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(B) Given,$OT = 13 \, cm$ and radius $OP = 5 \, cm$.Since the tangent at any point of a circle is perpendicular to the radius through the point of con

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$\frac{20}{3}$

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(B) Given,$OT = 13 \, cm$ and radius $OP = 5 \, cm$.Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OP \perp PT$.In right-angled $	riangle OPT$,by Pythagoras theorem:$PT^2 = OT^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144$.So,$PT = 12 \, cm$.Since the lengths of tangents drawn from an external point to a circle are equal,$PT = QT = 12 \, cm$.Also,$ET = OT - OE = 13 - 5 = 8 \, cm$.Let $PA = AE = x$. Then $AT = PT - PA = 12 - x$.In right-angled $	riangle AET$,by Pythagoras theorem:$AT^2 = AE^2 + ET^2$$(12 - x)^2 = x^2 + 8^2$$144 - 24x + x^2 = x^2 + 64$$24x = 144 - 64 = 80$$x = \frac{80}{24} = \frac{10}{3} \, cm$.Similarly,$EB = QB = \frac{10}{3} \, cm$.Therefore,$AB = AE + EB = \frac{10}{3} + \frac{10}{3} = \frac{20}{3} \, cm$.

In the figure,$O$ is the centre of a circle of radius $5 \, cm$. $T$ is a point such that $OT = 13 \, cm$ and $OT$ intersects the circle at $E$. If $AB$ is the tangent to the circle at $E$,find the length of $AB$ in $cm$.

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