Question

The diagonals of a quadrilateral PQRS, PR and SQ intersect at O. If QO = OS, then find the area of triangle PQR given PR = 10 cm and altitude ST = 5 cm.

• A. $30{\text{cm}}^2$
• B. $10{\text{cm}}^2$
• C. $15{\text{cm}}^2$
• D. $25{\text{cm}}^2$

Answer 1

The correct option is D $25{\text{cm}}^2$


$Ar\left(\Delta PQR\right)=Ar\left(\Delta PQO\right)+Ar\left(\Delta QRO\right)$

As PO and RO are the medians of $\Delta PQS$ and $\Delta QRS$ respectively, We have, $Ar\left(\Delta PQO\right)=Ar\left(\Delta PSO\right)=Ar\left(\Delta QRO\right)=Ar\left(\Delta SRO\right)$

So, $Ar\left(\Delta PQR\right)=Ar\left(\Delta PQO\right)+Ar\left(\Delta QRO\right)$
$=Ar\left(\Delta PSO\right)+Ar\left(\Delta SRO\right)$
$=Ar\left(\Delta PSR\right)$
$=\frac{1}{2}\times PR\times ST$
$\frac{1}{2}\times 10\times 5$
$=25{\text{cm}}^2$