"The ratio of areas of similar triangles is equal to the ratio of squares of corresponding side". Prove by using equilateral triangle scelane triangles.

Open in App

Solution

Ratio of area of similar $△$ is $=$ ratio of square of corresponding sides
$△ A B C sin △ P Q R$
$a r (A B C) = \frac{1}{2} \times B \times H = \frac{1}{2} \times B C \times A K$ ----- $(1)$
$a r (P Q R) = \frac{1}{2} \times B \times H = \frac{1}{2} \times Q R \times P L$ ----- $(2)$
Dividing $(1)$ and $(2)$
$\frac{a r (A B C)}{a r (△ P Q R)} = \frac{B C \times A M}{Q R \times P N}$ ------ $(3)$
In $△ A B K$ and $△ P Q L$
$∠ B = ∠ Q$ (angle of similar $△$ are equal )
$∠ M = ∠ N$ ( Both $90^{o}$ )
$△ A B K sin △ P Q L$ ( By $A A$ similarity )
$∴ \frac{A B}{P Q} = \frac{A M}{P N}$ {By $C P C T$ } ------ $(4)$
$\Rightarrow \frac{a r (△ A B C)}{a r (△ P Q R)} = \frac{B C}{Q R} \times \frac{A B}{P Q}$ ------- $(5)$ {from $(3)$ and $(4)$ }
$a r △ A B C sin △ P Q R$
$\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}$
$\Rightarrow \frac{a r (△ A B C)}{a r (△ P Q R)} = \frac{A B \times A B}{P Q \times P Q} = {(\frac{A B}{P Q})}^{2} = {(\frac{B C}{Q R})}^{2} = {(\frac{A C}{P R})}^{2}$

Suggest Corrections

Similar questions

Ratio of the areas of two similar triangles is not equal to the ratio of the squares of corresponding sides of the triangles.

Join BYJU'S Learning Program