Question

(i) Prove that: If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half of the parallelogram.

(ii) If the area of the parallelogram is 202 sq. units, find the area of the triangle which is on the same base as that of the parallelogram between the same parallel lines .

Answer 1

(i)

Given: $△$ ABP and $\square$ ABCD are on the same base AB.

To prove: A($△$ABP) $=\frac{1}{2}\times A(\square$ ABCD)

Proof:

Let $△$ ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC. Let “h” be the height of the triangle and the parallelogram.

Area of $△$ ABP $=\frac{1}{2}\times AB\times h...\left(i\right)$

Area of parallelogram ABCD = AB $\times$ h ...(ii)

From (i) and (ii)

Area ( $△$PAB) $=\frac{1}{2}\times$ area of parallelogram ABCD.

(ii) It is given that the area of the parallelogram is 202 sq. units,

So area of triangle $=\frac{1}{2}\times$ area of parallelogram

$\therefore$ Area of triangle $=\frac{1}{2}\times 202=101$ sq.units