Question

find the two parallel sides of trapezium whose area is 1.6 metre square altitude is 10 DM and one of the parallel side is longer than the other by 8dm.​

Answer 1

given :-

area of the trapezium = 1.6m²

we know that, 1m² = 100dm²

➡ 1.6m² = 160dm²

altitude of the trapezium = 10dm

ATQ,

one of the parallel sides of the trapezium is 8dm more than the other one.

let the smaller parallel side be x

therefore the longer parallel side = x + 8

formula to find the area of a trapezium = 1/2 × h(b1 + b2)

where h is the altitude and b1 and b2 are the parallel sides Respectively.

➡ 1/2 × h (b1 + b2) = 160dm²

➡ 1/2 × 10(x + x + 8) = 160dm²

➡ 5(2x + 8) = 160dm²

➡ 10x + 40 = 160dm²

➡ 10x = 160 - 40

➡ x = 120/10

➡ x = 12dm

hence, the parallel sides are :-

• b1 = x = 12dm

• b2 = x + 8 = 12 + 8 = 20dm

Answer 2

Let smaller parallel side = x

Longer parallel side = x+8

Area of trapezium = 1/2 × h(d1 + d2)

1/2 × 10(x + x + 8) = 160

5(2x + 8) = 160

10x + 40 = 160

10x = 160 - 40

x = 120/10 = 12dm

So,

Parallel sides are :

d1 = x = 12dm

d2 = x + 8 = 20dm