Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 18 - Area Of A Trapezium And A Polygon

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Rs Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 18 Area Of A Trapezium And A Polygon are provided here with simple step-by-step explanations. These solutions for Area Of A Trapezium And A Polygon are extremely popular among Class 8 students for Maths Area Of A Trapezium And A Polygon Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 207:

Answer:

$Area of a trapezium = \frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$

$= \{\frac{1}{2} \times (24 + 20) \times 15\} {cm}^{2} = (\frac{1}{2} \times 44 \times 15) {cm}^{2} = (22 \times 15) {cm}^{2} = 330 {cm}^{2}$

$Hence, the area of the trapezium is 330 {cm}^{2} .$

Page No 207:

Answer:

$Area of a trapezium = \frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$

$= \{\frac{1}{2} \times (38.7 + 22.3) \times 16\} {cm}^{2} = (\frac{1}{2} \times 61 \times 16) {cm}^{2} = (61 \times 8) {cm}^{2} = 488 {cm}^{2}$

$Hence, the area of the trapezium is 488 {cm}^{2} .$

Page No 207:

Answer:

$Area of a trapezium = \frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$
$= \{\frac{1}{2} \times (1 + 1.4) \times 0.9\} m^{2} = (\frac{1}{2} \times 2.4 \times 0.9) m^{2} = (1.2 \times 0.9) m^{2} = 1.08 m^{2}$
$Hence, the area of the top surface of the table is 1.08 m^{2} .$

Page No 207:

Answer:

$Let the distance between the parallel sides be x . Now, Area of trapezium = \{\frac{1}{2} \times (55 + 35) \times x\} {cm}^{2}$

$= (\frac{1}{2} \times 90 \times x) c m^{2} = 45 x c m^{2}$
$Area of the trapezium = 1080 {cm}^{2} (Given) ∴ 45 x = 1080 \Rightarrow x = \frac{1080}{45} \Rightarrow x = 24 cm Hence, the distance between the parallel sides is 24 cm .$

Page No 207:

Answer:

$Let the length of the required side be x cm . Now, Area of trapezium = \{\frac{1}{2} \times (84 + x) \times 26\} m^{2}$

$= (1092 + 13 x) m^{2}$

$Area of trapezium = 1586 m^{2} (Given) ∴ 1092 + 13 x = 1586 \Rightarrow 13 x = (1586 - 1092) \Rightarrow 13 x = 494 \Rightarrow x = \frac{494}{13} \Rightarrow x = 38 m Hence, the length of the other side is 38 m .$

Page No 207:

Answer:

$Let the lengths of the parallel sides of the trapezium be 4 x cm and 5 x cm, respectively . Now, Area of trapezium = \{\frac{1}{2} \times (4 x + 5 x) \times 18\} {cm}^{2}$
$= (\frac{1}{2} \times 9 x \times 18) c m^{2} = 81 x c m^{2}$
$Area of trapezium = 405 {cm}^{2} (Given) ∴ 81 x = 405 \Rightarrow x = \frac{405}{81} \Rightarrow x = 5 cm Length of one side = (4 \times 5) cm = 20 cm Length of the other side = (5 \times 5) cm = 25 cm$

Page No 207:

Answer:

$Let the lengths of the parallel sides be x cm and (x + 6) cm . Now, Area of trapezium = \{\frac{1}{2} \times (x + x + 6) \times 9\} {cm}^{2}$
$= (\frac{1}{2} \times (2 x + 6) \times 9) {cm}^{2} = 4.5 (2 x + 6) {cm}^{2} = (9 x + 27) {cm}^{2}$

$Area of trapezium = 180 {cm}^{2} (Given) ∴ 9 x + 27 = 180 \Rightarrow 9 x = (180 - 27) \Rightarrow 9 x = 153 \Rightarrow x = \frac{153}{9} \Rightarrow x = 17 Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, (17 + 6) cm .$

Page No 207:

Answer:

$Let the lengths of the parallel sides be x cm and 2 x cm . Area of trapezium = \{\frac{1}{2} \times (x + 2 x) \times 84\} m^{2}$
$= (\frac{1}{2} \times 3 x \times 84) m^{2} = (42 \times 3 x) m^{2} = 126 x m^{2}$

$Area of the trapezium = 9450 m^{2} (Given) ∴ 126 x = 9450 \Rightarrow x = \frac{9450}{126} \Rightarrow x = 75 Thus, the length of the parallel sides are 75 m and 150 m, that is, (2 \times 75) m, and the length of the longer side is 150 m .$

Page No 207:

Answer:

$Length of the side AB = (130 - (54 + 19 + 42)) m$
$= 15 m$
$Area of the trapezium - shaped field = \{\frac{1}{2} \times (AD + BC) \times AB\}$
$= \{\frac{1}{2} \times (42 + 54) \times 15\} m^{2} = (\frac{1}{2} \times 96 \times 15) m^{2} = (48 \times 15) m^{2} = 720 m^{2}$

$Hence, the area of the field is 720 m^{2} .$

Page No 207:

Answer:

$∠ ABC = 90 ° From the right ∆ ABC, we have : {AB}^{2} = ({AC}^{2} - {BC}^{2}) \Rightarrow {AB}^{2} = \{(41^{2}) - (40^{2})\} \Rightarrow {AB}^{2} = (1681 - 1600) \Rightarrow {AB}^{2} = 81 \Rightarrow AB = \sqrt{81} \Rightarrow AB = 9 cm ∴ Length AB = 9 cm Now, Area of the trapezium = \{\frac{1}{2} \times (AD + BC) \times AB\}$
$= (\frac{1}{2} \times (16 + 40) \times 9) {cm}^{2} = (\frac{1}{2} \times 56 \times 9) {cm}^{2} = (28 \times 9) {cm}^{2} = 252 {cm}^{2}$
$Hence, the area of the trapezium is 252 {cm}^{2} .$

Page No 207:

Answer:

$Let ABCD be the given trapezium in which AB ∥ DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm . Draw CL ⊥ AB and CM ∥ DA meeting AB at L and M, respectively . Clearly, AMCD is a parallelogram . Now, AM = DC = 10 cm MB = (AB - AM) = (20 - 10) cm = 10 cm Also, CM = DA = 13 cm Therefore, ∆ CMB is an isosceles triangle and CL ⊥ MB . L is the midpoint of B . \Rightarrow ML = LB = (\frac{1}{2} \times MB)$
$= (\frac{1}{2} \times 10) cm = 5 cm$

$From right ∆ CLM, we have : {CL}^{2} = ({CM}^{2} - {ML}^{2}) {cm}^{2} \Rightarrow {CL}^{2} = \{{(13)}^{2} - {(5)}^{2}\} {cm}^{2} \Rightarrow {CL}^{2} = (109 - 25) {cm}^{2} \Rightarrow {CL}^{2} = 144 {cm}^{2} \Rightarrow CL = \sqrt{144} cm \Rightarrow CL = 12 cm ∴ Length of CL = 12 cm Area of the trapezium = \{\frac{1}{2} \times (AB + DC) \times CL\}$
$= \{\frac{1}{2} \times (20 + 10) \times 12\} {cm}^{2} = (\frac{1}{2} \times 30 \times 12) {cm}^{2} = (15 \times 12) {cm}^{2} = 180 {cm}^{2}$
$Hence, the area of the trapezium is 180 {cm}^{2} .$

Page No 207:

Answer:

$Let ABCD be the trapezium in which AB ∥ DC, AB = 25 cm, CD = 11 cm, AD = 13 cm and BC = 15 cm . Draw CL ⊥ AB and CM ∥ DA meeting AB at L and M, respectively . Clearly, AMCD is a parallelogram . Now, MC = AD = 13 cm AM = DC = 11 cm \Rightarrow MB = (AB - AM)$
$= (25 - 11) cm = 14 cm$

$Thus, in ∆ CMB, we have : CM = 13 cm MB = 14 cm BC = 15 cm ∴ s = \frac{1}{2} (13 + 14 + 15) cm = \frac{1}{2} 42 cm = 21 cm (s - a) = (21 - 13) cm = 8 cm (s - b) = (21 - 14) cm = 7 cm (s - c) = (21 - 15) cm = 6 cm ∴ Area of ∆ CMB = \sqrt{s (s - a) (s - b) (s - c)}$
$= \sqrt{21 \times 8 \times 7 \times 6} {cm}^{2} = 84 {cm}^{2}$

$∴ \frac{1}{2} \times MB \times CL = 84 {cm}^{2} \Rightarrow \frac{1}{2} \times 14 \times CL = 84 {cm}^{2}$
$\Rightarrow CL = \frac{84}{7} \Rightarrow CL = 12 cm$
$Area of the trapezium = \{\frac{1}{2} \times (AB + DC) \times CL\}$
$= \{\frac{1}{2} \times (25 + 11) \times 12\} {cm}^{2} = (\frac{1}{2} \times 36 \times 12) {cm}^{2} = (18 \times 12) {cm}^{2} = 216 {cm}^{2}$

$Hence, the area of the trapezium is 216 {cm}^{2} .$

Page No 210:

Answer:

$Area of quadrilateral ABCD = (Area of ∆ ADC) + (Area of ∆ ACB)$
$= (\frac{1}{2} \times AC \times DM) + (\frac{1}{2} \times AC \times BL) = [(\frac{1}{2} \times 24 \times 7) + (\frac{1}{2} \times 24 \times 8)] {cm}^{2} = (84 + 96) {cm}^{2} = 180 {cm}^{2}$

$Hence, the area of the quadrilateral is 180 {cm}^{2} .$

Page No 210:

Answer:

$Area of quadrilateral ABCD = (Area of ∆ ABD) + (Area of ∆ BCD)$
$= (\frac{1}{2} \times BD \times AL) + (\frac{1}{2} \times BD \times CM)$
$= [(\frac{1}{2} \times 36 \times 19) + (\frac{1}{2} \times 36 \times 11)] m^{2} = (342 + 198) m^{2} = 540 m^{2}$

$Hence, the area of the field is 540 m^{2} .$

Page No 210:

Answer:

$Area of pentagon ABCDE = (Area of ∆ AEN) + (Area of trapezium EDMN) + (Area of ∆ DMC) + (Area of ∆ ACB)$
$= (\frac{1}{2} \times AN \times EN) + (\frac{1}{2} \times (EN + DM) \times NM) + (\frac{1}{2} \times MC \times DM) + (\frac{1}{2} \times AC \times BL) = (\frac{1}{2} \times AN \times EN) + (\frac{1}{2} \times (EN + DM) \times (AM - AN)) + (\frac{1}{2} \times (AC - AM) \times DM) + (\frac{1}{2} \times AC \times BL) = [(\frac{1}{2} \times 6 \times 9) + (\frac{1}{2} \times (9 + 12) \times (14 - 6)) + (\frac{1}{2} \times (18 - 14) \times 12) + (\frac{1}{2} \times 18 \times 4)] {cm}^{2} = (27 + 84 + 24 + 36) {cm}^{2} = 171 {cm}^{2}$

$Hence, the area of the given pentagon is 171 {cm}^{2} .$

Page No 210:

Answer:

$Area of hexagon ABCDEF = (Area of ∆ AFP) + (Area of trapezium FENP) + (Area of ∆ END) + (Area of ∆ CMD) + (Area of trapezium BCML) + (Area of ∆ ALB)$

$= (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times PN) + (\frac{1}{2} \times ND \times EN) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times LM) + (\frac{1}{2} \times AL \times BL) = (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times (PL + LN)) + (\frac{1}{2} \times (NM + MD) \times EN) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times (LN + NM)) + (\frac{1}{2} \times (AP + PL) \times BL) = (\frac{1}{2} \times 6 \times 8) + (\frac{1}{2} \times (8 + 12) \times (2 + 8)) + (\frac{1}{2} \times (2 + 3) \times 12) + (\frac{1}{2} \times 3 \times 6) + (\frac{1}{2} \times (6 + 8) \times (8 + 2)) + (\frac{1}{2} \times (6 + 2) \times 8) {cm}^{2} = (24 + 100 + 30 + 9 + 70 + 32) {cm}^{2} = 265 {cm}^{2}$

$Hence, the area of the hexagon is 265 {cm}^{2} .$

Page No 210:

Answer:

$Area of pentagon ABCDE = (Area of ∆ ABC) + (Area of ∆ ACD) + (Area of ∆ ADE)$
$= (\frac{1}{2} \times AC \times BL) + (\frac{1}{2} \times AD \times CM) + (\frac{1}{2} \times AD \times EM) = [(\frac{1}{2} \times 10 \times 3) + (\frac{1}{2} \times 12 \times 7) + (\frac{1}{2} \times 12 \times 5)] {cm}^{2} = (15 + 42 + 30) {cm}^{2} = 87 {cm}^{2}$

$Hence, the area of the pentagon is 87 {cm}^{2} .$

Page No 210:

Answer:

$Area enclosed by the given figure = (Area of trapezium FEDC) + (Area of square ABCF)$
$= [\{\frac{1}{2} \times (6 + 20) \times 8\} + (20 \times 20)] {cm}^{2} = (104 + 400) {cm}^{2} = 504 {cm}^{2}$

$Hence, the area enclosed by the figure is 504 {cm}^{2} .$

Page No 211:

Answer:

$We will find the length of AC . From the right triangles ABC and HGF, we have : {AC}^{2} = {HF}^{2} = \{{(5)}^{2} - {(4)}^{2}\} cm$
$= (25 - 16) c m = 9 c m$

$AC = HF = \sqrt{9} c m = 3 c m$
$Area of the given figure ABCDEFGH = (Area of rectangle ADEH) + (Area of ∆ ABC) + (Area of ∆ HGF)$
$= (Area of rectangle ADEH) + 2 (Area of ∆ ABC) = (AD \times DE) + 2 (Area of ∆ ABC) = \{(AC + CD) \times DE\} + 2 (\frac{1}{2} \times BC \times AC) = \{(3 + 4) \times 8\} + 2 (\frac{1}{2} \times 4 \times 3) {cm}^{2} = (56 + 12) cm = 68 {cm}^{2}$
$Hence, the area of the given figure is 68 {cm}^{2}$ .

Page No 211:

Answer:

$Let AL = DM = x cm LM = BC = 13 cm ∴ x + 13 + x = 23 \Rightarrow 2 x + 13 = 23 \Rightarrow 2 x = (23 - 13) \Rightarrow 2 x = 10 \Rightarrow x = 5 ∴ AL = 5 cm From the right ∆ AFL, we have : F L^{2} = A F^{2} - A L^{2} \Rightarrow F L^{2} = \{(13^{2}) - {(5)}^{2}\} \Rightarrow F L^{2} = (169 - 25) \Rightarrow F L^{2} = 144 \Rightarrow F L = \sqrt{144} \Rightarrow F L = 12 c m ∴ FL = BL = 12 cm Area of a regular hexagon = (A rea of t h e trapezium ADEF) + (A rea of t h e trapezium ABCD)$
$= 2 (A r e a o f t r a p e z i u m A D E F) = 2 \{\frac{1}{2} \times (A D + E F) \times F L\} = 2 \{\frac{1}{2} \times (23 + 13) \times 12\} c m^{2} = 2 (\frac{1}{2} \times 36 \times 12) c m^{2} = 432 c m^{2}$
$Hence, the area of the given regular hexagon is 432 {cm}^{2} .$

Page No 211:

Answer:

(b) 144 cm²
$Area of the trapezium = \{\frac{1}{2} \times (14 + 18) \times 9\} {cm}^{2}$
$= (\frac{1}{2} \times 32 \times 9) c m^{2} = 144 c m^{2}$

Page No 211:

Answer:

(c) 8 cm

$Let the distance between the parallel sides be x cm . Then, area of the trapezium = \{\frac{1}{2} \times (19 + 13) \times x\} {cm}^{2}$
$= (\frac{1}{2} \times 32 \times x) {cm}^{2} = 16 x {cm}^{2}$
$But it is given that the area of the trapezium is 128 {cm}^{2} .$
$∴ 16 x = 128 \Rightarrow x = \frac{128}{16} \Rightarrow x = 8 cm$

Page No 211:

Answer:

(a) 45 cm

$Let the length of the parallel sides be 3 x cm and 4 x cm, respectively . Then, area of the trapezium = \{\frac{1}{2} \times (3 x + 4 x) \times 12\} {cm}^{2}$
$= (\frac{1}{2} \times 7 x \times 12) c m^{2} = 42 x c m^{2}$

$But it is given that the area of the trapezium i s 630 {cm}^{2} .$

$∴ 42 x = 630 \Rightarrow x = \frac{630}{42} \Rightarrow x = 15 c m$
$L ength of the parallel sides = (3 \times 15) cm = 45 cm (4 \times 15) cm = 60 cm Hence, the sho rter of the parallel sides is 45 cm .$

Page No 211:

Answer:

(b) 23 cm

$Let the length of the parallel sides be x cm and (x + 6) cm, respectively . Then, area of the trapezium = \{\frac{1}{2} \times (x + x + 6) \times 9\} {cm}^{2}$
$= \{\frac{1}{2} \times (2 x + 6) \times 9\} {cm}^{2} = 4.5 (2 x + 6) {cm}^{2} = (9 x + 27) {cm}^{2}$
$But it is given that the area of the trapezium i s 180 {cm}^{2} . ∴ 9 x + 27 = 180 \Rightarrow 9 x = (180 - 27) \Rightarrow 9 x = 153 \Rightarrow x = \frac{153}{9} \Rightarrow x = 17 Therefore, the length of the parallel sides are 17 cm and (17 + 6) cm, which is equal to 23 cm . Hence, the length of the longer parallel side is 23 cm .$

Page No 211:

Answer:

(c) 80 cm²

$From the given trapezium, we find : D C = A L = 7 c m [since D A ⊥ A B a n d C L ⊥ A B] From the right ∆ CBL, we have : C L^{2} = C B^{2} - L B^{2} \Rightarrow C L^{2} = {(10)}^{2} - {(6)}^{2} \Rightarrow C L^{2} = 100 - 36 \Rightarrow C L^{2} = 64 \Rightarrow C L = \sqrt{64} \Rightarrow C L = 8 c m Area of the trapezium = \{\frac{1}{2} \times (7 + 13) \times 8\} {cm}^{2}$
$= (\frac{1}{2} \times 20 \times 8) {cm}^{2} = 80 {cm}^{2}$

Page No 212:

Answer:

$Let the base of the triangular field be 3 x cm and its height be x cm . Then, area of the triangle = (\frac{1}{2} \times 3 x \times x) m^{2}$
$= \frac{3 x^{2}}{2} m^{2}$
$But it is given that the area of the triangular field i s 1350 m^{2} .$
$∴ \frac{3 x^{2}}{2} = 1350 \Rightarrow x^{2} = (1350 \times \frac{2}{3}) \Rightarrow x^{2} = 900 \Rightarrow x = \sqrt{900} \Rightarrow x = 30 m$
$Hence, the height of the field is 30 m . I ts base = (3 \times 30) m = 90 m$

Page No 212:

Answer:

$Area of an equilateral traingle = (\frac{\sqrt{3}}{4} \times {(side)}^{2}) square units = (\frac{\sqrt{3}}{4} \times 6 \times 6) {cm}^{2}$
$= (\frac{\sqrt{3}}{4} \times 36) {cm}^{2} = 9 \sqrt{3} {cm}^{2}$
$Hence, the area of an equilateral triangle is 9 \sqrt{3} {cm}^{2}$ .

Page No 212:

Answer:

$Let ABCD be a rhombus whose diagonals AC and BD intersect at a point O . Let the length of the diagonal AC b e 72 cm and the side of the rhombus be x cm . P erimeter of the rhombus = 4 x cm But it is given that the perimeter of the rhombus i s 180 cm . ∴ 4 x = 180 \Rightarrow x = \frac{180}{4} \Rightarrow x = 45 Hence, the length of the side of the rhombus is 45 cm . We know that the diagonals of the rhombus bisect each other at right angles . ∴ A O = \frac{1}{2} A C \Rightarrow A O = (\frac{1}{2} \times 72) c m \Rightarrow A O = 36 c m From right ∆ AOB, we have : B O^{2} = A B^{2} - A O^{2} \Rightarrow B O^{2} = \{{(45)}^{2} - {(36)}^{2}\} \Rightarrow B O^{2} = (2025 - 1296) \Rightarrow B O^{2} = 729 \Rightarrow B O = \sqrt{729} \Rightarrow B O = 27 c m ∴ B D = 2 \times B O B D = (2 \times 27) c m B D = 54 c m Hence, the length of the other diagonal is 54 cm . Area of the rhombus = (\frac{1}{2} \times 72 \times 54) {cm}^{2}$
$= 1944 c m^{2}$

Page No 212:

Answer:

$Let the length of the parallel sides be x m and (x - 14) m . Then, area of the t rapezium = \{\frac{1}{2} \times (x + x - 14) \times 12\} m^{2}$
$= 6 (2 x - 14) m^{2} = (12 x - 84) m^{2}$
$But it is given that the area of the trapezium i s 216 m^{2} . ∴ 12 x - 84 = 216 \Rightarrow 12 x = (216 + 84) \Rightarrow 12 x = 300 \Rightarrow x = \frac{300}{12} \Rightarrow x = 25 Hence, the length of the parallel sides are 25 m and (25 - 14) m, which is equal to 11 m .$

Page No 212:

Answer:

$Let A B C D be a quadilateral . D iagonal, A C = 40 cm B L ⊥ A C, such that B L = 16 cm D M ⊥ A C, such that D M = 12 cm A rea of the quadilateral = (A rea of ∆ D A C) + (A rea of ∆ A C B)$

$= [(\frac{1}{2} \times A C \times D M) + (\frac{1}{2} \times A C \times B L)] {cm}^{2} = [(\frac{1}{2} \times 40 \times 12) + (\frac{1}{2} \times 40 \times 16)] {cm}^{2} = (240 + 320) {cm}^{2} = 560 {cm}^{2}$

$Hence, the area of the quadilateral is 560 {cm}^{2} .$

Page No 212:

Answer:

$Let the other side of the triangular field be x m . ∴ x^{2} = \{{(50)}^{2} - {(30)}^{2}\} \Rightarrow x^{2} = (2500 - 900) \Rightarrow x^{2} = 1600 \Rightarrow x = \sqrt{1600} \Rightarrow x = 40 ∴ A rea of the field = (\frac{1}{2} \times 30 \times 40) m^{2}$

$= 600 m^{2}$

Page No 212:

Answer:

(b) 56 cm²
$Area of the triangle = (\frac{1}{2} \times 14 \times 8) {cm}^{2}$
$= 56 {cm}^{2}$

Page No 212:

Answer:

(c) 20 m

$Let the height of the triangle be x m and its base be 4 x m respectively . Then, area of the triangle = (\frac{1}{2} \times 4 x \times x) m^{2}$
$= 2 x^{2} m^{2}$
$But, the area of the triangle i s 50 m^{2} . ∴ 2 x^{2} = 50 \Rightarrow x^{2} = \frac{50}{2} \Rightarrow x^{2} = 25 \Rightarrow x = \sqrt{25} \Rightarrow x = 5 ∴ L ength of its base = (4 \times 5) m = 20 m$

Page No 212:

Answer:

(b) 200 cm²

$Let A B C D be a quadilateral . D iagonal, A C = 20 cm B L ⊥ A C, such that B L = 8.5 cm D M ⊥ A C, such that D M = 11.5 cm A rea of the quadilateral = (A rea of ∆ D A C) + (A rea of ∆ A C B)$
$= [(\frac{1}{2} \times A C \times D M) + (\frac{1}{2} \times A C \times B L)] {cm}^{2} = [(\frac{1}{2} \times 20 \times 11.5) + (\frac{1}{2} \times 20 \times 8.5)] {cm}^{2} = (85 + 115) {cm}^{2} = 200 {cm}^{2}$

Page No 212:

Answer:

(b) 216 cm²

$Let A B C D be a rhombus whose diagonals A C and B D intersect at a point O . Let the length of the diagonal A C be 24 cm and the side of the rhombus be 15 cm . We know that the diagonals of the rhombus bisect each other at right angles . ∴ A O = \frac{1}{2} A C \Rightarrow A O = (\frac{1}{2} \times 24) cm \Rightarrow A O = 12 cm From right ∆ A O B, we have : B O^{2} = A B^{2} - A O^{2} \Rightarrow B O^{2} = \{{(15)}^{2} - {(12)}^{2}\} \Rightarrow B O^{2} = (225 - 144) \Rightarrow B O^{2} = 81 \Rightarrow B O = \sqrt{81} \Rightarrow B O = 9 cm ∴ B D = 2 \times B O B D = (2 \times 9) cm B D = 18 cm Hence, the length of the other diagonal is 18 cm . Area of the rhombus = (\frac{1}{2} \times 24 \times 18) {cm}^{2}$

$= 216 {cm}^{2}$

Page No 212:

Answer:

(b) 13 cm

$Let A B C D be a rhombus whose diagonals A C and B D intersect at a point O . Let the length of the diagonal A C b e 24 cm . A rea of the rhombus = (\frac{1}{2} \times A C \times B D) {cm}^{2} But the area of the rohmbus i s 120 {cm}^{2} (given) ∴ \frac{1}{2} \times A C \times B D = 120 or \frac{1}{2} \times 24 \times B D = 120 or 12 \times B D = 120 or B D = \frac{120}{12} = 10 cm OB = \frac{BD}{2} = \frac{10}{2} = 5 cm$

$And O A = \frac{A C}{2} = \frac{24}{2} = 12 cm Now, in right triangle A O B : {(A B)}^{2} = {(O A)}^{2} + {(O B)}^{2} or {(A B)}^{2} = 12^{2} + 5^{2} = 144 + 25 = 169 or A B = \sqrt{169} = 13 cm Therefore, each side of the rhombus is 13 cm .$

Page No 212:

Answer:

(d) 600 cm²

$Area of the trapezium = \{\frac{1}{2} \times (54 + 26) \times 15\} cm$ ²
$= (\frac{1}{2} \times 80 \times 15) {cm}^{2}$
=600 cm²

Page No 212:

Answer:

(b) 40 cm

$Let the length of the parallel sides be 5 x cm and 3 x cm, respectively . A rea of the trapezium = \{\frac{1}{2} \times (5 x + 3 x) \times 12\} cm$ ²
$= (\frac{1}{2} \times 8 x \times 12) cm$ ²=48x cm²

$But, the area of the trapezium i s 384 {cm}^{2} . 48 x = 384 \Rightarrow x = \frac{384}{48} = 8 L onger side = 5 x = 5 \times 8 = 40 cm$

Page No 212:

Answer:

$(i) Area of a triangle = \frac{1}{2} \times (B ase) \times (H eight) (ii) Area of a ∥ gm = (B ase) \times (H eight) (iii) Area of a trapezium = \frac{1}{2} \times (Sum of the parallel sides) \times (D istance between them) (iv) Area of a trapezium = \{\frac{1}{2} \times (14 + 18) \times 8\} {cm}^{2}$
$= (\frac{1}{2} \times 32 \times 8) c m^{2} = 128 c m^{2}$

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