Question 69

In $$\triangle$$ABC, P is a point on BC such that BP : PC = 4 : 5 and Q is the mid point of BP. Then ar($$\triangle$$ABQ) : ar($$\triangle$$ABC) is equal to:

Solution

From figure , we can observe that height of $$\triangle ABC and \triangle ABQ$$ are equal.

Area of $$\triangle ABQ$$ = $$\frac{1}{2} \times 2x \times h $$

Area of $$\triangle ABC$$ = $$\frac{1}{2} \times 9x \times h $$

$$\frac{area of \triangle ABQ}{area of \triangle ABC}$$ = 2:9

So, the answer would be option c) 2:9

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