"
">

In Fig., $PQ$ is a chord of length $8\ cm$ of a circle of radius $5\ cm$ and center $O$. The tangents at $P$ And $Q$ intersect at point $T$. Find the length of $TP$.

"

Academic Mathematics NCERT Class 10

Given: Length of the chord, $PQ=8\ cm$ and the radius of the circle, $OP=5\ cm$.

To do: To find the length of $TP$.

Solution:

Given radius, $OP=OQ=5\ cm$

Length of chord,

$PQ=8\ cm $

$OT \perp PQ$.

$\because$ Perpendicular drawn from the center of the circle to a chord bisect the chord.

$\therefore PM=MQ=4\ cm$

In right $\vartriangle OPM$,

$(OP)^{2}=(PM)^{2} +(OM)^{2}$

$\Rightarrow(5)^{2}=(4)^{2}+(OM)^{2}$

$\Rightarrow ( OM)^{2}=25-16=9$

$\Rightarrow OM=\sqrt{9}$

$\Rightarrow OM=3\ cm$

$\angle OPT=90^{o} $ [Radius is perpendicular to tangent at point of contact ]

In right $\vartriangle OPT$ ,

$( OT)^{2}=( PT)^{2}+( OP)^{2}$ ....................$( 1)$

In right $\vartriangle PTM$,

$( PT)^{2}=( TM)^{2} +( PM)^{2}$ ......................... $( 1)$

From equations $( 1)$ and $( 2)$,

$( OT)^{2}=( PT)^{2}+( OP)^{2}=( TM)^{2} +( PM)^{2} +( OP)^{2}$

$\Rightarrow ( TM+OM)^{2}=( TM)^{2} +( PM)^{2} +( OP)^{2}$

$\Rightarrow ( TM)^{2}+( OM)^{2}+2\times TM\times OM=( TM)^{2} +( PM)^{2} +( OP)^{2} $

$\Rightarrow (OM)^{2}+2\times TM\times OM=(PM)^{2} +(OP)^{2} $

$\Rightarrow (3)^{2}+2\times 3\times TM=(4)^{2}+(5)^{2}$

$\Rightarrow 9+6TM=16+25=41$

$\Rightarrow 6TM=41-9=32$

$\Rightarrow TM=\frac{32}{6}=\frac{16}{3}\ cm$

On subtituting $TM=\frac{16}{3}\ cm$ in equation $( 2)$,

$( PT)^{2}=( TM)^{2} +( PM)^{2}$

$\Rightarrow ( PT)^{2}=(\frac{16}{3})^{2} +(4)^{2}$

$\Rightarrow ( PT)^{2}=\frac{256}{9}+16$

$\Rightarrow ( PT)^{2}=\frac{256+144}{9}$

$\Rightarrow ( PT)^{2}=\frac{400}{9}$

$\Rightarrow PT=\sqrt{\frac{400}{9}}$

$\Rightarrow PT=\frac{20}{3} cm$

Therefore $PT=\frac{20}{3}\ cm$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

85 Views

Kickstart Your Career

Get certified by completing the course

Get Started

Advertisements