Question

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

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

• A. $\frac{PT}{TR}$
• B. $\frac{RT}{TP}$
• C. $\frac{PS}{SR}$
• D. $\frac{SR}{TP}$

Answer 1

The correct option is A $\frac{PT}{TR}$

Given: $ST\parallel QR$

Construct a perpendicular from S on PR, such that it meets PR at M

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

$\frac{A\left(△PST\right)}{A\left(△RST\right)}=\frac{\frac{1}{2}SM\times PT}{\frac{1}{2}SM\times RT}$

$\frac{A\left(△PST\right)}{A\left(△RST\right)}=\frac{PT}{RT}=\frac{PT}{TR}$