Question

In a circle of radius 5 cm, AB and AC are two equal chords such that $AB=AC=6$. Find the length of the chord BC

• A. 9.8 cm
• B. 9.6 cm
• C. 9.2 cm
• D. 7.3 cm

Answer 1

The correct option is B 9.6 cm

R.E.F image

Given circle of radius $=5cm$

$\Rightarrow OA=OB=OC=5cm$

and $AB=AC=6cm,BC=?$

Consider $\Delta$ le OAB & $\Delta$ le OAC $\Rightarrow OC=OB\Rightarrow$ Both triangle are congruent

OA is common

$\therefore \angle OAC=\angle OAB$

Draw perpendicular from center O to AB at D

In $\Delta$le OAB, $\angle ODB=\angle ODB={90}^{\circ }\Rightarrow AD=DB=\frac{AB}{2}=3cm$

$OA=OB$ as D bisector AB

Applying Pythagoras theorem in $\Delta$ le $ODB\Rightarrow O{D}^2+D{B}^2=O{B}^2$

$\Rightarrow OD=4cm$

Area of $\Delta$ $OAB=\frac{1}{2}bh=\frac{1}{2}\times AB\times OD=\frac{1}{2}\times 6cm\times 4=12c{m}^2$

Let 'E' be the point where OA meets BC

It can be observed that $\angle AEC=\angle AEB={90}^{\circ },$ further E bisector BC

$\therefore BE=EC=\frac{BC}{2}$

$\therefore$ Area of $\Delta$ le OAB, considering OA as base (b) & BE at height (h)

$\Rightarrow 12sqcm=\frac{1}{2}\times OA\times BE$

$\Rightarrow BE=\frac{24}{5}cm$

$\therefore BC=2\times BE=\frac{48}{5}=9.6cm$