Class 10 Triangles

### Topic to be covered

☼ Areas of Similar Triangles .

### Areas of Similar Triangles

● You have learnt that in two similar triangles, the ratio of their corresponding sides is the same.
● You know that area is measured in square units. So, you may expect that this ratio is the square of the ratio of their corresponding sides. This is indeed true and we shall prove it in the next theorem.

color{blue}{"Theorem 6.6 :"} The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Proof : We are given two triangles ABC and PQR such that Δ ABC ~ Δ PQR (see Fig. 6.42).

●We need to prove that {ar(ABC)}/{ar(PQR)} = ((AB)/(QR))^2 = ((CA)/(RP))^2

● For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles.

Now ar(ABC) = 1/2 BC xx AM

and ar (PQR) = 1/2 QR xx PN

So {ar(ABC)}/{ar(PQR)} = color{red}{(1/2 xx BC xx AM)/(1/2 xx QR xx PN) = (BC xx AM)/(QR xx PN)} ...............(1)

Now, in Δ ABM and Δ PQN,

∠ B = ∠ Q (As Δ ABC ~ Δ PQR) and ∠ M = ∠ N (Each is of 90°)

So, Δ ABM ~ Δ PQN (AA similarity criterion)

Therefore, (AN)/(PN) = (AB)/(PQ) ................(2)

Also, Δ ABC ~ Δ PQR (Given)

so (AB)/(PQ) = (BC)/(QR) = (CA)/(RP) .............(3)

Therefore {ar(ABC)}/{ar(PQR)} = (AB)/(PQ) xx (AM)/(PN) [From (1) and (3)]

 = (AB)/(PQ) xx (AB)/(PQ) [From (2)]

 = ((AB)/(PQ))^2

● Now using (3), we get color{red}{{ar(ABC)}/{ar(PQR)} = ((AB)/(PQ))^2 = ((BC)/(QR))^2 = ((CA)/(RP))^2}
Q 3169480315

In Fig. 6.43, the line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal
areas. Find the ratio (AX)/(AB)
Class 10 Chapter 6 Example 9
Solution:

We have XY ∥ AC (Given)

So, ∠ BXY = ∠ A and ∠ BYX = ∠ C (Corresponding angles)
Therefore, Δ ABC ~ Δ XBY (AA similarity criterion)

so {ar(ABC)}/{ar(XBY)} = ((AB)/(XB))^2 ...............(1)

Also, ar (ABC) = 2 ar (XBY) (Given)

so {ar(ABC)}/{ar(XBY)} = 2/1 ..............(2)

Therefore, from (1) and (2),

((AB)/(XB))^2 = 2/1 , i.e. (AB)/(XB) = sqrt2/1

(XB)/(AB) = 1/sqrt2

or  1- (XB)/(AB) = 1- 1/sqrt2

or (AB - XB)/(AB) = ( sqrt2-1)/sqrt2

i.e. (AX)/(AB) = ( sqrt2-1)/sqrt2 = (2-sqrt2)/2