Question

In Figure , $\overline{EG}||\overline{DH}$, and the lengths of segments $\overline{DE}$ and $\overline{EF}$ are as marked. If the area of $△EFG$ is a, calculate the area of $△DFH$ in terms of $a$.

• A. $\frac{4a}{5}$
• B. $\frac{16a}{25}$
• C. $\frac{16a}{20}$
• D. $\frac{25a}{16}$

Answer 1

The correct option is D $\frac{25a}{16}$

Given: $EG\parallel DH$

By basic proportionality theorem,

$\frac{ED}{EF}=\frac{GH}{GF}$

So, $\frac{EF+ED}{EF}=\frac{GF+GH}{GF}$ ....By componendo

$\therefore \frac{FD}{EF}=\frac{FH}{GF}=\frac{5}{4}$ ....(I)

In $△FEG$ and $△FDH$

$\angle F$ is the common angle

And, $\frac{EF}{ED}=\frac{GF}{GH}$ ....From (I)

$\therefore △FEG\sim △FDH$ ....S.A.S test of similarity

By theorem on area of similar triangles, we get

$\frac{A\left(△FDH\right)}{A\left(△FEG\right)}={\left(\frac{FD}{EF}\right)}^2$

$\Rightarrow \frac{A\left(△FDH\right)}{a}={\left(\frac{5}{4}\right)}^2$

$\Rightarrow A\left(△FDH\right)=\frac{25}{16}a$