question_answer

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If\[PA=12\text{ }cm,\text{ }QC=QD=3\text{ }cm\], then find \[PC+PD\].

Answer:

Given,   \[PA=PB=12\text{ }cm\]

[Tangent from an external point]

\[AC=CQ=3\text{ }cm\]

\[BD=QD=3\text{ }cm\]

[Tangent from external point]

So, \[PC+PD\]

\[=(PA-AC)+(PB-BD)\]

\[=(12-3)+(12-3)\]

\[=9+9=18\text{ }cm\]