Question

In the figure given below, diameter $AB$ and $CD$ of a circle meet at $P.PT$ is a tangent to the circle at $T$. $CD=7.8cm,PD=5cm.PB=4cm$. Find $AB$ (in $cm$)

Answer 1

Since chord $CD$ and tangent at point $T$ intersect each other at $P$,
$\therefore PC\times PD=P{T}^2.....\left(1\right)$
Since chord $AB$ and tangent at point $T$ intersect each other at $P$.
$\therefore PA\times PB=P{T}^2....\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, $PC\times PD=PA\times PB....\left(3\right)$
Given: $CD=7.8cm;PD=5cm,PB=4cm$.
$\therefore PA=PB+AB=4+AB,PC=PD+CD=5+7.3=12.8cm$.
Putting these values in eq. $\left(3\right)$
$12.8\times 5=(4+AB)\times 4$
$\Rightarrow 4+AB=\frac{12.8\times 5}{4}$
$\Rightarrow 4+AB=16$
$\Rightarrow AB=12cm$.
Hence, $AB=12cm$.