Question

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

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

Answer 1

Given: $ST\parallel QR$
Construct a perpendicular from Q on ST such that it meets ST on M and COnstruct a perpendicular from R on ST such that it meets ST at N.
Hence, $QM\perp ST$ and $RN\perp ST$
Since, $ST\parallel QR$, perpendicular distance between two parallel lines is same
hence, $QM=RN$
Now, Area of triangle = $\frac{1}{2}base\times height$
$\frac{A(△QST}{A\left(△RST\right)}=\frac{\frac{1}{2}QM\times ST}{\frac{1}{2}RN\times ST}$
$\frac{A(△QST}{A\left(△RST\right)}=\frac{1}{1}$ (Since, $QM=RN$)