## 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 cm ^{2}. 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 m ^{2} 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 cm ^{2}. 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 cm ^{2} 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 m ^{2} 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 = 90 ^{o}, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.**

Solution: From the right, ∆ ABC,

∴ AB^{2} = (AC^{2} – BC^{2})

⇒ AB^{2} = (41)^{2} – (40)^{2}

⇒ AB^{2} = (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 – **

1096 kese aya in question no 5.

1092 kese aya in question no 5.

U are mad

I want and short Number 11 and Number 12 math

Ques5.solution:-

1586=½(84+x)×26

1586=(84+x)×26/2

1586/26×2=84+x

122=84+x

x=122-84

x=38m=ans.