A circle is inscribed in a ^@ \triangle ABC ^@, touching ^@ BC, CA, ^@ and ^@ AB ^@ at points ^@ P, Q, ^@ and ^@ R ^@ respectively. If ^@ AB = 14 \space cm, AQ = 11 \space cm ^@ and ^@ CQ = 9 \space cm ^@ then find the length of ^@ BC ^@.
Answer:
^@ 12 \space cm ^@
- We know that the lengths of tangents drawn from an external point to a circle are equal.
Thus, @^ \begin{aligned} & AR = AQ = 11 \space cm \\ & BP = BR \\ \text{ and } & CP = CQ = 9 \space cm \end{aligned} @^ - We see that ^@ BR = AB - AR = 14 \space cm - 11 \space cm = 3 \space cm. ^@
Thus, ^@ BP = BR = 3 \space cm. ^@ - Therefore, ^@ BC = BP + CP = 3 \space cm + 9 \space cm = 12 \space cm. ^@