Question

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

• A. 4.8 cm
• B. 10.8 cm
• C. 9.6 cm
• D. none of these

Answer 1

The correct option is C

9.6 cm

Consider the triangles OAB and OAC are congruent as

AB=AC

OA is common

OB = OC = 5cm .

So $\angle$OAB = $\angle$OAC

Draw OD perpendicular to AB

Hence AD = AB/2 = $\frac{6}{2}$ = 3 cm as the perpendicular to the chord from the center bisects the chord.

In $△$ADO

${OD}^2$= ${AO}^2$ – ${AD}^2$

OD2 = $5^2$ – $3^2$

OD = 4 cm

So Area of OAB = $\frac{1}{2}$ AB x OD = $\frac{1}{2}$ 6 x 4 = 12 sq. cm. …..(i)

Now AO extended should meet the chord at E and it is middle of the BC as ABC is an isosceles with AB= AC

Triangles AEB and AEC are congruent as

AB =AC

AE common,

∠OAB = ∠ OAC.

Therefore triangles being congruent, ∠AEB = ∠AEC = ${90}^{\circ }$

Therefore BE is the altitude of the triangle OAB with AO as base.

Also this implies BE =EC or BC =2BE

Therefore the area of the $△$ OAB

= $\frac{1}{2}$ AO$\times$BE = $\frac{1}{2}$5$\times$BE = 12 sq. cm as arrived in eq (i).

BE = 12 $\times$ $\frac{2}{5}$ = 4.8cm

Therefore BC = 2BE = 2$\times$4.8 cm = 9.6 cm.