer) 54 A Square and an equilateral triangle have the same perimeter. If the diagonal of the Square is 1 cm, then the area of the Triangle is. Let x be the side of the Square Diagonal (d) = side of a Square 1 = x => x = 1 Perimeter of the Square as well as of the Equilateral Triangle = 4 1 => 48 cm Side of the Equilateral Triangle =>3a = 48 => a = 16 cm From above Triangle AD = AC CD = 16 8 =56 64 AD = 8 3

46 Area of the Triangle ABC = 1 BC AD = = 64 3 cm (answer) 55 Three Circles of equal radius a cm touch each other. The area of the shaded region is From above figure, AB = BC = CA = a cm Area of Equilateral Triangle ABC = 3 4 4a = 3 a Area of 3 Sectors of θ = 60 0 = 3 πa = πa Area of shaded region = area of ABC Area of 3 Sectors = 3 a πa = ( 3 π ) a cm (answer) 56