In the given figure, P and Q are the mid-points of AC and AB, respectively. Also, PG = GR and HQ = HR. Calculte the ratio of area of $Δ P Q R$ to area of $Δ A B C .$

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Solution

Construction: Join P and H.

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

Also, Area $(Δ P Q H) = \frac{1}{4} A r e a (Δ A B C) . . . (I)$

Area (quad. PGHQ) $= \frac{3}{4} A r e a (Δ P Q R) - - - (I I)$

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

Also, Area $(Δ P G H) = \frac{1}{2}$ Area $(Δ P Q H) - - - - (I I I)$

[Since they have same height but base $(Δ P G H) = \frac{B a s e (Δ P Q H)}{2}$ or $G H = \frac{1}{2} P Q]$

Area (quad. PGHQ)=Area $(Δ P Q H)$ +Area $(Δ P G H)$

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

$\frac{3}{4}$ Area $(Δ P Q R) = \frac{1}{4}$ Area $(Δ A B C) + \frac{3}{8}$ Area $(Δ P Q R)$

Simplifying, we get.

$\frac{A r e a (Δ P Q R)}{A r e a (Δ A B C)} = \frac{2}{3}$

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