MCQ 2011 Mark
If $\triangle \text{ABC} \sim \triangle \text{PQR}$, then $y+z$ equals
- A$2+\sqrt{3}$
- ✓$4+3 \sqrt{3}$
- C$4+\sqrt{3}$
- D$3+4 \sqrt{3}$
Answer
View full question & answer→Correct option: B.
$4+3 \sqrt{3}$
Given, $\triangle \text{ABC} \sim \triangle \text{PQR}$
$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$
$\Rightarrow \frac{z}{3}=\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow \frac{z}{3}=\frac{8}{6}$ and $\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow z=\frac{24}{6}=4$ and $y=\frac{24 \sqrt{3}}{8}=3 \sqrt{3}$
$\Rightarrow y+z=4+3 \sqrt{3} .$
$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$
$\Rightarrow \frac{z}{3}=\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow \frac{z}{3}=\frac{8}{6}$ and $\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow z=\frac{24}{6}=4$ and $y=\frac{24 \sqrt{3}}{8}=3 \sqrt{3}$
$\Rightarrow y+z=4+3 \sqrt{3} .$
