Question

A circle touches the side BC of A ABC at P and touches AB and AC produced at 0 and R respectively as shown in the adjoining figure if AO -5 cm sternal find the perimeter of SABC​

Answer 1

Step-by-step explanation:

hope this helps you sweetheart

Answer 2

Answer:

GIVEN: A circle touching the side BC of AA BC at P and AB, AC produced at Q and R. To Prove: AQ = 1/2 (Perimeter of AA BC)

PROOF:

AQ = AR

B Q = BP

[From A]. (1)

[From B]...(2)

CP = CR. [From C].(3)

Lengths of tangents drawn from an external point to a circle are equal.

Perimeter of AA BC = AB + BC + CA Perimeter of AA BC=AB+ (BP+ PC) + (AR-CR)

Perimeter of AA BC = (AB + B Q) + (PC) + (AQ - PC)

[From EQ 1,2 & 3, AQ = AR, B Q = BP, CP = CR]

Perimeter of AA BC = AQ + AQ = 2AQ

Perimeter of AA BC = 2AQ

AQ = 1/2 (Perimeter of AA BC)

Hence, AQ is half the perimeter of A ABC.

HOPE THIS WILL HELP YOU.