A circle is inscribed in a right-angled triangle ABC, right-angled at B.

Q) A circle is inscribed in a right-angled triangle ABC, right-angled at B. If BC = 7 cm and AB = 24 cm, find the radius of the circle

Ans: Let’s first draw a diagram for the given question:

By Pythagorus theorem,

AC = $\sqrt{AB^2 + BC^2}$

= $\sqrt {(24)^2 + (7)^2}$

= $\sqrt {576 + 49} = \sqrt {625}$

= 25 cm

Method 1:

OP = OQ (radius of a circle)

BP = BQ (tangents from external point)

Since OPBQ is a square, therefore, BP = OQ = r

and BQ = OP = r

Now ∵ BQ = r ∴ CQ = 7 – r

∵ CQ = CR (tangents from external point)

∴ CR = 7 – r …. (i)

Similarly, ∵ BP = r, ∴ AP = 24 – r

∵ AP = AR (tangents from external point)

∴ AR = 24 – r …… (ii)

Since, AC = AR + CR

∴ 25 = (24 – r) + (7 – r)

∴ 25 = 31 – 2 r

∴ 2 r = 31 – 25 = 6

∴ r = 3 cm

Method 2: The radius of in-circle of a right angled triangle is given by:

r = $\frac {p + b - h}{2}$

= $\frac{24 + 7 - 25}{2}$

= $\frac{6}{2}$

= 3 cm

Note: Method 1 is for CBSE exams. Method 2 is for quick calculations in competitive exams.

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