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### 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}`

● 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

areas. Find the ratio `(AX)/(AB)`

Class 10 Chapter 6 Example 9

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`