In this video, let's see how we can use similarity in area related problems.
We have two triangles, ABC and pick you up and say it's given a triangle, ABC similar to Triangle
Epicure.
If this is given, then the ratio of area of triangle ABC to the area of Triangle Epicure will be equal
to the square of the ratio of corresponding sites.
That is, I can write this as equal to a B squared by P Q Square because these are corresponding sites.
Right.
And this is also equal to C Square by QR Square or X square by square.
You can see B.C. and you are corresponding sites and AC and B are also corresponding sites.
For example, if the sides of Triangle ABC are three, five and seven and if the sides of Triangle P,
Q are six then and fourteen and let this area be A1 and let this area be A2, then even by A2 is equal
to three by six, the whole square, which is equal to one by four.
Now it's very easy to know how this comes about, right.
We know that area of a triangle is equal to of.
Into base.
In due height.
Now, if we take the ratio of two areas, this half will get cancelled, therefore we can write even
by a two is equal to base one.
Into height, one divided by base, two into height to a base one by base two.
This is the ratio of sides, right?
And in similar triangles, the ratio of heights also will be equal to the ratio of sides.
So I can write each one by X to as again, base one by base two.
And that leaves me with even by a two is equal to be run by B to the whole square.
Which is what we have written over here, right?
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
equal to AC bypass the whole square because the ratio of corresponding sites in the case of similar
triangles is equal.