Question

Pq is a chord of length 16 cm of a circle of radius 10cm. the tangents at p and q intersect at apoint t. find length of tp

Answer 1

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Answer 2

Answer:

From the figure,

PQ = 16 cm ,

QR = RP = PQ/2 = 16/2 = 8 cm

i ) In ∆ORQ , <ORQ = 90°

/* By Phythagorean theorem,

OQ²= OR² + QR²

=> 10² =OR² + 8²

=> 100 - 64 = OR²

=> OR = √36 = 6 cm

ii ) ∆ORQ ~ ∆OQT

$\frac{10}{6+x}=\frac{6}{10}$

$\implies 100 =6(6+x)$

$\implies 100 - 36 = 6x$

$\implies 64 = 6x$

$\implies \frac{64}{6}= x$

$\implies x = \frac{32}{3}$

$iii ) In ∆OQT , <OQT = 90°,</p><p><strong>TQ²</strong><strong> </strong><strong>=</strong><strong> </strong><strong>OT²</strong><strong> </strong><strong>-</strong><strong> </strong><strong>OQ²</strong></p><p>[tex]=\left(6+\frac{32}{3}\right)^{2} - 10^{2}$

$= \left(\frac{18+32}{3}\right)^{2} - 10^{2}$

$= \left(\frac{50}{3}\right)^{2} - 100$

$= \frac{2500}{9}- 100$

$= \frac{2500-900}{9}$

$= \frac{1600}{9}$

$\implies TQ = \frac{40}{3}\:cm$

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