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In fig.2, PQ is a chord of length 8cm of a circle of radius 5cm and centre O.

The tangents at P and Q intersect at point T. Find the length of TP.

Answers (1)

In $\triangle TPQ,$

TP=TQ

So, $\triangle TPQ,$ is an isoceles triangle.

Here, OT is an angle bisector of $\angle QTP$ ,

So $OT\perp PQ$ (altitudes of isoceles triangle are same)

$OT\perp PQ$

So, PM=MQ

PM=MQ=1/2(PQ)=8/2=4cm.

In right angle triangle OMP,

$H^2=L^2+B^2$

$OP^2=PM^2+OM^2$

$5^2=4^2+OM^2$

$25=16+OM^2$

$OM^2=9$

$OM=3$

Now, in $\triangle PMT,$

$TP^2=TM^2+PM^2$

$TP^2=16+PM^2$ ...(1)

Since TP is a tangent

$\angle OPT =30$

In right angle triangle OPT,

$OT^2=5^2+TP^2$

$(TM+MO)^2=25+TP^2$

$(TM+3)^2=25+TP^2$

$3^2+TM^2+2(3)TM=25+TP^2$

Putting value of TP from eqn(1)

$9+TM^2+6TM=25+16+TM^2$

Putting value of TM in eqn(1)

$TP^2=16+\left ( \frac{16}{3} \right )^{2}$

$TP^2=16+\frac{256}{9}$

$TP^2=\frac{400}{9}$

$TP=\frac{20}{3}$

Hence the length of $TP=\frac{20}{3}$ .

Posted by

Safeer PP

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