Question

The radii of two concentric circles are 13 cm and 8 cm respectively ab is the diameter of the bigger circle and bd is a tangent to the smaller circle touching it hd and intersecting the large circle it on producing find the length of ap

Answer 1

check the attachment

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}$