Question

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

Answer 1

In the given circle,

$PQ=PR$ (Tangents from the external point $P$)

$XB=BQ$ (Tangents from the external point $B$)

$XC=CR$ (Tangents from the external point $C$)

Perimeter of $△PCB$ is $PC+CB+PB$ but $CB=CX+XB$,

Therefore, perimeter of $△PCB$ is $PC+CX+XB+PB$ but $CX=CR$ and $XB=BQ$

Therefore, perimeter of $△PCB$ is $PC+CR+PB+BQ$ but $PC+CR=PR$ and $PB+BQ=PQ$

Therefore, perimeter of $△PCB$ is $PR+PQ$ but $PQ=PR=7$ cm [As length of the tangents drawn from an external point are equal]

Hence, perimeter of $△PCB$ is $7+7=14$ cm.