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 equilateral triangles described on the two sides of a right-angled triangle are together equal to the equilateral triangle on the hypotenuse in terms of their area.

Accepted Answer

Let a right triangle $△ A B C$ in which $∠ A$ is right angle and $A C = y, A B = x$

Three equilateral triangles $△ A E C, △ A F B$ and $△ C B D$ are drawn on the three sides of $△ A B C$ . Again let area of triangles made on $A C, A S$ and $B C$ are $A_{1}, A_{2}$ and $A_{3,}$ respectively.

Now, in $△ C A B$

$B C^{2} = A C^{2} + A B^{2} B C^{2} = y^{2} + x^{2} B C = \sqrt{y^{2} + x^{2}}$

Area of equilateral triangle is $\frac{\sqrt{3}}{4} {(s i d e)}^{2}$

Therefore, Area of equilateral triangle $A E C$

$A_{1} = \frac{\sqrt{3}}{4} {(A C)}^{2} A_{1} = \frac{\sqrt{3}}{4} y^{2}$

and area of equilateral triangle $A F B$

$A_{2} = \frac{\sqrt{3}}{4} {(A B)}^{2} = \frac{\sqrt{3}}{4} \sqrt{y^{2} + x^{2}} = \frac{\sqrt{3}}{4} (y^{2} + x^{2}) = \frac{\sqrt{3}}{4} y^{2} + \frac{\sqrt{3}}{4} x^{2} = A_{1} + A_{2}$

Hence, proved

Prove that the equilateral triangles described on the two sides of a right-angled triangle are together equal to the equilateral triangle on the hypotenuse in terms of their area.

Important Questions on Triangles

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