Question

In Figure , $\overline{QR}=\overline{RS}$. If the area of $△RST$ is $\frac{c}{2}$, what is the area of $△QSP$?

• A. $c\sqrt{2}$
• B. $c\sqrt{3}$
• C. $c$
• D. $2c$
• E. $3c$

Answer 1

Answer: (D)

$2c$

$QR=RS$ ....Given

So, $SR:SQ=1:2$ ...(I)

$\angle RTP+\angle RTS={180}^o$ ...Angles in linear pair

$\therefore {126}^o+\angle RTS={180}^o$

$\therefore \angle RTS={54}^o=\angle SPQ$ ...(II)

In $△RST$ and $△QSP$

$\angle S$ is the common angle.

$\therefore \angle RTS=\angle SPQ$ ... from (II)

$△RST\sim △SPQ$ ....AAA test of similarity

So, $\frac{A\left(△RST\right)}{A\left(△SPQ\right)}={\left(\frac{RS}{SQ}\right)}^2$ ....Theorem on ratio of areas of similar triangles

$\Rightarrow \frac{A\left(△RST\right)}{A\left(△SPQ\right)}={\left(\frac{1}{2}\right)}^2$

$\Rightarrow \frac{\frac{c}{2}}{A\left(△SPQ\right)}=\frac{1}{4}$

$A\left(△SPQ\right)=2c$ sq. units