Question 87

Points P and Q lie on side AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio of areas of triangle APQ: triangle ABC is 16:25. Then the ratio of AP:PB is?

Solution

PQ is parallel to BC angle A is common, thus $$\triangle$$ APQ $$\sim$$ $$\triangle$$ ABC

In two similar triangles, the ratio of their areas is equal to the square of the ratio of their sides.

=> $$(\frac{AP}{AB})^2 = \frac{ar (\triangle APQ)}{ar (\triangle ABC)}$$

=> $$\frac{AP}{AB} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

Also, PB = AB - AP = 5 - 4 = 1

=> $$\frac{AP}{PB} = \frac{4}{1}$$ = 4 : 1

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