Question

AB is a chord of length 24cm of circle of radius 13cm the tangents at A&B intersect at C.Find the length of AC

Answer 1

the length of AC= 31.2

Answer 2

Answer:

The length of $AC=31.2cm$.

Step-by-step explanation:

Let 0 be the center of the circle and OB and DC intersect at D.

Given,

Chord $AB = 24 cm$,

Radius $OB - OA - 13 cm$.

To find the length of AC.

Step 1

Draw $O P \perp A B$

$&\text { In } \triangle O P B, O P \perp A B$

$&\Rightarrow A P=P B\end{aligned}$

[Perpendicular from the center of chord bisect the chord]

$=\frac{1}{2} A B=12$

then, $O B^{2}=O P^{2}+P B^{2}$

Substituting,

$\Rightarrow(13)^{2}=O P^{2}+P B^{2}$

$\Rightarrow 169=O P^{2}+(12)^{2}$

$\Rightarrow O P^{2}=169-144=25$

$\Rightarrow O P=5 \mathrm{~cm}$

Step 2

In $\triangle B P C, B C^{2}=x^{2}+B P^{2}$[ By Pythagoras theorem]

$B C^{2}=x^{2}+144$

In $\Delta O B C, O C^{2}=O B^{2}+B C^{2}$

$\Rightarrow(x+5)^{2}=(13)^{2}+B C^{2}$

$\Rightarrow x=\frac{288}{10}=28.8 \mathrm{~cm}$

Step 3

Put the value of x in (i), and we get

$B C^{2}=x^{2}+144$

$=\frac{(144)^{2}}{25}+144$

Equating, we get

$\Rightarrow B C=31.2$

$\Rightarrow A C=B C=31.2 \mathrm{~cm}$

Therefore, the length of $AC=31.2cm$.