Question

AB is a chord of length 24 cm of
circle of radius 13 cm The tangents
of A and B intersect at c. find length Ac.

Answer 1

Answer:

Let O the center of the circle and AB and OC intersect at D.

Since OC is the perpendicular bisector of AB.

$\therefore \: ab = bd = \frac{24}{2} \\ \\ = 12cm$

In right triangle OAP

${oa}^{2} = {ad}^{2} + {od}^{2} \\ \\ = {od}^{2} = {oa}^{2} - {ad}^{2} \\ \\ = {(13cm)}^{2} - {(12cm)}^{2} \\ \\ = 169 {cm}^{2} - {144cm}^{2} = {25cm}^{2}$

$= od \: = \sqrt{25 {cm}^{2} } \\ \\ = 5cm$

In triangle OAC and ADC

${ac}^{2} + {ad}^{2} = {dc}^{2} \\ \\ {oc}^{2} + {oa}^{2} = {ac}^{2} \\ \\ = {oc}^{2} + {oa}^{2} = ( {ad}^{2} + {dc}^{2}) \\ \\ = (cd + od) ^{2} = {13cm}^{2} + {(12cm)}^{2} + {y}^{2} \\ \\ = {y}^{2} + 10y \: cm + 25cm^{2} = {313cm}^{2} + {y}^{2} \\ \\ = 10y = (313 - 25)cm = 288cm \\ \\ y = \frac{288}{10} cm \\ \\ = \frac{144}{5}cm$

CD=28.2cm^2

$= {(12cm)}^{2} + ({ \frac{144}{5})}cm^{2} \\ \\ = {144cm}^{2} + \frac{20736}{25} {cm}^{2} \\ \\ = \frac{3600 {cm}^{2} + 20736 {cm}^{2} }{25} \\ \\ = \frac{24336}{5} {cm}^{2} \\ \\ = \frac{156}{5} {cm} \\ \\ = 31.2cm$

So 31.2 cm is the answer

Hope it helps you from my side