Question

Construct a $\angle$ABC = 60${}^{\circ }$ . Mark a point D on BC such that BD = 6 cm. Find the radius of the circle which touches AB at B and also pass through D.

• A. 2.44 cm
• B. 4.44 cm
• C. 5.48 cm
• D. None of these

Answer 1

The correct option is A

2.44 cm

$\angle$ABC = 60${}^{\circ }$ is drawn.

BE is drawn perpendicular to AB at B.

BD = 6 cm is taken from BC.

Perpendicular bisector of BD is drawn which intersects BE at O.

Taking O as centre and OB as radius, a circle is drawn which passes through B and D.

Consider triangle BOD; ON is the perpendicular bisector of BD.




$cos\left({30}^{\circ }\right)=\frac{BN}{OB}$

$⟹\frac{\sqrt{3}}{2}=\frac{3}{OB}$

$⟹OB=\frac{6}{\sqrt{3}}=3.46$