myhelper: RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Test Paper 18

Test Paper 18

Page-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.

Sol :

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

Question 2:

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

Answer 2:

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

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 3:

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

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 4:

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

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 5:

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

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

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

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 6:

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

$= 600 m^{2}$

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 7:

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

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 8:

(c) 20 m

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

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 9:

(b) 200 cm²

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

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 10:

(b) 216 cm²

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

$= 216 {cm}^{2}$

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 11:

(b) 13 cm

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

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

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 12:

(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

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 13:

(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

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

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 14:

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

RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Test Paper 18