ABCD is a rectangle with O as any point in its interior. If $a r (Δ A O D) = 3 c m^{2}$ and $a r (Δ B O C) = 6 c m^{2},$ then find area of rectangle ABCD.

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Solution

Given: ABCD is a rectangle, and O is any point in its interior.
Area $(Δ A O D) = 3$ sqcm and Area $(Δ B O C) = 6$ sq cm.
From O draw a line EF parallel to AB intesecting AD at E and BC at F.
Therefore $O E ⊥ A D$ and $O F ⊥ B C .$
Let OE =x and OF =y
OE +OF=EF =AB
Therefore x+y=AB ....(1)
Area $(Δ A O D)$ + Area $(Δ B O C) = 3 + 6 = 9$ sq. cm.
$\frac{1}{2} \times A D \times O E + \frac{1}{2} \times O F \times B C = 9 s q . c m \frac{1}{2} [A D \times O E + O F \times B C] = 9 s q . c m A D \times O E + O F \times A D = 9 \times 2 = 18 s q . c m$
Since BC =AD [opposite sides of a rectangle]
$\Rightarrow A D (O E + O F) = 18 s q . c m A D \times E F = 18 s q . c m$
So, EF =AB
$\Rightarrow$ $A D \times A B = 18 s q . c m$
Area of the rectangle ABCD = $A B \times A D = 18 s q . c m$

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ABCD is a rectangle with O as any point in its interior. If ar(ΔAOD)=3cm2 and ar(ΔBOC)=6cm2, then find area of rectangle ABCD.

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