Question

AC and BC are two equal chords of a circle with diameter AB forming a $\Delta ABC$ as shown in the figure. If the equal chords have length 20 cm, then the area of the circle is equal to

• A. $50\pi c{m}^2$
• B. $100\pi c{m}^2$
• C. $200\pi c{m}^2$
• D. $150\pi c{m}^2$

Answer 1

The correct option is C

$200\pi c{m}^2$

In the given above figure, triangle ABC is isosceles (AC = BC) given

$\angle ACB={90}^{\circ }$ (angle subtended by diameter AB on circumference)

$\angle A+\angle B+{90}^{\circ }={180}^{\circ }$ (sum of angles of a triangle)

$2x+{90}^{\circ }={180}^{\circ }(\angle A=\angle B$. say x as triangle is isosceles)

$x={45}^{\circ }$

Now, angle of tirangle ABC are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$

$\Rightarrow \sin \left(45\right):\sin \left(45\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{\sqrt{2}}:\frac{1}{\sqrt{2}}:1$

$\Rightarrow 1:1:\sqrt{2}$

So, sides AC, CB, AB will be in ratio $1:1:\sqrt{2}$

So, the corresponding sides will be calculated as

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ AC& & CB& & AB\\ \downarrow & & \downarrow & & \downarrow \\ 20cm& & 20cm& & 20\sqrt{2}cm\end{array}$

Diameter $AB=20\sqrt{2}cm$, radius $=10\sqrt{2}cm$

Area $=\pi r^2=\pi (10\sqrt{2}{)}^2=200\pi c{m}^2$