The triangle ABC is given.

If CO : OB = 2 : 5 and Area of Triangle AOC = $44 c m^{2}$

Find Area of Triangle AOB.

$88 c m^{2}$

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$22 c m^{2}$

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$110 c m^{2}$

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$100 c m^{2}$

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Solution

The correct option is C
$110 c m^{2}$

Since OA is the common height for both the triangles

Ratio of Areas = Ratio of Bases

$\frac{C O}{O B}$ = $\frac{A r e a o f t r i a n g l e A O C}{A r e a o f t r i a n g l e A O B}$

$\frac{A r e a o f t r i a n g l e A O C}{A r e a o f t r i a n g l e A O B}$ = 2 : 5

Area of triangle AOC = 2x = $44 c m^{2}$

$\Rightarrow$ x = $22 c m^{2}$

$∴$ Area of triangle AOB = 5x = 5 $\times$ 22 = $110 c m^{2}$

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