Question

If a triangle and a parallelogram are on the same base and between the same parallels, then:

• A. area of triangle = $\frac{1}{4}$area of parallelogram
• B. area of triangle = $\frac{1}{3}$area of parallelogram
• C. area of triangle = $\frac{1}{2}$area of parallelogram
• D. area of triangle = area of parallelogram

Answer 1

The correct option is C

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

Let $△$ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC.

To prove that, $ar\left(△PAB\right)=\frac{1}{2}ar\left(ABCD\right)$

Draw BQ parallel to AP to obtain another parallelogram ABQP. Now parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC.

Therefore, ar (ABQP) = ar (ABCD)

But $△$PAB $\cong$ $△$BQP (Diagonal PB divides parallelogram ABQP into two congruent triangles.)

So, ar ($△$PAB) = ar ($△$BQP)

Therefore, $ar\left(△PAB\right)=\frac{1}{2}ar\left(ABQP\right)$

This gives $ar\left(△PAB\right)=\frac{1}{2}ar\left(ABCD\right)$