Question : A circle is inscribed in a triangle ABC. It touches sides AB, BC, and AC at points R, P, and Q, respectively. If AQ = 2.6 cm, PC = 2.7 cm, and BR = 3 cm, then the perimeter (in cm) of the triangle $\triangle \mathrm{ABC}$ is:
Option 1: 16.6
Option 2: 33.2
Option 3: 30
Option 4: 28
Correct Answer: 16.6
Solution :
The circle is inscribed in a triangle ABC. It touches sides AB, BC, and AC at points R, P, and Q, respectively.
AQ = 2.6 cm, PC = 2.7 cm, BR = 3 cm
Concept Used:
The lengths of two tangents drawn on a circle from an outside point are the same.
Calculation:
AQ and AR are two tangents on the circle from point A
⇒ AQ = AR = 2.6 cm
BP and BR are two tangents on the circle from point B
⇒ BR = BP = 3 cm
CP and CQ are two tangents on the circle from point C
⇒ CP = CQ = 2.7 cm
The perimeter of the triangle is AB + BC + CA
= AR + RB + BP + PC + CQ + QA
= 2.6 + 3 + 3 + 2.7 + 2.7 + 2.6
= 16.6
$\therefore$ The perimeter of the triangle is 16.6 cm.
Hence, the correct answer is 16.6.
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