Question

In $\Delta ABC$, which is circumscribed by a circle ,$\angle A={45}^{\circ }$, BC =6 cm. The diameter of the circumcircle is equal to

• A. 3 cm
• B. 6 cm
• C. $3\sqrt{2}cm$
• D. $6\sqrt{2}cm$

Answer 1

The correct option is D

$6\sqrt{2}cm$

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

$\angle BDC={45}^{\circ }$ (Angle formed by same segment BC)

$\angle BCD={90}^{\circ }$ (diameter BD subtends ${90}^{\circ }$ at circumference C.)

Now, in triangle BDC, angles are ${45}^{\circ },{45}^{\circ }{90}^{\circ }$

Hence, ratio of sides of triangle BDC will be $1:1:\sqrt{2}$

corresponding side of triangle can be calculated as

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

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

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

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ BC& & CD& & BD\\ \downarrow & & \downarrow & & \downarrow \\ 6& & 6& & 6\sqrt{2}\end{array}$

So, diameter $BD=6\sqrt{2}cm$