Short Note

*A *is a point at a distance 13 cm from the centre* O* of a circle of radius 5 cm . *AP* and *AQ *are the tangents to the circle at *P* and *Q* . If a tangent* BC* is drawn at a point *R* lying on the minor arc *PQ* to intersect *AP* at *B * and *AQ* at *C* , find the perimeter of the \[∆\]*ABC* .

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#### Solution

A is a point 13 cm from centre O. AP and AQ are the tangents to the circle with centre O.

AP = AQ

In ∆APO,

\[{AP}^2 + {OP}^2 = {AO}^2 \]

\[ \Rightarrow {AP}^2 = {AO}^2 - {OP}^2 \]

\[ \Rightarrow {AP}^2 = {13}^2 - 5^2 \]

\[ \Rightarrow {AP}^2 = 169 - 25 = 144\]

\[ \Rightarrow AP = 12 cm\]

Now In ∆ABC,

Perimeter = AB + BC + AC

= AB + BR + RC + AC

= AB + BP + CQ + AC ( BR = BP, RC = CQ)

= AP + AQ

= 12 + 12

= 24 cm

Is there an error in this question or solution?

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