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 radius of the circle is 5 cm. find the length of the equal chords.

• A. $10\sqrt{2}cm$
• B. 10 cm
• C. $5\sqrt{2}cm$
• D. 5 cm

Answer 1

The correct option is C

$5\sqrt{2}cm$

Given radius AO = 5 cm, then diameter AB =10 cm

In $\Delta ACB$

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

$\angle A=\angle B$ (ABC is an isosceles traingle, AC = BC)

$\angle A+\angle B+{90}^{\circ }={180}^{\circ }$

$x+x+{90}^{\circ }={180}^{\circ }\Rightarrow x={45}^{\circ }$

Now, angle of triangle 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, BC, AB will be in the ratio $1:1:\sqrt{2}$

The corresponding sides can be calculated as.

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ x& :& x& :& x\sqrt{2}\\ AC& & CB& & AB\\ \downarrow & & \downarrow & & \downarrow \\ 5\sqrt{2}& & 5\sqrt{2}& & 10\end{array}$

($x\sqrt{2}=10,\Rightarrow x=\frac{10}{\sqrt{2}}=5\sqrt{2}$)

So, the chord AC =CB $=5\sqrt{2}cm$