Question

In ∆ABC, angleB=90. A circle touches all the sides of ∆ABC. If AB=16, BC=30, then find the radius of the circle.
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Answer 1

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Given:

AB = 5 cm, BC = 12 cm

Using Pythagoras theorem,

AC²=AB²+BC²

= 5²+12²

= 25+144

= 169

AC=13.

We know that two tangents drawn to a circle from the same point that is exterior to the circle are of equal lengths.

So, AM=AQ=a

Similarly MB=BP=b and PC=CQ=c

We know

AB=a+b=5

BC=b+c=12 and

AC=a+c=13

Solving simultaneously we get a=3,b=2 and c=10

We also know that the tangent is perpendicular to the radius

Thus OMBP is a square with side b.

Hence the length of the radius of the circle inscribed in the right angled triangle is 2cm.

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Answer 2

ANSWER: (AC)^2=(AB)^2 + (BC)^2