Question

In the adjoining figure, $\Delta QPR$ is right-angled at Q in which QR = 6 cm and PQ = 7 cm. Find the area of $\Delta QSR$, given that PS is parallel to QR.

• A. $21c{m}^2$
• B. $20c{m}^2$
• C. $10c{m}^2$
• D. $11c{m}^2$

Answer 1

The correct option is A $21c{m}^2$


Area of right angled triangle PQR
= $\frac{1}{2}$ x Base x Height
=$\frac{1}{2}$ x QR x PQ
=$\frac{1}{2}$ x 6 x 7
$=21c{m}^2$

$\Delta QPR$ and $\Delta QSR$ lie on same base QR and are between same parallels hence, their areas are equal.
Area $\Delta QPR$ = Area $\Delta QSR$
$\therefore Areaof\Delta QSR=21c{m}^2$