Question

A right angled isosceles triangle is inscribed in a circle of radius $r$. What is the area of the remaining portion of the circle?

• A. $\frac{\pi r^2}{2}$
• B. $(\pi -\frac{1}{2})r^2$
• C. $(\pi -1)r^2$
• D. $(\pi -2)r^2$

Answer 1

The correct option is C $(\pi -1)r^2$

Area of reaming portion$=$area of circle$=$area of triangle
Radius of circle$=r$, Diameter$=2r$
Now, ${\left(2r\right)}^2=a^2+a^2{4r}^2={2a}^2\sqrt{2r}=a$
$\therefore$ area of reaming portion$={\pi r}^2-\frac{1}{2}\times a\times a={\pi r}^2-\frac{1}{2}\times \sqrt{2r}\times \sqrt{2r}={\pi r}^2-r^2=r^2(\pi -1)$