Question

In figure $\mathrm{PQ},\mathrm{PR}$ and $\mathrm{BC}$ are the tangents to the circle. $\mathrm{BC}$ touches the circle at $X$. If $\mathrm{PQ}=7\mathrm{cm}$ find the perimeter of $∆\mathrm{PBC}$.

Answer 1

Find the perimeter of $∆\mathrm{BPC}.$

Given: $\mathrm{PQ}=7\mathrm{cm}$

Since tangents drawn from external points are equal.

$$\begin{gathered} \therefore \mathrm{PQ}=\mathrm{PR}(\mathrm{Tangents}\mathrm{from}\mathrm{point}\mathrm{P}) \\ \mathrm{BQ}=\mathrm{BX}(\mathrm{Tangents}\mathrm{from}\mathrm{point}\mathrm{B}) \\ \mathrm{CX}=\mathrm{CR}(\mathrm{Tangents}\mathrm{from}\mathrm{point}\mathrm{C}) \end{gathered}$$

Perimeter of $∆\mathrm{PBC}$,

$$\begin{gathered} \mathrm{Perimeter}=\mathrm{PB}+\mathrm{BC}+\mathrm{PC} \\ =\mathrm{PB}+(\mathrm{BX}+\mathrm{XC})+\mathrm{PC}(\because \mathrm{BC}=\mathrm{BX}+\mathrm{XC}) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\mathrm{PB}+\mathrm{BQ}+\mathrm{CR}+\mathrm{PC}(\because \mathrm{BX}=\mathrm{BQ},\mathrm{XC}=\mathrm{CR}) \\ =\mathrm{PQ}+\mathrm{PC}(\because \mathrm{PB}+\mathrm{BQ}=\mathrm{PQ},\mathrm{PC}+\mathrm{CR}=\mathrm{PR}) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=7+7(\mathrm{PQ}=7\mathrm{cm}) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=14\mathrm{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Hence, the perimeter of $∆\mathrm{BPC}$ is $14\mathrm{cm}.$