Question 89
A circle is inscribed in $$\triangle ABC$$ , touching AB, BC and AC at the points P, Q and respectively. If AB - BC = 4 cm, AB - AC = 2 cm and the perimeter of $$\triangle ABC$$ = 32 cm, then PB + AR is equal to:
Solution
Perimeter = 32 cm
AB + BC + AC = 32 cm ---(1)
AB - BC = 4 cm ---(2)
AB - AC = 2 cm ---(3)
On eq(1) + (2) + (3),
3AB = 38
AB = PB + AR = 38/3
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