Question

In the adjoining figure $AB=2AD=a.P$ is the mid-point of $AD$. The area of the shaded region is

• A. $\frac{1}{6}{a}^2$
• B. $\frac{1}{2}{a}^2$
• C. $\frac{1}{4}{a}^2$
• D. $\frac{1}{3}{a}^2$

Answer 1

The correct option is C $\frac{1}{4}{a}^2$

Given $2AD=a$ then $AD=\frac{a}{2}$.

Since $P$ is the mid point of $AD$ then $AP=PD=\frac{a}{4}$.

Area of shaded region $=\Delta APB+\Delta PDC$.

Now, $\Delta APB\cong \Delta PDC$. [ $S-A-S$ rule]

Now $\Delta APB=\frac{1}{2}\times AP\times AB=\frac{a^2}{8}$

So area of shaded region is $2\times \frac{a^2}{8}=\frac{a^2}{4}$.