Question

In the figure $segST\parallel sideQR$. Find the following ratio

$\frac{A\left(△PST\right)}{A\left(△QST\right)}$

• A. $\frac{PS}{QS}$
• B. $\frac{QS}{PS}$
• C. $\frac{PT}{QT}$
• D. $\frac{QT}{PT}$

Answer 1

The correct option is A $\frac{PS}{QS}$

Given: $△PQR$, $ST\parallel QR$

Construction: Draw a perpendicular from $T$ on $PQ$ to meet $PQ$ at $M$

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

Thus, $\frac{A\left(△PST\right)}{A\left(△QST\right)}=\frac{\frac{1}{2}PS\times TM}{\frac{1}{2}QS\times TM}$

= $\frac{PS}{QS}$