myhelper: Exercise 18C

Exercise 18C

Question 1

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If $m \| n$ and $\angle 1=20^{\circ}$, find the measures of angles $\angle 2, \angle 3, \angle 4, \angle 5, \angle 6, \angle 7$ and $\angle 8$.

Question 2

2. Find the value of each one of the lettered angles giving reasons.

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Question 3

Find the value of each one of the lettered angles giving reasons.

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Question 4

Find each of the lettered angles giving reasons.

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Question 5

Find each of the lettered angles giving reason.

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Multiple Choice Question (MCQ)
Tick (✔) the correct option.

Question 6

6. In the given figure, the arms of two angles are parallel. If $\angle A B C=55^{\circ}$, then $\angle P Q R$ equals

(a) $125^{\circ}$

(b) $35^{\circ}$

(d) None of these

Exercise 18C

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

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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 1:

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

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

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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 2:

(c) 8 cm

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

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

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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 3:

(a) 45 cm

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

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

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

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

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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 4:

(b) 23 cm

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

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

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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 5:

(c) 80 cm²

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

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

Exercise 18C

Page-211

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

Question 1:

Answer 1:

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

Question 2:

Answer 2:

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

Question 3:

Answer 3:

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

Question 4:

Answer 4:

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

Question 5:

Answer 5:

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