Question

A circle is drawn touching the sides of ∆ ABC at P,Q,R if AP+BC=13cm,

then find perimeter of ∆ ABC.​

Answer 1

Answer:

In △ABC

AB, BC and AC touches the circle at the point P, Q and R respectively.

∴ BC+AP=AC+PB

⟹ 13+7=AC+PB

⟹ AC=20−PB

Perimeter of △ABC=AB+BC+AC

=AP+PB+13+20−PB

=7+13+20=40

∴ Perimeter of △ABC is 40 cm.