Question : A circle touches the side BC of a $\triangle A B C$ at P and also touches AB and AC produced at Q and R, respectively. If the perimeter of $\triangle A B C=26.4 \mathrm{~cm}$, then the length of AQ is:
Option 1: 17.6 cm
Option 2: 13.2 cm
Option 3: 15.4 cm
Option 4: 8.8 cm
Correct Answer: 13.2 cm
Solution :
Given: Perimeter of triangle ABC = 26.4 cm
Perimeter of triangle ABC = AB + BP + PC + AC
We know that,
AQ = AR, BP = BQ, CP = CR as they are tangents.
⇒ 26.4 = AB + BQ + CR + AC
⇒ 26.4 = AQ + AR
The lengths of two tangents drawn from an external point to a circle are equal.
Thus, AQ = AR
⇒ 26.4 = AQ + AQ
⇒ 26.4 = 2AQ
⇒ AQ = $\frac{26.4}{2}$
⇒ AQ = 13.2 cm
Hence, the correct answer is 13.2 cm.
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