Question

In the given figure, P and Q are the mid-points of AC and AB. Also, PG =GR and HQ=HR. What is the ratio of area of $\Delta PQR$ to area of $\Delta ABC?$

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• A. $\frac{3}{5}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C

$\frac{2}{3}$

Construction: Join P and H.

PQ||BC and $PQ=\frac{1}{2}BC$ [Mid point theorem].

Also, Area $\left(\Delta PQH\right)=\frac{1}{4}Area\left(\Delta ABC\right)...\left(I\right)$

Area (quad. PGHQ) $=\frac{3}{4}Area\left(\Delta PQR\right)---\left(II\right)$

[Since GH||PQ and GH $=\frac{1}{2}PQ]$

Also, Area $\left(\Delta PGH\right)=\frac{1}{2}$ Area $\left(\Delta PQH\right)----\left(III\right)$

[Since they have same height but base $\left(\Delta PGH\right)=\frac{Base\left(\Delta PQH\right)}{2}$ or $GH=\frac{1}{2}PQ]$

Area (quad. PGHQ)=Area $\left(\Delta PQH\right)$ +Area $\left(\Delta PGH\right)$

From (I),(II) and (III).

$\frac{3}{4}$ Area $\left(\Delta PQR\right)=\frac{1}{4}$ Area $\left(\Delta ABC\right)+\frac{3}{8}$ Area $\left(\Delta PQR\right)$

Simplifying, we get.

$\frac{Area\left(\Delta PQR\right)}{Area\left(\Delta ABC\right)}=\frac{2}{3}$