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Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

Answers (1)

Let PQR is a right angle triangle which is right angle at point Q.

$PQ = b, QR = a$

Three semicircles are drawn on the sides of $\Delta PQR$ having diameters PQ, QR and PR respectively.

Let $x_{1}, x_{2}$ and $x_{3}$ are the areas of semicircles respectively.

To prove:- $x_{3}=x_{1}+x_{2}$

Proof: In $\Delta PQR$ use Pythagoras theorem we get

$PR^{2}=PQ^{2}+QR^{2}$

$PR^{2}=a^{2}+b^{2}$

$PR^{2}=\sqrt{a^{2}+b^{2}}$

Now area of semi-circle drawn on side PR is

$x_{3}=\frac{\pi }{2}\left ( \frac{PR}{2} \right )^{2}$ $\left [ \therefore \text {area of semicircle=}\frac{\pi r^{2}}{2} \right ]$

$=\frac{\pi}{2}\left ( \frac{\sqrt{a^{2}+b^{2}}}{2} \right )^{2}$ $\left ( \therefore PR=\sqrt{a^{2}+b^{2}} \right )$

$=\frac{\pi}{2}\times\frac{\left ( a^{2}+b^{2} \right )}{4}$

$x_{3}=\frac{\pi}{8}\left ( a^{2}+b^{2} \right )$

area of semi-circle drawn a side QR is

$x_{2}=\frac{\pi }{2}\left ( \frac{QR}{2} \right )^{2}$

$=\frac{\pi }{2}\left ( \frac{a}{2} \right )^{2}$

$=\frac{\pi}{2}\times \frac{a^{2}}{4}$

$x_{2}=\frac{\pi}{8}a^{2}\; \; \; \; \; \; \; ....(2)$

area of semicircle drawn on side PQ is

$x_{1}=\frac{\pi }{2}\left ( \frac{PQ}{2} \right )^{2}$

$x_{1}=\frac{\pi }{2}\left ( \frac{b^{2}}{4} \right )\Rightarrow \frac{\pi}{8}b^{2}\; \; \; \; \; \; \; \; ....(3)$

add equation (2) and (3) we get

$x_{2}+x_{1}=\frac{\pi }{8}a^{2}+\frac{\pi}{8}b^{2}$

$x_{2}+x_{1}=\frac{\pi }{8}\left ( a^{2}+b^{2} \right )=x_{3}$

Hence $x_{2}+x_{1}=x_{3}$

Hence proved

Posted by

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