myhelper: RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Exercise 18A

Exercise 18A

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Q1 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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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 1:

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

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

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

Q2 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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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 2:

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

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

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

Q3 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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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 3:

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

Q4 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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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 4:

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

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

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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 5:

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

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

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

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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 6:

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

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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 7:

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

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

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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 8:

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

$Area of the trapezium = 9450 m^{2}$

$(Given) ∴ 126 x = 9450 \Rightarrow x = \frac{9450}{126} \Rightarrow x = 75$

Thus, the length of the parallel sides are 75 m and 150 m, that is, 2×75 m, and the length of the longer side is 150 m.

Q9 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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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 9:

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

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

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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 10:

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

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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 11:

$Let ABCD be the given trapezium in which AB ∥ DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm .$

$Draw CL ⊥ AB and CM ∥ DA meeting AB at L and M, respectively .$

$Clearly, AMCD is a parallelogram .$

$Now, AM = DC = 10 cm MB = (AB - AM) = (20 - 10) cm = 10 cm$

$Also, CM = DA = 13 cm$

$Therefore, ∆ CMB is an isosceles triangle and CL ⊥ MB . L is the midpoint of B .$

$\Rightarrow ML = LB = (\frac{1}{2} \times MB)$
$= (\frac{1}{2} \times 10) cm = 5 cm$

$From right ∆ CLM, we have : {CL}^{2} = ({CM}^{2} - {ML}^{2}) {cm}^{2}$

$\Rightarrow {CL}^{2} = \{{(13)}^{2} - {(5)}^{2}\} {cm}^{2}$

$\Rightarrow {CL}^{2} = (109 - 25) {cm}^{2}$

$\Rightarrow {CL}^{2} = 144 {cm}^{2}$

$\Rightarrow CL = \sqrt{144} cm$

$\Rightarrow CL = 12 cm$

$∴ Length of CL = 12 cm$

$Area of the trapezium = \{\frac{1}{2} \times (AB + DC) \times CL\}$
$= \{\frac{1}{2} \times (20 + 10) \times 12\} {cm}^{2} = (\frac{1}{2} \times 30 \times 12) {cm}^{2} = (15 \times 12) {cm}^{2} = 180 {cm}^{2}$
$Hence, the area of the trapezium is 180 {cm}^{2} .$

Q12 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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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 12:

$Let ABCD be the trapezium in which AB ∥ DC, AB = 25 cm, CD = 11 cm, AD = 13 cm and BC = 15 cm .$

$Draw CL ⊥ AB and CM ∥ DA meeting AB at L and M, respectively . Clearly,$

$AMCD is a parallelogram .$

$Now, MC = AD = 13 cm AM = DC = 11 cm$

$\Rightarrow MB = (AB - AM)$
$= (25 - 11) cm = 14 cm$

$Thus, in ∆ CMB, we have : CM = 13 cm MB = 14 cm BC = 15 cm$

$∴ s = \frac{1}{2} (13 + 14 + 15) cm = \frac{1}{2} 42 cm = 21 cm (s - a) = (21 - 13) cm = 8 cm (s - b) = (21 - 14) cm = 7 cm (s - c) = (21 - 15) cm = 6 cm$

$∴ Area of ∆ CMB = \sqrt{s (s - a) (s - b) (s - c)}$
$= \sqrt{21 \times 8 \times 7 \times 6} {cm}^{2} = 84 {cm}^{2}$

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

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

RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Exercise 18A

Exercise 18A

Page-207

Q1 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q2 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q3 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q5 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q6 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q7 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q8 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q9 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q10 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q11 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Q12 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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