Question

Can sm1 solve this??

Answer 1

Radius= 13cm

AB= 24cm

NM= 17cm

We know,

The perpendicular line from centre bisects the chord.

$am = \frac{1}{2} ab$

$am = 12cm$

In ∆OAM

${h}^{2} = {p}^{2} + {b}^{2}$

${13}^{2} = {p}^{2} + {12}^{2}$

$169 = {p}^{2} + 144$

$169 - 144 = {p}^{2}$

$25 = {p}^{2}$

$\sqrt{25} = p$

$p = 5cm$

OM= 5cm

Now,

ON= NM - OM

= 17 - 5

= 12cm

Construction: Join OC

In ∆ONC

${h}^{2} = {p}^{2} + {b}^{2}$

${13}^{2} = {12}^{2} + {b}^{2}$

$169 = 144 + {b}^{2}$

$169 - 144 = {b}^{2}$

$25 = {b}^{2}$

$\sqrt{25} = b$

$b = 5cm$

CN= 5cm

We know,

Perpendicular line from centre to chord, bisects the chord.

CD= 2CN

$= 2 \times 5$

$= 10cm$

Therefore, the length of chord CD is 10cm