Question

The radii of two concentric circles are 13 cm and 8 cm respectively. ab is a diameter of the bigger circle. bd is a tangent to the smaller circle touching it at

d. find the length ad.

Answer 1

Correction in question - Since length of AD that is radius os already given, I'm assuming that I have to find length of tangent BD.


$\text{Given}\\ AB = 13 cm \\ AD = 8 cm \\ \\ \text{To find} ~:~ \text{Length of BD} \\ \text{ we know that the angle formed between the radius and tangent of a cirlce is} ~90^{\circ}~~\\ \implies \angle ADB = 90^{\circ} \\ \implies \triangle \text{ADB is an right angled triangle.}\\ \text{We can find side BD by Pythagorean theorem}\\ \implies BD^2 = AB^2 - AD^2 \\ \implies BD^2 = 13^2 - 8^2 \implies BD^2 = 169-64\\ \implies BD^2 = 105 \\ \implies \sqrt{BD^2} = \sqrt{105}\\ \implies BD = 10.24 cm$

Answer 2

Let BD produced meet the larger circle at E.

Join OD and AE.

So, $\sf{OD\:\perp\:BDE}$

and $\sf{\angle{AEB}=90°}$

$\longrightarrow$$\sf{\angle{ODB}=\angle{AEB}=90°}$

But these corresponding angles

$\therefore$ $\sf{OD||AE}$

Now, in ∆BEA,AE = 2OD ....(mid - point theorem)

= $\sf{2 × 8 = 16cm}$

In right ∆ODB,

$\longrightarrow$$\sf{ {BD}^{2} = {OB}^{2} - {OD}^{2} }$

$\longrightarrow$$\sf{ {(13)}^{2} - {(8)}^{2} }$

$\longrightarrow$$\sf{169-64=105}$

$\longrightarrow$ $\sf\red{BD=√105}$

$\longrightarrow$$\sf\red{DE=√105}$

Now, in right ∆AED,

$\longrightarrow$$\sf{ {AD}^{2} = {AE}^{2} - {ED}^{2} }$

$\longrightarrow$$\sf{{AD}^{2} = {(16)}^{2} + {( \sqrt{105}) }^{2}}$

$\longrightarrow$$\sf{{AD}^{2} = 256 + 105}$

$\longrightarrow$$\sf{ {AD}^{2} = 361}$

$\longrightarrow$$\sf{AD = \sqrt{361} = 19cm}$