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In Fig. 5, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents drawn at P and Q intersect at T. Find the length of TP.

Answers (1)

Given :

$PQ=8\; cm$

$OQ=OP=\text{radius}=5\; cm$

Solution :

$TP=TQ$ (Tangents from external point have equal length)

$\Rightarrow \bigtriangleup TPQ$ is an isosceles triangle because $TP=TQ$

$\Rightarrow$ OT is the bisector of $\angle PTQ.$ (Angle bisector and altitude are same in the isosceles triangle)

Hence $OT \perp PQ$

$\Rightarrow$ Since $OT \perp PQ$ , hence $PR=QR$

Also $PQ=PR+QR=2PR$

$\Rightarrow PR=\frac{PQ}{2}=\frac{8}{2}=4\; cm$

In a right-angled triangle $\bigtriangleup O\!RP$ ,

$OP^2=PR^2+OR^2$ (Pythagoras theorem)

$\Rightarrow 5^2=4^2+OR^2$

$\Rightarrow O\!R^2=5^2-4^2$

$\Rightarrow O\!R=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$

$\Rightarrow O\!R=3\;cm$

Let PT be $x$ .

In right angled $\bigtriangleup P\!RT$ ,

$(PT)^2=(PR)^2+(RT)^2$ (By Pythagoras theorem)

$x^2=4^2+RT^2$

$x^2=16+RT^2\;\;\;\;\;\;\;\;-(i)$

Since TP is a tangent,

$O\!P \perp T\!P$ ( tangent at any point of the circle is perpendicular to the radius through point of contact )

$\therefore \angle OPT = 90^{\circ}$

In right angled $\bigtriangleup O\!PT$

$(OT)^2=(OP)^2+(TP)^2$

$(OT)^2=5^2+x^2$

$\Rightarrow (O\!R+RT)^2=5^2+x^2$

$\Rightarrow (3+RT)^2=5^2+x^2$

$\Rightarrow 3^2+RT^2+6RT=25+x^2$

$\Rightarrow 9+RT^2+6RT=25+x^2$

$\Rightarrow 9+RT^2+6RT=25+RT^2+16 \;\;\; -(\text{from (1) }x^2=16+RT^2)$

$\Rightarrow RT^2-RT^2+6RT=25+16-9$

$\Rightarrow 6RT=32$

$\Rightarrow RT=\frac{32}{6}=\frac{16}{3}$

From (i):

$x^2=16+RT^2$

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

$x=\sqrt{16+ \frac{256}{9} }$

$x= \frac{\sqrt{16 \times 9+256}}{\sqrt{9}}$

$x= \frac{\sqrt{144+256}}{3}$

$x= \frac{\sqrt{400}}{3} =\frac{20}{3}$

Hence,

$PT=x=\frac{20}{3}$

Posted by

Safeer PP

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