In Fig. 10.80, the perimeter of ΔABC is
Given:
AQ = 4 cm
BR = 6 cm
PC = 5 cm
Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
By the above property,
AR = AQ = 4 cm (tangent from A)
BR = BP = 6 cm (tangent from B)
CP = CQ = 5 cm (tangent from C)
Now,
Perimeter of ∆ABC = AB + BC + CA
⇒ Perimeter of ∆ABC = AR + RB + BP + PC + CQ + QA
[∵ AB = AR + RB
BC = BP + PC
CA = CQ + QA]
⇒ Perimeter of ∆ABC = 4 cm + 6 cm + 6 cm + 5 cm + 5 cm + 4 cm
⇒ Perimeter of ∆ABC = 30 cm
Hence, Perimeter of ∆ABC = 30 cm