Question

Circle with centre o is inscribed in triangle pqr touches the sides pq qr and pr

Answer 1

Answer:

x

Step-by-step explanation:

yz

Answer 2

Answer:

Step-by-step explanation:

Since lenth from an exterior point to a circle are equal.

Therefore PX=PZ.....1

                QY=QX......2

                RZ=RY........3

Adding equation 1,2 and 3 ,we get

PX+QY+RZ = PZ+QX+RY

Now,

      Perimeter of triangle PQR =PQ+PR+QR

= (PX+XQ)+(QY+YR)+(PZ+ZR)

= (PX+PZ)+(QX+QY)+(RZ+RY)

= 2PX+2QY+2RZ

= 2(PX+QY+RZ)

Therefore,PX+QY+RZ = XQ+YR+ZP =2/1​(perimeter of triangle PQR)