Question

Draw a circle with radius $3$ cm and center at point O. Draw two radii OA and OB of the circle drawn such that $\angle AOB={60}^0$, then join A and B, then find the measure of the length AB

• A. $9$ cm
• B. $6$ cm
• C. $3$ cm
• D. $18$ cm

Answer 1

Answer: (C)

$3$ cm

Construction- A circle with centre O and of radius $=3$ cm $=$OA is drawn.

At O a ${60}^o$ angle OAB is made.

OB intersect the cicumference at B.

$m\left(AB\right)=3$ cm.

Proof- In $\Delta$ AOB we have $OA=OB$ (radii of the sme circle).

So $\Delta$ AOB is isosceles. i.e $\angle OAB=\angle OBA$

$⟹\angle OAB+\angle OBA=2\angle OAB$ or $2\angle OBA$.

$\therefore$ By angle sum propety of triangles, we have $\angle AOB+\angle OAB+\angle OBA=\angle AOB+2\angle OAB={180}^o$

$⟹{60}^o+2\angle OAB={180}^o$

$⟹\angle OAB={60}^o=\angle OBA.$

$\therefore \Delta$ AOB is equlateral.

So $AB=OA=OB=3$ cm