Question

In the given figure, area of $△AID$ = _____ $\times$ area of parallelogram ABCD

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

Answer 1

The correct option is B

$\frac{1}{2}$

Join AC. AC will be the diagonal of the parallelogram ABCD

Now AC divides the parallelogram into two triangles of equal area

Hence, area of $△ACD$ = area of $△ABC$

Now, area of $△ACD$ + area of $△ABC$ = Area of parallelogram ABCD

Hence, area of $△ACD$ = area of $△ABC$ = $\frac{1}{2}\times$ Area of parallelogram ABCD

Since $△AID$ and $△ACD$ lie between the same parallels AD and BI, and on the same base AD

Area of $△AID$ = Area of $△ACD$

Area of $△AID$ = $\frac{1}{2}\times$ Area of parallelogram ABCD

In decimals, it would be Area of triangle AID = $0.5\times$ Area of parallelogram ABCD