Question 51

A circle is inscribed in a triangle ABC. It touches the sides AB. BC and AC at the points R. P and Q respectively. If AQ = 4.5 cm. PC = 5.5 cm and BR = 6 cm then the perimeter of the triangle ABC is:

Solution

We know that length of the tangents drawn from an external point are equal.

Here if external point is A , AQ = AR => AR = 4.5 cm

If external point is C , PC =CQ, CA =5.5 cm

If external point is B , PB =BR, PB =6 cm

Perimeter of $$\triangle ABC$$ = 4.5 + 4.5 + 6 + 6 + 5.5 + 5.5 = 32 cm

So, the answer would be option c)32 cm.

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