Mathematics
In the given figure, a circle touches the side BC of ΔABC at P and AB and AC produced at Q and R respectively. If AQ = 15 cm, find the perimeter of ΔABC.
Answer
We know that,
Tangents from an exterior point to a circle are equal in length.
From A,
⇒ AQ = AR ….(1)
From B,
⇒ BQ = BP ….(2)
From C,
⇒ CP = CR …..(3)
Perimeter of triangle ABC = AB + BC + CA
= AB + (BP + PC) + (AR - CR)
= (AB + BP) + PC + (AQ - CP) …[From equation (1) and (3)]
= (AB + BQ) + PC + (AQ - CP) …[From equation (2)]
= AQ + PC + AQ - PC
= 2AQ
= 2(15)
= 30 cm.
Hence, perimeter of triangle ABC = 30 cm.
Related Questions
In the given figure, quadrilateral ABCD is circumscribed. The circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively. If AP = 9 cm, BP = 7 cm, CQ = 5 cm and DR = 6 cm, find the perimeter of quadrilateral ABCD.
In the given figure, the circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively. If AB = 11 cm, BC = x cm, CR = 4 cm and AS = 6 cm, find the value of x.
In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 40°, find ∠AQB and ∠AMB.
In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50°, find :
(i) ∠AOB
(ii) ∠OAB
(iii) ∠ACB
