Question

In a circle of radius 5 cm, AB and AC are two chords of 6 cm each. Then the length of the chord BC is

• A. $9.6$ cm
• B. $8.4$ cm
• C. $8.8$ cm
• D. $9.2$ cm

Answer 1

The correct option is A

$9.6$ cm

Let AB and AC be two equal chords , of 6 cm length, of a circle whose centre is O.

Join BC and draw AD$\perp$BC and join D to O.

Now, AB = AC = 6 cm and AD$\perp$BC.

$\therefore$ BD = DC

$⟹$AD is the perpendicular bisector of BC.

Also, centre O will lie on this perpendicular bisector. Join OB and OD.

Now let BD = DC = x and OD = y

In $△$OBD, $\angle D={90}^{\circ }$

$\therefore O{B}^2=B{D}^2+O{D}^2$

$⟹25=x^2+y^2$ - - - - - (i)

Again in $△$ABD, $\angle D={90}^{\circ }$

$\therefore A{B}^2=B{D}^2+A{D}^2$

$⟹36=x^2+(5-y{)}^2$ - - - - - (ii)

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

$36=25-y^2+(5-y{)}^2$

$⟹y=\frac{7}{5}$ cm

Putting $y=\frac{7}{5}$ in equation (i),

$25=x^2+\frac{49}{25}$

$⟹x=4.8cm$

Now, $BC=2\times x=2\times 4.8=9.6$ cm