3. AREA OF TRIANGLES AND QUADRILATERALS

Transcription

1 3. AREA OF TRIANGLES AND QUADRILATERALS. Introduction Area of a plane Figure could be understood as the space occupied by a closed Figure on a plane. The floor of our rooms is a plane surface. The surface of the black board is a plane surface and so on. When we say that the room is bigger than some other room, we basically mean that the Area/Volume of the room is bigger. When we say that the black board is bigger, we basically mean that the writing area of the board is bigger. Figure 3. Suppose some planar object is kept on a surface, it occupies certain portion of the surface, area is nothing but the measurement of the portion occupied on the plane. The diagram above shows a plane represented by the rectangle and two closed irregular plane shapes represented by A and B. Clearly the area occupied by A is greater than the area occupied by B. In this chapter we would be discussing about the area of some simple standard triangles and quadrilaterals.. Area of a Triangle with Given Base and Height From your earlier classes, you know that: Area of a triangle = Base Corresponding Height Any side of the triangle may be taken as base and the length of perpendicular from the opposite vertex to the base is the corresponding height. In given figure, Figure 3.

2 3. Area of Triangles and Quadrilaterals Area of ABC = BC AD sq. units.. Area of a Right Angled Triangle When the triangle is right angled, we can directly apply the above mentioned formula by using two sides containing the right angle as base and height. In given figure, Area of ABC = BC AB sq. units. Figure 3.3. Area of an Equilateral Triangle Let ABC be an equilateral triangle with side a and AD be the perpendicular from A on BC. Then, D is a the mid-point of BC i.e. BD = In right-angled ABD, by Pythagoras theorem, we have AD = AB BD a a 3 3 AD = a = a = a AD = a So, area of ABC = BC AD = a a = a. 4 Figure Area of equilateral triangle with side a units = a sq. units. 4.3 Area of an Isosceles Triangle Let ABC be an isosceles triangle with AB = AC = a and BC = b, and AD be the perpendicular from A on BC. Then, D is the mid-point of BC, i.e. BD = b. In right-angled ABD, by Pythagoras theorem we have: AD = AB BD b b 4a b AD = a = a = 4 4 AD = 4a b Figure 3.5

3 Foundation for Mathematics 3.3 So, area of ABC 4a b = BC AD = b = b 4a b. 4 Area of isosceles ABC with AB = AC = a units and BC = b units = b 4a b 4 sq. units. 3. Area of a Triangle By Using Heron s Formula In a scalene triangle, if the length of each side is given but its height is not known and cannot be obtained easily. In such cases, we take the help of Heron s formula or Hero s formula to find the area. 3. Heron s formula If a, b, c denote the lengths of the sides of a triangle ABC. Then, Area of ABC = s(s a)(s b)(s c) where s= a + b + c, is the semi-perimeter of ABC. PLANCESS CONCEPTS Proof of Heron s Formula: We use the formula Area of a triangle = Base Corresponding Height to prove Heron s Formula If we consider the triangle as shown in the diagram, we have the base but we do not have the height, so our first step would be to find the height (in terms of a, b and c) and then try to prove Heron s Formula Step : Drop a perpendicular from any vertex on the opposite side having length equal to c (as shown Figure 3.6 in the diagram). The length of the altitude is not known (let it be equal to k). The altitude divides the base into two pieces of unknown length. If we call one piece as h, then the other will be c - h. Step : We can solve for h and k in terms of a, b, and c by applying the Pythagorean Theorem for right triangles and obtaining two equations in two unknowns. The two relations we get are h + k = a (c h ) + k = b

4 3.4 Area of Triangles and Quadrilaterals On subtracting these two equations we get h (c h) = a b h c + ch h = a b h = (a b + c ) / (c) Putting this value of h into the first equation and solving for k gives us k = [a (a b + c ) / (4c )] Step 3: Area, = base height = [(ac) (/ 4)(a b + c ) ] = [4(ac) (a b + c ) ] 4 = [4(ac) a + (ab) b (ac) c + (bc) ] 6 = [((ab) + (ac) + (bc) ) a b c ] 6 = [(a+ b+ c)(a+ b c)(a b+ c)(b+ c a)] 6 a+ b+ c a+ b+ c a+ b+ c a+ b+ c = a b c = [s(s a)(s b)(s c)] Nitin Chandrol NTSE Scholar Illustration : Find the area of a triangle, two sides of which are 0 cm and 3 cm and the perimeter is 34 cm. Solution: Let a, b, c be the sides of the given triangle and s be its perimeter such that a = 0 cm, b = 3 cm and s =34 cm i.e. s=7 cm Now, a + b + c = s c =34 c =. s a =7 0 = 7, s - b =7 3 =4 and s c =7 =6 Hence, Area of given triangle = s(s a)(s b)(s c) = = 53.44cm [NCERT] Illustration : The sides of a triangle are in the ratio 5 : 7 : 8 and its perimeter is 400 m. Find its area. Solution: Let us take the sides of the triangle as 5x, 7x and 8x because the ratio of the sides is given to be 5: 7: 8. Also, we are given that 5x + 7x + 8x = 400 0x = 400 x = 0. Hence, the lengths of the three sides are 5 0 m, 7 0 m, 8 0 m. i.e., 00 m, 40 m, 60 m.

5 Foundation for Mathematics 3.5 Now, s = 00 m Area of the triangle = 00 (00 00) (00 40) (00 60)m = m = m Illustration 3: The perimeter of a right triangle is 88 cm and its hypotenuse measures 30 cm. Find the lengths of other sides and calculate its area. Verify the result using Heron s formula. [NCERT] Solution: Perimeter of a right triangle = 88 cm; Hypotenuse = 30 cm Sum of the other two sides = = 58 cm Let one side be x, then the other side is (58 x) cm. In a right angle ABC, by Pythagoras we have AC = AB + BC (30) = (58 x) + x 6900 = x 36x + x 6900 = x 36x x 36x = 0 x 36x = 0 x 58x = 0 x 3x 6x = 0 x(x 3) 6(x 3) = 0 (x 3) (x 6) = 0 Either x = 3 cm or x = 6 cm (i) When x = 3 cm BC = 3 cm and AB = 58 3 = 6 cm (ii) When x = 6 cm BC = 6 cm and AB = 58 6 = 3 cm Hence, the three sides of the triangle are 30 cm, 3 cm and 6 cm Area of right triangle = BC AB = 3 6 = 06 cm Verification by Heron s Formula we have a = 6 cm, b = 30 cm, c = 3 cm s = = 44 cm s a = 44 6 = 8 cm, s b = = 4 cm, s c = 44 3 = cm Area of ABC = s(s a)(s b)(s c) = = 06 cm Illustration 4: Calculate the area of the shaded portion of the as shown in figure. Solution: In right ADB, by Pythagoras Theorem AB = AD + DB = (4) + (3) =

6 3.6 Area of Triangles and Quadrilaterals = 600 cm AB = 600 cm = 40 cm Now, a = 40 cm, b = 96 cm and c = 04 cm a+ b+ c 40cm+ 96cm+ 04cm 40cm s = = = = 0cm Area of ABC = s(s a)(s b)(s c) = 0(0 40)(0 96)(0 04) = = 90cm And area of ABD = AD BD = 4 3 = 384 cm Area of the shaded portion of the triangle = = 536 cm. 4. Applications of Heron s Formula Heron s formula can be applied to find the area of a quadrilateral by dividing the quadrilateral into two triangular parts. If we join any of the two diagonals of the quadrilateral, then we get two triangles. Area of each triangle is calculated and the sum of two areas is the area of the quadrilateral. 4. Rhombus A Rhombus is a quadrilateral whose all the sides are equal and opposite sides are parallel. The diagonals of a rhombus bisect each other at right angle. In figure, ABCD is a rhombus with side a and AC and BD are the two diagonals and AC = d, BD = d, then Perimeter = 4a Area = Area of ABD + Area of CDB Figure 3.7 = BD OA + BD OC = BD [OA+OC] = BD AC = d d or product of diagonals. Area of Rhombus = product of diagonals 4. Trapezium A Trapezium is a quadrilateral which has one pair of opposite sides parallel. In figure, ABCD is a trapezium with AB CD, AB = b, BC = c, CD = a, DA = d and the height of the trapezium ABCD is h, then Figure 3.8

7 Foundation for Mathematics 3.7 Perimeter = a + b + c + d Area = Area of AMD + Area of rectangle MNCD + Area of CNB = y h + a h + x h= h(x + y) + ah = h [x + y + a] = h [(x + y + a) + a] = h [b + a] = (a + b) h = (Sum of the parallel sides) Distance between them. Area of Trapezium = (Sum of the parallel sides) Distance between them 4.3 Quadrilateral A figure, which has four sides and four vertices. Generally a quadrilateral is denoted by the symbol rectangle. In figure, ABCD is a quadrilateral in which AC is a diagonal and AC divides quadrilateral ABCD in two triangle ADC and ABC. Also, h and h are the altitudes of the triangle ADC and ABC respectively. Area = Area of ADC + Area of ABC = AC h + AC h = AC (h + h ) = Figure 3.9 one diagonal (Sum of heights of those triangles whose base is former diagonal) Area of Quadrilateral = is former diagonal) one diagonal (Sum of heights of those triangles whose base Illustration 5: In the Fig. shown, PQRS is a field in the form of a quadrilateral whose sides are indicated in the figure. If SPQ = 90, find the area of the field. Solution: Clearly, SPQ is a right-angled triangle. Therefore, SQ = SP + PQ [Using Pythagoras Theorem] SQ = SQ = m = 674 = 8 m For SPQ, b = 80 m and h = 8 m Therefore, Area of SPQ = b h= 80 8 = 70 m

8 3.8 Area of Triangles and Quadrilaterals For SRQ, we have s = = = 84 m. A = Area of SRQ = 84 (84 56) (84 30) (84 8) = = 5406 = 504m Hence, Area of field =A +A = (70+504) m =4 m Theorem : Parallelograms on the same base or equal base and between the same parallels are equal in area. Given: Two gm ABCD and gm ABEF on the same base AB and between the same parallels AB and FC. To prove: area ( gm ABCD) = area ( gm ABEF) Proof: In ADF and BCE, we have AD = BC [Opposite sides of a gm ] AF = BE [Opposite sides of a gm ] DAF = CBE [ AD BC and AF BE] [Angle between AD and AF = angle between BC and BE] ADF BCE [By SAS] Figure 3.0 Area ( ADF) = area ( BCE)... (i) Area ( gm ABCD) = area (ABED) + area ( BCE) = area (ABED) + area ( ADF) [Using (i)] = area ( gm ABEF). Hence, area ( gm ABCD) = area ( gm ABEF). Hence Proved. NOTE: A rectangle is also parallelogram. Theorem : The area of parallelogram is the product of its base and the corresponding altitude. Given: A gm ABCD in which AB is the base and AL is the corresponding height. To prove: Area ( gm ABCD) = AB AL. Construction: Draw BM DC so that rectangle ABML is formed. Proof: gm ABCD and rectangle ABML are on the same base AB and between the same parallel lines AB and LC. Figure 3. area ( gm ABCD) = area(rectangle ABML) = AB AL. area of a gm = base height. Hence Proved. Theorem 3: Parallelograms on equal bases and between the same parallels are equal in area. Given: Two gm ABCD and PQRS with equal base AB and PQ and between the same parallels, AQ

9 Foundation for Mathematics 3.9 and DR. To prove: area ( gm ABCD) = area ( gm PQRS). Construction: Draw AL DR and PM DR. Proof: AB DR, AL DR and PM DR AL = PM. Area ( gm ABCD) = AB AL = PQ PM [ AB = PQ and AL = PM] Figure 3. = area ( gm PQRS). Hence Proved. Theorem 4: Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Given: Two triangles ABC and PCB on the same base BC and between the same parallel lines BC and AP. To prove: area ( ABC) = area ( PBC) Construction : Through B, draw BD CA intersecting PA produced in D and through C, draw CQ BP, intersecting line AP in Q. Proof: We have, BD CA And, BC DA [By construction] [Given] Quad. BCAD is a parallelogram. Similarly, Quad. BCQP is a parallelogram. Figure 3.3 Now, parallelogram BCQP and BCAD are on the same base BC, and between the same parallels. ar( gm BCQP) = ar( gm BCAD)...(i) We know that the diagonals of a parallelogram divides it into two triangles of equal area. ar( PBC= ar( gm BCQP) And ar( ABC) = ar( gm BCAD)...(ii)...(iii) Now, ar( gm BCQP) = ar( gm BCAD) ar( gm BCAD) = ar( gm BCQP) Hence, ar( ABC) = ar( PBC) Hence Proved. [From (i)] [Using (ii) and (iii)] Illustration 6: Find the area of a trapezium whose parallel sides 50 cm, 6 cm and other two sides are 30 cm.

10 3.0 Area of Triangles and Quadrilaterals Solution: Let PQRS be the given trapezium in which PQ = 50 cm, RS = 6 cm, QR = 30 cm and PS = 30 cm. Draw RT PS. Now, PSRT is a parallelogram in which PS RT and PT RS. PT = RS = 6 cm and QS = PQ - PT = 50 6 = 4 cm In QRT, we have s = = 4 cm Area of QRT = s(s a)(s b)(s c) = 4(4 30)(4 30)(4 4) 4 8 = 7 cm...(i) Let h be the height of QRT, then Area of QRT = (Base Height) = 4 h = h...(ii) From (i) and (ii), we have, h = 7 h = 6 cm Clearly, the height of trapezium PQRS is same as that of QRT. Area of trapezium = (PQ + RS) h Area of trapezium = (50 + 6) 6 cm = 8 cm Illustration 7: A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 5 cm, 56 cm and 60 cm, and the parallelogram stands on the base 56 cm, find the height of the parallelogram Solution: Semi perimeter of OQR, s = = 84 cm Area of ABC = s(s a)(s b)(s c) ( ) ( ) ( ) = = 344cm Area of parallelogram = Area of triangle [Given] h PQ = 344 h 56 cm= 344 cm h = 344 = 4 cm 56 Illustration 8: A field is in the shape of a tapezium whose parallel sides are 50 m and 0 m. The non-parallel sides are 8 m and 6 m. Find the area of the field.

11 Foundation for Mathematics 3. Solution: Through R draw RT SP and draw RU PQ In QRT, we have, s = = 4 m Area of QRT = s(s a)(s b)(s c) ( ) ( ) ( ) = m Now, area of QRT = 336 m Base Altitude = 336 m h = 336 h = 30 Distance between parallel sides of trapezium = m Area of parallelogram, PTRS = base height = m = 56 = 448 m Area of trapezium PQRS = Area of parallelogram PTRS + Area of QRT = = 784 m.

12 3. Area of Triangles and Quadrilaterals SUMMARY Area of a triangle = Base Height 3 Area of an equilateral triangle = a sq. units. 4 Area of an isosceles triangle = b 4a b 4 sq.units. (where a is the length of equal sides and b is the third side.) Heron s formula: Area of ABC = s(s a)(s b)(s c) where s = a + b + c (semi-perimeter) Area of a rhombus = product of diagonals Area of a trapezium = (Sum of the parallel sides) Distance between them. Area of a quadrilateral = (Length of any diagonal) (Sum of the heights from the other two vertices on this diagonal)

13 Foundation for Mathematics 3.3 SOLVED EXAMPLES Example : A traffic signal board, indicating SCHOOL AHEAD is an equilateral triangle with sides of length a. Find the area of the signal board, using Heron s Formula. If its perimeter is 80 cm. Solution: The traffic signal is in the form of an equilateral triangle having side s length a cm its perimeter is 80 cm. i.e. a + a + a = 80 3a = 80 a = 60 s = 80; s = 90 Area of triangle using Heron s Formula = s(s a)(s b)(s c) = 90( 90 60)( 90 60)( 90 60) = = cm (i) o Example : A park, in the shape of a quadrilateral PQRS, has R = 90, PQ = 8 m, QR = 4 m, RS = 0 m, and PS = 6 m. How much area does it occupy? Solution: Given, a quadrilateral PQRS in which PQ = 8m, QR = 4 m, RS = 0m, and PS = 6m. We divide the quadrilateral in two triangular region PQS and QRS. Now, In QRS it is right angle at R. We have QS = QR + RS [By Pythagoras theorem] = = = 676 m QS = 676 = 6m Area of QRS = base height = 0 4 = 0 m Now, for triangle PQS, we have a = 8 m, b = 6m, c = 6m a + b + c s = = = 30m Area of PQS = s(s a)(s b)(s c) = 30 ( 30 8) ( 30 6) ( 30 6) = = 4 35 m Area of quadrilateral PQRS = Area of QRS + area of PQS = ( )m

14 3.4 Area of Triangles and Quadrilaterals Example 3: PQR is a triangle in which S is the mid-point of QR and T is the mid-point of PS. Prove that the area of QTS = area of PQR. 4 Solution: Given: A PQR in which S is the mid-point of QR and T is the mid-point of PS. To prove: ar ( QTS) = ar ( PQR) 4 Proof: PS is median of PQR ar ( PQS) = ar( PSR) ( Median of a divides it into two s of equal area) = ar ( PQR) Again, QT is a median of PQS ar ( QTP) = ar( QTS) ( Median of a divides it into two s of equal area) Area of QTS = area of PQR = 4. Area of PQR Example 4: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar( APB) ar( CPD) = ar( APD) ar ( BPC). Solution: Given: In a quadrilateral ABCD, diagonals AC and BD intersect each other at P. To Prove: ar( APB) ar( CPD) ar ( APD) ar ( BPC). Proof: From A and C, draw perpendiculars AE and CF respectively to BD. ( DP) ( CF) (PB)(AE) ar( APB) ar( CPD) = (DP)(AE) (PB)(CF) And ar( APD) ar( BPC) = = (PB) (AE) (DP) (CF)..() 4 From () and (), ar( APB) ar( CPD) ar ( BPC). Hence proved. = (PB) (AE) (DP) (CF)..() 4 Example 5: A room is half as long as it is broad. The cost of carpeting the room at 3.70 per m is 00 and the cost of papering the walls at per m is 300. If door and windows occupy 0m, find the dimensions of the room. x 3x Solution: Let the breadth of the room be x m. Then, Length = x+ m = m 3x 3x Area of the room = x m = m

15 Foundation for Mathematics 3.5 Cost of carpeting the room at the rate of 3.70 per m = 3x 3.70 = 00 [ cost of carpeting = 00 (given)] x = = = Thus, breadth = 6m and length x = 6 6 = 6 + m = 9m 3x 4 Let the height of the room be h meters. Then, Area of 4 walls = (length + breadth) height = (6 + 9) h m = 30 h m Area of door and windows = 0 m Area to be papered = Area of 4 walls Area of door and windows = (30 h 0)m Cost of papering walls at per m = (30h 0) (30h 0) = (30h 0) = 30h 0 = h = 60 h = = Height = 5.33 m Hence, the dimensions of the room are: Length = 9m, breadth = 6m and height = 5.33 m Example 6: If the medians of a PQR intersect at V, show that ar( PVQ) = ar( QVR) = 3 ar( PQR). Solution: Given: A PQR its medians PS, QT and RU intersect at V. To prove: ar( PVQ) = ar( PVR) = ar( QVR) = 3 ar( PQR). Proof: A median of triangle divides it into two triangles of equal area. In PQR, PS is the median. ar( PQS) = ar( PRS) In VQR, VS is the median. ar( VQS) = ar( VRS) From (i) and (ii), we get...(i)...(ii)

16 3.6 Area of Triangles and Quadrilaterals ar( PQS) - ar( VQS) = ar( PRS) -ar( VRS) a( PVQ) = ar( PVR). Similarly, ar( PVQ) = ar( PVR) =ar( QVR)...(iii) But, ar(pqr) = ar( PVQ) + ar( PVR) + ar( QVR) = 3 ar( PVQ) [Using (iii)] ar( PVQ) = 3 ar( PQR). Hence, ar( PVQ) = ar( PVR) = ar QVRC) = 3 ar( PQR). Example 7: Triangles PQR and SQR are on the same base QR; with P, S on opposite sides of the line QR, such that ar( PQR) = ar( SQR). Show that QR bisects PS. Solution: Construction: Draw PA QR and SB QR. Proof: ar( PQR) = ar( SQR) [Given] QR PA QR SB = PA = SB...(i) Now in s OPA and OSB PA = SB [From (i)] PAO = SBO [Each = 90 ] POA = BOS [Vert. opp. s] OPA = OSB [Third angles of the triangles] OPA OSB [By ASA] OP = OS [By cpct] i.e., QR bisects PS. Hence Proved. Example 8: The diagonals of a parallelogram PQRS intersect in O. A line through O meets PQ is A and the opposite side RS in B. Show that Ar (quadrilateral PABS) = ar(parallelogram PQRS). Solution: PR is a diagonal of the parallelogram PQRS. ar( PRS) = ar(pqrs)...(i) Now, in s POA and ROB, PO = RO Diagonals of parallelogram bisect each other. POA = ROB [Vert. opp. s]

17 Foundation for Mathematics 3.7 OPA = ORB [Alt. Int. s] PQ RS and transversal intersects them POA ROB [ASA] ar( POA) = ar( ROB) Adding ar(quad. POBS) to both sides of (ii), we get ar(quad. POBS) + ar( POA) = ar(quad. POBS) + ar( ROB) ar(quad. PABS) = ar( PRS) = ar( gm PQRS) (using (i)) Hence Proved....(ii) Example 9: In figure, T is any point on median PS of a PQR. Show that ar(pqt) = ar(prt). Solution: Construction: From P draw PV QR and from T draw TU QR. Proof: ar( PQS) = QS PV ar( PSR) = SR PV But, QS = SR [ S is the mid-point of QR, PS being the median] ar( PQS) = ar( PSR) Again, ar( TQS)= QS TU ar( TSR) = SR TU But, QS = SR ar( TQS) = ar( TSR)...(ii) Subtracting (ii) from (i), we get ar( PQS) - ar( TQS) = ar( PSR) - ar( TSR) ar( PQT) = ar( PRT). Hence Proved....(i)

18 3.8 Area of Triangles and Quadrilaterals EXERCISE For School Examinations Fill in the Blanks Directions: Complete the following statements with an appropriate word/term to be filled in the blank space(s). Q.. of a solid is the amount of space enclosed by the boundary of a flat object. Q.. The volume of a rectangular solid measuring m by 50 cm by 0.5 m is cm 3 Q.3. A sphere has only Surface and that is curved. Q.4. A right circular cone is generated by revolving a right angled triangle about one of the sides containing the. Q.5. When a right angled triangular lamina is revolved about one of its sides (other than hypotenuse), then the solid so formed is called a. Q.6. Volume of a cylinder is three times the volume of a on the same base and of the same height. True / False Directions: Read the following statements and write your answer as true or false. Q.7. Area of a quadrilateral whose sides and one diagonal are given, can be calculated by dividing the quadrilateral into two triangles and using the Heron s formula. True False Q.8. Sides of a triangle are in the ratio of : 7 : 5 and its perimeter is 540 cm. Its area 8000 cm Q.9. True False The sides of a quadrilateral taken in order are 5m, m, 4m and 5m. If the angle between the first two sides be 90 o, its area 4m. True False Q.0. In triangle, sum of two side is always greater than third side. True False Q.. Triangle with sides 0, 0 and 0 cm is not possible. True False Q.. If the diagonals of a quadrilateral divide it into four triangles which are equal in area, then the quadrilateral must be a parallelogram. True False

19 Foundation for Mathematics 3.9 Q.3. The areas of the triangles having a common side are proportional to their altitude to the common side. True False Q.4. If triangles of equal area have a common base, then their vertices must lie on a line parallel to the base True False Q.5. If P is any point in the interior of a rectangle ABCD, then Area ( PAB) + Area ( PCD) = Area ( PBC)+ Area ( PDA). True False Q.6. A median of a triangle divides it into two triangles of equal areas. True False Match the Following Columns Directions: Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II Q.7. Column I Column II (A) ar ( QST ) (p) ( PUS) (B) ar ( QST) (q) ar( PQR) 4 (C) ar ( PQR) (r) ar( PUR) 8 (D) ar ( QUT) (s) ar ( QTR) (E) ar ( UTS) (t) ar( QPT) Short Answer Questions Directions: Give answer in two to three sentences. Q.8. An isosceles triangle has perimeter 60 cm and each of the equal sides is 4 cm. Find the area of the triangle. Q.9. An umbrella is made by stitching triangular pieces of cloth of two different colour (see figure), each piece measuring 40 cm, 00 cm and 00 cm. How much cloth of each colour required for the umbrella?

20 3.0 Area of Triangles and Quadrilaterals Long Answer Questions Directions: Give answer in four to five sentences. Q.0. Find the area of a quadrilateral PQRS in which PQ = 6 cm, QR = 8 cm, RS = 8 cm, SP = 0 cm and PR = 0 cm. Q.. A kite in the shape of a square with a diagonal 64 cm and an isosceles triangle of base 6 cm and sides cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it? Q.. A floral design on a floor is made up of 6 tiles which are triangular, the sides of the triangle being 0 cm, 0 cm and 30 cm (see figure). Find the cost of polishing the tiles at the rate of 50/- per cm.

21 Foundation for Mathematics 3. EXERCISE For Competitive Examinations Multiple Choice Questions Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Q.. A regular hexagon has a side 8 cm. Its perimeter and area are (a) 35 cm, 8 3 cm (b) 48 cm, 96 3 cm (c) 40 cm, cm (d) 36 cm, 54 3 cm Q.. In ABC if D is a point on BC and divides it in the ratio 3 : 5 i.e. If BD : DC = 3 : 5 then ar( ADC): ar( ABC) = (a) 3 : 5 (b) 3 : 8 (c) 5 : 8 (d) 8 : 3 Q.3. Points D and E lie on the lines AB and AC respectively of a triangle ABC, such that AD : BD = : and AE : EC = : ADE and trapezium DECB have their areas in the ratio of : (a) : 4 (b) : 8 (c) : 9 (d) : Q.4. The Fig. formed by joining the consecutive mid-points of any rhombus is always: (a) A square (b) A rhombus (c) A parallelogram (d) None of these Q.5. X and Y are respectively two points on the sides DC and AD of the parallelogram ABCD. The area of ABX is equal to: (a) 3 area of BYC (b) Area of BYC (c) area of BYC (d) area of BYC Q.6. BD is a median of a triangle ABC. F is a point on AB such that CF intersects BD at E and BE = ED. If BF = 5 cm, BA is equal to: (a) 0 (b) (c) 5 (d) 7 Q.7. D and E are the mid points of the sides AB and AC of a triangle ABC respectively. Then the area of the triangles ADE and ABC are in the ratio: (a) : (b) : 3 (c) : 4 (d) : 3

22 3. Area of Triangles and Quadrilaterals Q.8. AD is the median of a triangle ABC. If area of triangle ADC = 5 cm, then ar ( ABC) is (a) 5 cm (b).5 cm (c) 30 cm (d) 37.5 cm Q.9. The base BC of triangle ABC is divided at D so that BD = DC area of ABD = (a) of the ar ( ABC) 3 (b) of the ar ( ABC) (c) 4 of the ar ( ABC) (d) 6 of the ar ( ABC) Q.0. In the parallelogram ABCD, the side AB is produced to point X, so that BX = AB. The line DX cuts BC at E. Area of AED = (a) ar ( CEX) (b) ar ( CEX) (c) ar ( CEX) (d) 3 ar ( CEX) Passage Based Questions Directions: Study the given passage (s) and answer the following questions. Passage I Radha made a picture of an aeroplane with coloured paper as shown in figure. Q.. Area of region I is (a).5 cm (b) cm (c) 5 cm (d) 3 cm Q.. Area of region II is (a) 6 cm (b) 5 cm (c) 6.5 cm (d) 7 cm Q.3. Area of region III and find the total area of the paper used. (a) cm and 9.3 cm (b) 3 cm and 7.3 cm (c) cm and.3 cm (d).3 cm and 9.3 cm

23 Foundation for Mathematics 3.3 Assertion and Reason Directions: Each of these questions contains an Assertion followed by Reason. Read them carefully and answer the questions on the basis of following options. You have to select the one that best describes the two statements. a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. c) If Assertion is correct but Reason is incorrect. d) If Assertion is incorrect but Reason is correct. Q.4. Assertion: An edge of a cube measures r cm. If the largest possible right circular cone is cut out of this cube, then the volume of the cone is r 3 6 π 3 Reason: Area of an equilateral triangle is given by (side). 4 Q.5. Assertion: In a cylinder, if radius is halved and height is doubled. The volume will be halved. Reason: In a cylinder, radius is doubled and height is halved, curved surface area will be same. Q.6. Assertion: The total surface area of a cone whose radius is r r and slant height l is ( π )r l + 4 Reason: Total surface area of cone is π r(l+ r) where r is radius l is the slant height of the cone. Subjective Questions Directions: Answer the following questions. Q.7. Find the percentage increase in the area of a triangle if its each side is doubled. Q.8. In figure. ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meet CD produced at Q. Prove that ar (ABCDE) = ar (APQ)

24 3.4 Area of Triangles and Quadrilaterals SOLUTIONS EXERCISE For School Examinations Fill in the Blanks. Volume. 50,000 cm 3 3. One 4. Right angle 5. Right circular cone 6. Cone True / False 7. True 8. False 9. True 0. True. True. True 3. False 4. True 5. True 6. True Match the Following Columns 7. A = q; B = t; C = s; D = p; E = r (A) Let PQ = QR = RP = x x QS = = ST = QT We have, ar ( PQR) = 3 x 4 ar ( QST) = 3 x 3 = x ar( QST) = ar ( PQR) 4 (B) Given PQR and QST are equilateral triangles. o PRQ = SQT = 60 QT PR ar ( QPT) = ar ( QTR) ar ( QPT) = ar( QST) (TS is median of TQR ) ar( QST) = ar ( QPT)

25 Foundation for Mathematics 3.5 (C) Since TS is median of QTR ar( QTR) = ar ( QST) ar( QTR) = ar ( PQR) 4 (from part (A) above) ar( QTR) = ar ( PQR) (D) Since, PQR and QST are equilateral triangles o PQR = 60 and QST = 60 PQR = QST o PQ ST ar( QTS) = ar( PTS) ar( QTS) ar( TUS) = ar( PTS) ar( TUS) ar( QUT) = ar( PUS) (E) Let the altitude of PQS be h. h ar ( UTS) =.FD. h ( altitude of QTS = ) ar ( PUR) =.UR.h = (US+ RS)h = (US + UQ)h = (US + US)h ( QU = US) = US.h = 8ar( UTS) ar( UTS) = ar( PUR) 8 Short Answer Questions 8. Let the length of unequal side be x cm. Then perimeter = 30 cm. x = 60 x + 48 = 60 x = cm We have, s = 60 cm s = 30cm

26 3.6 Area of Triangles and Quadrilaterals Hence, area = s(s a)(s b)(s c) = 30 (30 4) (30 4) (30 ) = = 36 5 cm 9. Sides of one triangular piece of cloth are of lengths a = 40 cm, b = 00 cm and c = 00 cm. Let s be the semi-perimeter of the triangular piece. Then, s = a + b + c s = s = 0 = Area of one triangular piece = s(s a)(s b)(s c) = 0 (0 40) (0 00) (0 00) cm = cm = cm Area of cloth of each colour = cm = cm Long Answer Questions 0. Let PQRS in which PQ = 6 cm, QR = 8 cm, RS = 8 cm, SP = 0 cm and PR = 0 cm Let PR be the diagonal which divides quadrilateral in two triangles PQR and PRS. For PQR, we have, a = 8 cm, b = 0 cm, c = 6 cm Since, a + c = b Therefore, ABC is right angled with o Q = 90. Area of right angled triangle ABC = Base Height = 6 8 = 4 cm But, for PRS, the sides are a = 8 cm, b = 0 cm, c = 0 cm a+ b+ c We have s = s = = = 4cm Area of the PRS = s(s a)(s b)(s c) (By Heron s formula) = 4(4 8)(4 0)(4 0) = 4(6)(4)(4) = 8 cm = 36.8 cm (approx..) = cm (approx..) Area of the quadrilateral PQRS = Area of PQR + Area of PRS = 4cm cm = 60.8 cm. (approx.).. Area of paper of shade I= dd where d and d denotes the diagonals cm = =

27 Foundation for Mathematics 3.7 Area of paper of shad II = 04 cm For paper of shade III, sides are given as a = 6 cm, b = cm, c = cm S = = 0 Area of paper of shade III = s(s a)(s b)(s c) = 0(0 6)(0 )(0 ). Let the sides of one tile are given as a = 0 cm, b = 0 cm, c = 6 cm a + b + c We know s = = = 8cm Area of one tile = s(s a)(s b)(s c) (By Heron s formula) = 8(8 0)(8 0)(8 6) = 8(8)(8)() = 89.8cm Area of 6 tiles = = 436.8cm cost of polishing the tiles at the rate of 50 paise per cm = p = 7840 = = 0(4)(8)(8) = 3 5 = 7.68 cm

28 3.8 Area of Triangles and Quadrilaterals EXERCISE For Competitive Examinations Multiple Choice Questions. (d) Side = 6 cm Perimeter of regular hexagon = 8 6 = 48 cm The three diagonal divides the hexagonal in six. Congruent equilateral triangle with side 8cm. Area of one such triangle = s(s a)(s b)(s c) = ( 8)( 8)( 8) = = 6 3 cm Area of the regular hexagon = 6 Area of equilateral triangles = = 96 3cm. (c) 3. (a) 4. (c) 5. (b) 6. (c) 7. (c) 8. (c) AD is median of ABC ar( ABD) = ar( ADC) = ar( ABC) 9. (a) 0. (a) Passage Based Questions. (a). (c) 3. (d) For Triangular area I, we have a = 5 cm, b = 5 cm, c = cm as the sides of the triangle a+ b+ c s =, = = = 5.5cm Area of shaded region I = s(s a)(s b)(s c) = 5.5(5.5 5)(5.5 5)(5.5 ) = 5.5(.5)(4.5) = (.5) (5.5)(4.5) = (.5) (.5)()(.5)(9) = (.5)(.5)(3) = 0.75 = 0.75(3.3) (approx.) =.5 cm (approx..), Area of shaded region

29 Foundation for Mathematics 3.9 II = Area of rectangle = 6.5 = 6.5 cm For Area of shaded region III Area of region III = Area of parallelogram ABCD +Area of BCE = + = = = (Approx) cm = = (approx..) 4 Thus, area of region IV = Area of right angled \ 6.5 = = 4.5cm Similarly, 6.5 Area V = = 4.5cm Total area of the paper used = Area + Area II + Area III + Area IV + Area V =.5 cm cm +.3 cm cm cm = 9.3 cm. (approx.). Assertion and Reason 4. (d) Assertion is false but reason is correct. 5. (b) Both Assertion and Reason is correct. 6. (a) Assertion : T.S.A. = π r(l+ r) r r r = π l + = π r l + 4 Subjective Questions 7. Let a, b, c be the sides of the old triangle and s be its semi-perimeter. Then, s = (a + b + c) 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 's be the area of the old and new triangles respectively. Then, ' = s(s a)(s b)(s c) ' = s'(s' a)(s' b)(s' c) ' = s(s a)(s b)(s c) = 4 s(s a)(s b)(s c) = 4

30 3.30 Area of Triangles and Quadrilaterals Increase in the area of triangle = ' = 4 = 3 Hence, percentage increase in area 8. As BP AC and AD EQ ar( ABC) = ar( APC) 3 = 00 = 300 % [ s on the same base and between the same parallels]. (i) AD EQ [Given] ar( ADE) = ar( ADQ) [ s on the same base and between the same parallels].(ii) ar( ACD) = ar( ACD).(iii) Adding (i), (ii) and (iii) ar( ABC) + ar( ADE) + ar( ACD) = ar( APC) + ar( ADQ) + ar( ACD) ar( ABCD) = ar( APQ) Hence proved.