Question

In a right-angled triangle, the lengths of the sides containing the right angle are a and b. With the midpoint of each side as centre, three semicircular areas are drawn outside the triangle. The total area enclosed is

• A. $\frac{\pi }{2}(a^2+b^2)+(a+b{)}^2$
• B. $\frac{\pi }{2}(a^2+b^2)+\frac{1}{2}ab$
• C. $\frac{\pi }{2}(a^2+b^2)+\frac{1}{8}{(a+b)}^2$
• D. $\frac{\pi }{4}(a^2+b^2)+ab$

Answer 1

The correct option is B $\frac{\pi }{2}(a^2+b^2)+\frac{1}{2}ab$

sol:

(I) Area of semicircle $\left(\widehat{ODA}\right)D=OA=a$

$=\frac{OA}{2}=\frac{a}{2}$ [radius]

$=\frac{\pi r^2}{2}=\frac{\pi }{2}{\left(\frac{a}{2}\right)}^2=\frac{\pi a^2}{8}$

(II) Area of semicricle$\left(\widehat{OEB}\right)=\frac{\pi r^2}{2}=\frac{\pi }{2}{\left(\frac{b}{2}\right)}^2$

(III) Area of semicircle $\left(\widehat{OEA}\right)=\frac{\pi }{2}(a^2+b^2)$

$=\frac{\pi (a^2+b^2)}{2}$

(IV) Area of triangle $\left(\Delta OAB\right)=\frac{1}{2}\times a\times b$

$=\frac{ab}{2}$

Adding all areas$=I+II+III+IV$

$=\frac{\pi a^2}{8}+\frac{\pi b^2}{8}+\frac{\pi (a^2+b^2)}{2}+\frac{ab}{2}$

$=\frac{\pi }{4}(a^2+b^2+a^2+b^2)+\frac{ab}{2}$.