Question

In the figure, AB and AC are two equal chords of a circle of radius 5 cm. If AB = AC = 6 cm, find the length of chord BC.

• A. 10.6 cm
• B. 9.6 cm
• C. 8.6 cm
• D. 7.6 cm

Answer 1

The correct option is B

9.6 cm

Let O be the centre of the circle. Join OB.

Let AD be the bisector of $\angle$BAC, meeting BC at D.

In $\Delta$BAD and $\Delta$CAD, we have

STATEMENT REASON

AB= AC Given

$\angle$BAD= $\angle$CAD By construction

AD= AD Common

$\therefore$ $\Delta$BAD $\cong$ $\Delta$CAD by SAS

$\angle$BDA = $\angle$CDA By c.p.c.t.

But, $\angle$BDA + $\angle$CDA = 180° since BDC is a straight line

$\therefore$ $\angle$BDA = $\angle$CDA = 90°.

This shows that AD is the perpendicular bisector of BC and so, it when produced passes through O.

∴ OB = OA = 5 cm.

Let OD = x cm and BD = y cm.

Then, AD = (OA - OD) = (5 - x) cm.

In right-angled $\Delta$ODB and $\Delta$ADB, we get

${OB}^2$ = ${OD}^2$ + ${BD}^2$ and ${AB}^2$ = ${AD}^2$ + ${BD}^2$

$5^2$ = $x^2$+ $y^2$ and $6^2$ = ${(5-x)}^2$ + $y^2$

$y^2$ = 25 - $x^2$ ...(i) and $y^2$ = 36 - ${(5-x)}^2$ …… (ii)

From (i) and (ii), we get :

25 – $x^2$ = 36 - ${(5-x)}^2$

10x = 14,

x = 1.4 Substituting x = 1.4 in (i), we get

$y^2$ = 25 - ${\left(1.4\right)}^2$ = 23.04

y = $\sqrt{23.04}$ = 4.8 cm.

∴ BD = 4.8 cm.

∴ BC = 2xBD = (2x4.8)cm = 9.6cm.

Hence, the length of chord BC = 9.6 cm.