# RS Aggarwal Class 8 Math Eighteenth Chapter Area of a Trapezium and a Polygon Exercise 18A Solution

## EXERCISE 18A

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

Solution: Let the length of the other parallel side be x m. Therefore, the length of the other parallel sides is 38 m.

(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.

Solution: Let the length of the sides be 4x and 5x cm. Therefore the length of the sides are 20 cm and 25 cm.

(7) The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer that the other by 6 cm, find the two parallel sides.

Solution: Let the lengths of the parallel sides be x cm and (x + 6) cm. Therefore lengths of the parallel sides are 17 cm and (17 + 6) = 23 cm.

(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.

Solution: Let the lengths of the parallel sides be x m and 2x m. Therefore lengths of the parallel sides are 75 m and (75 × 2) = 150 m.

(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.

Solution: Length of the side = [130 – (54 + 19 + 42)] = 15 m

∴ Area of the trap. – shaped field, (10) In the given figure, ABCD is a trapezium in which AD BC, ABC = 90o, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.

Solution: From the right, ∆ ABC,

∴ AB2 = (AC2 – BC2)

⇒ AB2 = (41)2 – (40)2

⇒ AB2 = (1681 – 1600) = 81 (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.

Solution: Let ABCD be the given trapezium in which AB ∥ DC, AB = 20 cm, DC = 10 cm

AD = BC = 13 cm.

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

Clearly, AMCD is a ∥ gm.

∴ AM = DC = 10 cm

MB = (AB – AM) = (20 – 10) = 10 cm

Now, CM = DA = 13 cm and CB = 13 cm.

∴ ∆CMB is an isosceles triangle and CL ⊥ MB.

⇒ L is the midpoint of MB. (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.

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

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

Clearly, AMCD is a ∥ gm.

∴ AM = DC = 11 cm.

MB = (AB – AM) = (25 – 11) = 14 cm

From ∆CMB, we have:

CM = 13 cm

MB = 14 cm

BC = 15 cm For more exercise solution, Click below –

Updated: May 30, 2022 — 2:34 pm

1. 1096 kese aya in question no 5.

2. 1092 kese aya in question no 5.