Question

In the given figure, AF is parallel to CD, ED is parallel to FC and AD is parallel to BI. J is the mid point of AI. What will be the ratio of $\frac{\text{Area of trapezium AJCB}}{\text{Area of parallelogram EFCD}}$?

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{2}{5}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D

$\frac{3}{4}$

Fron the given figure, AF is parallel to CD, ED is parallel to FC, and BI is parallel to AD.

ABCD and EFCD are parallelograms with the same base CD and between same parallels AF and CD.

Thus, area of parallelogram ABCD = area of parallelogram EFCD ------------- (1)

$△AID$ and parallelogram ABCD lie on the same base AD and between the same parallels AD and BI

Hence, $Areaof△AID=\frac{1}{2}\times AreaofparallelogramABCD$

Now, J is the mid point of AI

Hence, DJ is a median of $△AID$

Thus, area of $△ADJ$ = area of $triangleAIJ$ = $\frac{1}{2}\times Areaof△AID$

Thus, area of $△ADJ$ = $\frac{1}{4}AreaofparalleogramABCD$

Now, area of trapezium AJCB + Area of $△ADJ$ = Area of parallelogram ABCD

Area of trapezium AJCB + $\frac{1}{4}AreaofparalleogramABCD$ = Area of parallelogram ABCD

Area of trapezium AJCB = $\frac{3}{4}$ Area of paralleogram ABCD}

From (1), we have area of parallelogram ABCD = area of parallelogram EFCD

Thus, Area of trapezium AJCB = $\frac{3}{4}$ Area of paralleogram EFCD

Or,

$\frac{\text{Area of trapezium AJCB}}{\text{Area of parallelogram EFCD}}=\frac{3}{4}$