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

Equilateral triangles are drawn on the three sides of a right-angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

Accepted Answer

A right-angled triangle $A B C$ is right angle at $∠ B$ . Equilateral triangles $△ P A B, △ Q B C$ and $△ R A C$ are drawn on the sides $A B, B C$ and $C A$ respectively.

To prove: $a r (△ P A B) + a r (△ B C Q) = a r (△ R A C)$

Proof: Since, $△ P A B, △ Q B C$ and $△ R A C$ are equilateral. Therefore, they are equiangular and hence similar.

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$∴ \frac{a r (△ P A B)}{a r (△ R A C)} = \frac{A B^{2}}{A C^{2}}$ ________________ $(1)$

And $\frac{a r (△ B C Q)}{a r (△ R A C)} = \frac{B C^{2}}{A C^{2}}$ ________________ $(2)$

Adding equation $(1)$ and $(2)$ , we get,

$\frac{a r (△ B C Q)}{a r (△ R A C)} + \frac{a r (△ P A B)}{a r (△ R A C)} = \frac{B C^{2}}{A C^{2}} + \frac{A B^{2}}{A C^{2}}$

$\Rightarrow \frac{a r (△ B C Q) + a r (△ P A B)}{a r (△ R A C)} = \frac{B C^{2} + A B^{2}}{A C^{2}}$

$\Rightarrow \frac{a r (△ B C Q) + a r (△ P A B)}{a r (△ R A C)} = \frac{A C^{2}}{A C^{2}}$ (Using Pythagoras theorem)

$\Rightarrow \frac{a r (△ B C Q) + a r (△ P A B)}{a r (△ R A C)} = 1$

$\Rightarrow a r (△ B C Q) + a r (△ P A B) = a r (△ R A C)$

Hence, proved.

Equilateral triangles are drawn on the three sides of a right-angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

Important Questions on Similar Triangles

Learn and Prepare for any exam you want !