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In this video, let's see how we can use similarity in area related problems.

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We have two triangles, ABC and pick you up and say it's given a triangle, ABC similar to Triangle

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Epicure.

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If this is given, then the ratio of area of triangle ABC to the area of Triangle Epicure will be equal

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to the square of the ratio of corresponding sites.

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That is, I can write this as equal to a B squared by P Q Square because these are corresponding sites.

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Right.

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And this is also equal to C Square by QR Square or X square by square.

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You can see B.C. and you are corresponding sites and AC and B are also corresponding sites.

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For example, if the sides of Triangle ABC are three, five and seven and if the sides of Triangle P,

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Q are six then and fourteen and let this area be A1 and let this area be A2, then even by A2 is equal

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to three by six, the whole square, which is equal to one by four.

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Now it's very easy to know how this comes about, right.

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We know that area of a triangle is equal to of.

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Into base.

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In due height.

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Now, if we take the ratio of two areas, this half will get cancelled, therefore we can write even

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by a two is equal to base one.

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Into height, one divided by base, two into height to a base one by base two.

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This is the ratio of sides, right?

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And in similar triangles, the ratio of heights also will be equal to the ratio of sides.

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So I can write each one by X to as again, base one by base two.

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And that leaves me with even by a two is equal to be run by B to the whole square.

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Which is what we have written over here, right?

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If you take B.C. and QR as the base and that is equal to A, B by P, Q, the whole square, and that's

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equal to AC bypass the whole square because the ratio of corresponding sites in the case of similar

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triangles is equal.