Question

Construct a $\angle$ABC = 30${}^{\circ }$. Mark a point D on BC such that BD = 6 cm. Find the radius of the circle to touch the line AB at B and also pass through D.

• A. 5.5 cm
• B. 2.5 cm
• C. 3.464 cm
• D. 6.4 cm

Answer 1

The correct option is C

3.464 cm

$\angle$ABC = 30${}^{\circ }$ 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 pass through B and D.

In triangle BOD

OB = OD

Therefore, $\angle OBD=\angle ODB={60}^{\circ }$

So,$\angle BOD={180}^{\circ }-\angle OBD-\angle ODB$

$\angle BOD={60}^{\circ }$

All three angles of triangle BOD are equal. So it is a equilateral triangle.

Therefore, OB = OB = BD = 6cm