A is point at a distance 26 cm from the centre O of a circle of radius 10 cm. AP and AQ are the tangents to the circle at the point of contacts 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, then the perimeter of $$\triangle$$ABC is:

Given, A is point at a distance 26 cm from the centre O

Radius of the circle = 10 cm

AP and AQ are tangents to the circle at the point of contacts P and Q

$$\Rightarrow$$ AP$$\bot\ $$OP and AQ$$\bot\ $$OQ

The length of tangents to the circle from an external point are equal.

$$\Rightarrow$$  AP = AQ .................(1)

Similarly, BP = BR and CQ = CR .......(2)

In $$\triangle$$OPA,

OP$$^2$$ + AP$$^2$$ = AO$$^2$$

$$\Rightarrow$$  10$$^2$$ + AP$$^2$$ = 26$$^2$$

$$\Rightarrow$$  100 + AP$$^2$$ = 676

$$\Rightarrow$$  AP$$^2$$ = 576

$$\Rightarrow$$  AP = 24 cm

Perimeter of $$\triangle$$ABC = AB + BC + AC

= AB + BR + CR + AC

= AB + BP + CQ + AC [From (2)]

= AP + AQ

= AP + AP [From (1)]

= 2AP

= 2(24)

= 48 cm

Hence, the correct answer is Option A