Name: Areas of Similar Triangles
Uploaded: 2019-07-19
Duration: 6 min 54 s
Description: The corresponding sides of similar triangles are proportional. Are the ratios of areas of similar triangles the same as that of their sides? Watch this video...

Question

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Accepted Answer

Let the two triangles $A B C$ and $P Q R$ are such that $∆ A B C ~ ∆ P Q R$ .

To prove: $\frac{ar (A B C)}{ar (P Q R)} = {(\frac{A B}{P Q})}^{2} = {(\frac{B C}{Q R})}^{2} = {(\frac{C A}{R P})}^{2}$ .

Construction: Altitudes $A M$ and $P N$ of the triangles $∆ A B C$ and $∆ P Q R$ are drawn .

Now, $ar (A B C) = \frac{1}{2} B C \times A M$

And, $ar (P Q R) = \frac{1}{2} Q R \times P N$ .

so, $\frac{ar (A B C)}{ar (P Q R)} = \frac{\frac{1}{2} \times B C \times A M}{\frac{1}{2} \times Q R \times P N} = \frac{B C \times A M}{Q R \times P N} . . . . . (1)$

Now, in $∆ A B M$ and $∆ P Q N$ .

$∠ B = ∠ Q$ (As $Δ A B C ~ ∆ P Q R$ )

And, $∠ M = ∠ N = 90 °$

So, $∆ A B M ~ ∆ P Q N$ (AA similarity criterion)

Therefore, $\frac{A M}{P N} = \frac{A B}{P Q} . . . . . (2)$

Also, $∆ A B C ~ ∆ P Q R$

So, $\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{C A}{R P} \dots (3)$

Therefore, $\frac{ar (A B C)}{ar (P Q R)} = \frac{A B}{P Q} \times \frac{A M}{P N}$ $[from (1) and (3)]$

$= \frac{A B}{P Q} \times \frac{A B}{P Q}$

$= {(\frac{A B}{P Q})}^{2}$ $[From (2)]$

Now using $(3),$ we get

$\frac{ar (A B C)}{ar (P Q R)} = {(\frac{A B}{P Q})}^{2} = {(\frac{B C}{Q R})}^{2} = {(\frac{C A}{R P})}^{2}$ .

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Important Questions on Triangles

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