In the given figure, △ABC is right-angled at B, in which AB = 15 cm and BC = 8 cm. A circle with center O has been inscribed in △ABC. Calculate the value of x, the radius of the inscribed circle. [4 MARKS]
Concept: 1 Mark
Application: 3 Marks
Let the inscribed circle touch the sides AB, BC and CA at P, Q and R respectively.
Applying Pythagoras theorem on right △ABC, we have
AC2=AB2+BC2=(15)2+(8)2=(225+64)=289
⇒AC=√289=17cm.
Clearly, OPBQ is a square.
[∠OPB=90∘,∠PBQ=90∘,∠OQB=90∘ and OP=OQ=xcm]
BP=BQ=xcm.
Since the tangents to a circle from an exterior point are equal in length,
we have AR = AP and CR = CQ.
Now, AR = AP = (AB - BP) = (15 – x) cm
CR = CQ = (BC – BQ) = (8 - x) cm.
AC=AR+CR⇒17=(15–x)+(8–x)⇒2x=6⇒x=3.
Hence, the radius of the inscribed circle is 3 cm.