Question

In the given triangle ABC, $\angle A={45}^{\circ }$, BC = 9 cm as shown in the figure. The diameter of the circumcircle is equal to

• A. $6\sqrt{2}cm$
• B. 18 cm
• C. $9\sqrt{2}cm$
• D. 9 cm

Answer 1

The correct option is C

$9\sqrt{2}cm$

Draw line BD as diameter passing through center O and join DC

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

$\angle BDC=\angle BAC={45}^{\circ }$ (angle subtanded by same chord BC)

In $\Delta ADC$, angles are ${45}^{\circ },{45}^{\circ },{90}^{\circ }.$

So, the ratio of sides will be $1:1:\sqrt{2}$ corresponding sides of triangle can be calculated as

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

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

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

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ BC& & CD& & BD\\ \downarrow & & \downarrow & & \downarrow \\ 9& & 9& & 9\sqrt{2}\end{array}$

Hence diameter $BD=9\sqrt{2}cm$