In Fig. the perimeter of ΔABC isA. 30 cm B. 60 cm C. 45 cm D. 15 c

1.	In Fig. the perimeter of ΔABC isA. 30 cm B. 60 cm C. 45 cm D. 15 cm
Answer» Answer is A. 30 cm 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

In Fig. the perimeter of ΔABC isA. 30 cm B. 60 cm C. 45 cm D. 15 cm

Answer»

Answer is A. 30 cm

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

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