**Question:**

If AB, AC, PQ are tangents in the given figure and AB = 5 cm, find the perimeter of Δ APQ.

**Solution:**

We have been asked to find the perimeter of the triangle *APQ*.

Therefore,

Perimeter of Δ*APQ* is equal to* AP + AQ + PQ*

By looking at the figure, we can rewrite the above as follows,

Let the Perimeter of Δ*APQ *be *P*. So *P*= *AP + AQ + PX + XQ*

From the property of tangents we know that when two tangents are drawn to a circle from the same external point, the length of the two tangents will be equal. Therefore we have,

*PX =PB*

*XQ =QC*

Replacing these in the above equation we have,

*P *=*AP + AQ + PB + QC*

From the figure we can see that,

*AP + PB = AB*

*AQ + QC = AC*

Therefore, we have, *P*= *AB + AC*

It is given that *AB* = 5 cm.

Again from the same property of tangents we know that that when two tangents are drawn to a circle from the same external point, the length of the two tangents will be equal. Therefore we have,

*AB = AC*

Therefore,

*AC* = 5 cm

Hence,

*P *= 5 + 5=10

Thus the perimeter of triangle *APQ* is 10 cm.