Question

One angle of a $\Delta ABC$ which is circumscribed by a circle, has one of the angles as ${135}^{\circ }$ and its opposite side is 5 cm as shown in the figure. The diameter of its circumcircle is equal to

• A. $10\sqrt{2}cm$
• B. $20\sqrt{2}cm$
• C. $5\sqrt{2}cm$
• D. $5\sqrt{3}cm$

Answer 1

The correct option is C

$5\sqrt{2}cm$

In $\Delta ABC$

$\angle B={135}^{\circ },AC=5cm$

Draw diameter AD and join CD to get quadrilateral ABCD.

$\angle ADC={180}^{\circ }-{135}^{\circ }={45}^{\circ }$ (ABCD is a cyclic quadrilateral )

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

Angles of $\Delta ADC$ are ${45}^{\circ },{45}^{\circ },{90}^{\circ }.$

The corresponding sides can be calculated as

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

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

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

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ x& :& x& :& x\sqrt{2}\\ \downarrow & & \downarrow & & \downarrow \\ AC& & CD& & AD\\ \downarrow & & \downarrow & & \downarrow \\ 5cm& & 5cm& & 5\sqrt{2}cm\end{array}$

Hence AD = $5\sqrt{2}cm$ is the diameter.