In parallelogram ABCD- P is mid-point of AB- CP and BD intersect each other at point O- If area of - POB - 40 cm2- find Area of - ABC and Parallelogram ABCD-

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Area of Triangle

In parallelog...

Question

In parallelogram ABCD. P is mid-point of AB. CP and BD intersect each other at point O. If area of $Δ P O B = 40 c m^{2}$ , find Area of $Δ$ ABC and Parallelogram ABCD.

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Solution

Area $(Δ P O B) = 40 {c m}^{2}$
Construct OE perpendicular to PB and OF perpendicular to CA.
$∴$ area of $Δ P O B = \frac{1}{2} O E \times P B = 40 {c m}^{2}$
Similarly,
$a r (D O C) = \frac{1}{2} O F \times D C . Δ P E 0 \sim Δ C F O \frac{O E}{O F} = \frac{P O}{C O} = \frac{P E}{C F} Δ P O B \sim Δ D O C ∴ \frac{P O}{C O} = \frac{P B}{D C} = \frac{1}{2}$
(since $P$ is midpoint of $A B$ )
$∴ \frac{a r (P O B)}{a r (D O C)} = \frac{\frac{1}{2} O E \times P B}{\frac{1}{2} O F \times D C} = \frac{O E}{O F} \frac{P B}{D C} = {(\frac{P O}{C O})}^{2} = \frac{1}{4} ∴ a r (D O C) = 40 \times 4 = 160 c m^{2}$
Let $a r (B O C) = y ∴ a r (A B C) = a r (B D C) = a r (B O C) + a r (D O C) = (y + 160) {c m}^{2}$
we know that area of two triangles between two parallel lines and same base are same.
$a r (P D C) = a r (B D C) a r (P O D) + 160 = 160 y a r (P O D) = y a r (A P D) = a r (A B D) - a r (P O D) - a r (P O B) a r (A P D) = 160 + y - y - 40 = 120 {c m}^{2} a r (A P D) = a r (P B C) = 40 + y ∴ 40 + y = 120 ∴ y = 80 {c m}^{2} ∴ a r (B O C) = 80 {c m}^{2} a r (A B C) = a r (B C D) = 160 + 80 = 24080 {c m}^{2} a r (A B C D) = 2 a r (A B C) = 480 {c m}^{2}$

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