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If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r = (a + b - c)/2

Solution:

Given, a, b and c are the sides of a right triangle where c is the hypotenuse.

The radius of the circle is r

We have to prove that r = (a + b - c)/2

Let ABC be a right triangle.

The sides AB, BC and AC are c, a and b respectively.

The centre of circle is O and the radius is r

Let the circle touch BC at M, AC at N and AB at P.

We know that the radius of a circle is perpendicular to the tangent at the point of contact.

So, OM ⟂ CB and ON ⟂ AC

If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r = (a + b - c)/2

From the figure,

We observe that OMCN is a square.

OM = CM = CN = NO = r (radius of the circle)

We know that tangents through an external point to a circle are equal.

So, CM = CN

BM = BP

AN = AP

Considering AN = AP

We know that AP = AB - BP

Also, AN = AC - CN

So, AC - CN = AB - BP

b - r = c - BM

From the figure,

BM = BC - CM

BM = a - r

So, b - r = c - (a - r)

b - r = c - a + r

By grouping,

b = c - a + r + r

b = c - a + 2r

2r = b - c + a

r = (b - c + a)/2

Therefore, it is proved that r = (b - c + a)/2

✦ Try This: AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10

NCERT Exemplar Class 10 Maths Exercise 9.3 Sample Problem 2

If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r = (a + b - c)/2

Summary:

If a, b, c are the sides of a right triangle where c is the hypotenuse, it is proven that the radius r of the circle which touches the sides of the triangle is given by r = (a+b-c)/2

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