Question

In the given figure, area of $△AED$ = _____ $\times$ 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 DB. DB will be the diagonal of the parallelogram ABCD

Now DB divides the parallelogram into two triangles of equal area

Hence, area of $△ADB$ = area of $△DCB$

Now, area of $△ADB$ + area of $△DCB$ = Area of parallelogram ABCD

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

Since $△AED$ and $△ADB$ lie between the same parallels AD and CE, and on the same base AD

Area of $△AED$ = Area of $△ADB$

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