Question

A circle passes through the three vertices of an isosceles triangle that has sides of length 3 and a base of length 2. The area of the circle is

• A. $\frac{9\pi }{4}$
• B. $\frac{81\pi }{32}$
• C. $\frac{27\pi }{16}$
• D. $\frac{5\pi }{2}$

Answer 1

The correct option is B $\frac{81\pi }{32}$

Extend segment AD to R on Circle O .
By the Pythagorean Theorem
$A{D}^2=3^2-1^2$
$AD=2\sqrt{2}$
$△ADC$ is Similar to $△ACR$ so
$\frac{2\sqrt{2}}{3}=\frac{3}{2r}$
Which gives us
$2r=\frac{9}{2\sqrt{2}}=\frac{9\sqrt{2}}{4}$
therefore
$r=\frac{9\sqrt{2}}{8}$
The area of the circle is $\pi r^2={\left(\frac{9\sqrt{2}}{8}\right)}^2$ = $\frac{81}{32}\pi$