RS Aggarwal 2018 Solutions for Class 8 Math Chapter 18 - Area Of A Trapezium And A Polygon

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Page No 207:

Question 1:

Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.

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:

Question 2:

Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.

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:

Question 3:

The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.

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:

Question 4:

The area of a trapezium is 1080 cm². If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.

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:

Question 5:

A field is in the form of a trapezium. Its area is 1586 m² and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.

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:

Question 6:

The area of a trapezium is 405 cm². Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.

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:

Question 7:

The area of a trapezium is 180 cm² 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.

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:

Question 8:

In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m² and the perpendicular distance between the two parallel sides is 84 m, find the length of the longer of the parallel sides.

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:

Question 9:

The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find the area of the field.

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:

Question 10:

In the given figure, ABCD is a trapezium in which AD||BC, ∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.

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:

Question 11:

The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.

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:

Question 12:

The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

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:

Question 1:

In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ⊥ AC and DM ⊥ AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.

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:

Question 2:

In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19 m and CM = 11 m. Find the area of the field.

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:

Question 3:

Find the area of pentagon ABCDE in which BL ⊥ AC, DM ⊥ AC and EN ⊥ AC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.

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:

Question 4:

Find the area of hexagon ABCDEF in which BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.

Answer:

$Area of hexagon ABCDEF = (Area of ∆ AFP) + (Area of trapezium FENP) + (Area of ∆ ALB)$
$= (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times PN) + (\frac{1}{2} \times ND \times EN) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times LM) + (\frac{1}{2} \times AL \times BL) = (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times (PL + LN)) + (\frac{1}{2} \times (NM + MD) \times CM) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times (LN + NM)) + (\frac{1}{2} \times (AP + PL) \times BL) = [(\frac{1}{2} \times 6 \times 8) + (\frac{1}{2} \times (8 + 12) \times (2 + 8)) + (\frac{1}{2} \times (2 + 3) \times 12) + (\frac{1}{2} \times 3 \times 6) + (\frac{1}{2} \times (6 + 8) \times (8 + 2)) + (\frac{1}{2} \times (6 + 2) \times 8)] {cm}^{2} = (24 + 100 + 30 + 9 + 70 + 32) {cm}^{2} = 265 {cm}^{2}$

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

Page No 210:

Question 5:

Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

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:

Question 6:

Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.

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:

Question 7:

Find the area of given figure ABCDEFGH as per dimensions given in it.

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:

Question 8:

Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.

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:

Question 1:

Tick (✓) the correct answer:
The parallel sides of a trapezium measure 14 cm and 18 cm and the distance between them is 9 cm. The area of the trapezium is
(a) 96 cm²
(b) 144 cm²
(c) 189 cm²
(d) 207 cm²

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:

Question 2:

Tick (✓) the correct answer:
The lengths of the parallel sides of a trapezium are 19 cm and 13 cm and its area is 128 cm². The distance between the parallel sides is
(a) 9 cm
(b) 7 cm
(c) 8 cm
(d) 12.5 cm

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:

Question 3:

Tick (✓) the correct answer:
The parallel sides of a trapezium are in the ratio 3 : 4 and the perpendicular distance between them is 12 cm. If the area of the trapezium is 630 cm², then its shorter of the parallel sides is
(a) 45 cm
(b) 42 cm
(c) 60 cm
(d) 36 cm

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:

Question 4:

Tick (✓) the correct answer:
The area of a trapezium is 180 cm² and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, the length of the longer of the parallel sides is
(a) 17 cm
(b) 23 cm
(c) 18 cm
(d) 24 cm

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:

Question 5:

Tick (✓) the correct answer:
In the given figure, AB||DC and DA ⊥ AB. If DC = 7 cm, BC = 10 cm, AB = 13 cm and CL ⊥ AB, the area of trap. ABCD is
(a) 84 cm²
(b) 72 cm²
(c) 80 cm²
(d) 91 cm²

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:

Question 1:

The base of a triangular field is three times its height and its area is 1350 m². Find the base and height of the field.

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:

Question 2:

Find the area of an equilateral triangle of side 6 cm.

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:

Question 3:

The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm. Find the length of the other diagonal and the area of the rhombus.

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:

Question 4:

The area of a trapezium is 216 m² and its height is 12 m. If one of the parallel sides is 14 m less than the other, find the length of each of the parallel sides.

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:

Question 5:

Find the area of a quadrilateral one of whose diagonals is 40 cm and the lengths of the perpendiculars drawn from the opposite vertices on the diagonal are 16 cm and 12 cm.

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:

Question 6:

A field is in the form of a right triangle with hypotenuse 50 m and one side 30 m. Find the area of the field.

Answer:

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

$= 600 m^{2}$

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Question 7:

Mark (✓) against the correct answer:
The base of a triangle is 14 cm and its height is 8 cm. The area of the triangle is
(a) 112 cm²
(b) 56 cm²
(c) 122 cm²
(d) 66 cm²

Answer:

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

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Question 8:

Mark (✓) against the correct answer:
The base of a triangle is four times its height and its area is 50 m². The length of its base is
(a) 10 m
(b) 15 m
(c) 20 m
(d) 25 m

Answer:

(c) 20 m

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

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Question 9:

Mark (✓) against the correct answer:
The diagonal of a quadrilateral is 20 cm in length and the lengths of perpendiculars on it from the opposite vertices are 8.5 cm and 11.5 cm. The area of the quadrilateral is
(a) 400 cm²
(b) 200 cm²
(c) 300 cm²
(d) 240 cm²

Answer:

(b) 200 cm²

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

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Question 10:

Mark (✓) against the correct answer:
Each side of a rhombus is 15 cm and the length of one of its diagonals is 24 cm. The area of the rhombus is
(a) 432 cm²
(b) 216 cm²
(c) 180 cm²
(d) 144 cm²

Answer:

(b) 216 cm²

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

$= 216 {cm}^{2}$

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Question 11:

Mark (✓) against the correct answer:
The area of a rhombus is 120 cm² and one of its diagonals is 24 cm. Each side of the rhombus is
(a) 10 cm
(b) 13 cm
(c) 12 cm
(d) 15 cm

Answer:

(b) 13 cm

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

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

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Question 12:

Mark (✓) against the correct answer:
The parallel sides of a trapezium are 54 cm and 26 cm and the distance between them is 15 cm. The area of the trapezium is
(a) 702 cm²
(b) 810 cm²
(c) 405 cm²
(d) 600 cm²

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

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Question 13:

Mark (✓) against the correct answer:
The area of a trapezium is 384 cm². Its parallel sides are in the ratio 5 : 3 and the distance between them is 12 cm. The longer of the parallel sides is
(a) 24 cm
(b) 40 cm
(c) 32 cm
(d) 36 cm

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

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

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Question 14:

Fill in the blanks.
(i) Area of triangle = $\frac{1}{2} \times (. . . . . . . . .) \times (. . . . . . . . .) .$
(ii) Area of a ||gm = $\frac{1}{2} \times (. . . . . . . . .) \times (. . . . . . . . .) .$
(iii) Area of a trapezium = $\frac{1}{2} \times (. . . . . . . . .) \times (. . . . . . . . .) .$
(iv) The parallel sides of a trapezium are 14 cm and 18 cm and the distance between them is 8 cm. The area of the trapezium is ......... cm².

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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