Question

In the figure, $segPR\perp segBC,segAS\perp segBCandsegQT\perp segBC$, then find the ratio

$\frac{A\left(△BPR\right)}{A\left(△CQT\right)}$

• A. $\frac{BR\times PR}{TC\times QT}$
• B. $\frac{AR\times PR}{TC\times QT}$
• C. $\frac{CR\times PR}{TC\times QT}$
• D. None of these

Answer 1

The correct option is A $\frac{BR\times PR}{TC\times QT}$

Area of triangle = $\frac{1}{2}$ base $\times$ height

$\frac{A\left(△BPR\right)}{A\left(△CQT\right)}=\frac{\frac{1}{2}\times BR\times PR}{\frac{1}{2}\times QT\times CT}$

$\frac{A\left(△BPR\right)}{A\left(△CQT\right)}=\frac{BR\times PR}{QT\times CT}$