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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