In the given figure- PQ BC and AP - PB -1- 2- Find area - APQarea - ABC-

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Relation between Areas and Sides of Similar Triangles

In the given ...

Question

In the given figure, PQ || BC and AP : PB = 1 : 2. Find $\frac{area (∆ APQ)}{area (∆ ABC)}$ .

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Solution

GIVEN: In the given figure PQ || BC, and AP: PB = 1:2

TO FIND:

We know that according to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other side, then it divides the two sides in the same ratio.

Since triangle APQ and ABC are similar

Hence,

Now, it is given that .

So;

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

$\frac{Area (APQ)}{Area (ABC)} = {(\frac{AP}{AB})}^{2} = {(\frac{1}{3})}^{2} = \frac{1}{9}$

Hence we got the result

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In the given figure, PQ || BC and AP : PB = 1 : 2. Find area ∆APQarea ∆ABC.

In the given figure, PQ || BC and AP : PB = 1 : 2. Find $\frac{area (∆ APQ)}{area (∆ ABC)}$ .