In parallelogram $A B C D$ . $P$ is the mid-point of $A B$ . $C P$ and $B D$ intersect each other at point $O$ . If area of $Δ P O B = 40 c m^{2}$ . Find the area of $Δ B O C$ .

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Solution

We know that the ratio of the areas of the two similar triangles is equal to the square of the ratio of the corresponding sides therefore,

$\frac{A r (Δ P O B)}{A r (Δ C O D)} = {(\frac{P O}{C O})}^{2}$

$\Rightarrow A r (Δ C O D) = 40 \times 4 = 160 c m^{2}$
Let the area of triangle BOC be 'y'.

We know that area of two triangles between the two parallel lines are equal so,

$A r (Δ P D C) = A r (Δ B D C)$

$\Rightarrow$ $A r (Δ P O D) + 160 = 160 + y$

$\Rightarrow$ $A r (Δ P O D) = y$

Similarly,

$A r (Δ A D P) = A r (Δ B C P)$

$\Rightarrow$ $A r (Δ A D P) = 40 + y$

Now,

$2 A r (Δ B C D) = A r (A B C D)$

$\Rightarrow$ $2 (160 + y) = (160 + y) + (40 + y) (40 + y)$

$\Rightarrow$ $320 + 2 y = 240 + 3 y$

$\Rightarrow$ $3 y - 2 y = 320 - 240$

$\Rightarrow$ $y = 80$

Therefore,

$A r (Δ B O C) = 80 c m^{2}$

$\Rightarrow$ $A r (Δ P B C) = 120 c m^{2}$

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