Calculating the Radius of an Inscribed Circle in a Right Triangle

The problem asks us to find the radius of the circle inscribed in a right-angled triangle ΔABC. We are given that the triangle is right-angled at B, and the lengths of the two legs are BC = 15 cm and AB = 8 cm.

Understanding the Inscribed Circle and Inradius

An inscribed circle (or incircle) of a triangle is the largest circle that can be contained inside the triangle. It is tangent to all three sides of the triangle. The center of the inscribed circle is called the incenter, which is the intersection point of the angle bisectors of the triangle. The radius of the inscribed circle is called the inradius.

Steps to Find the Inradius

Step 1: Identify the Sides of the Right Triangle

In ΔABC, right-angled at B:

• The two legs are AB and BC. We are given AB = 8 cm and BC = 15 cm.
• The hypotenuse is the side opposite the right angle, which is AC.

Step 2: Calculate the Length of the Hypotenuse

Since ΔABC is a right-angled triangle, we can use the Pythagorean theorem to find the length of the hypotenuse AC.

According to the Pythagorean theorem, \((\text{Hypotenuse})^2 = (\text{Leg 1})^2 + (\text{Leg 2})^2\).

So, \(AC^2 = AB^2 + BC^2\)

Substituting the given values:

\(AC^2 = (8 \, \text{cm})^2 + (15 \, \text{cm})^2\)

\(AC^2 = 64 \, \text{cm}^2 + 225 \, \text{cm}^2\)

\(AC^2 = 289 \, \text{cm}^2\)

To find AC, we take the square root of 289:

\(AC = \sqrt{289} \, \text{cm}\)

\(AC = 17 \, \text{cm}\)

So, the length of the hypotenuse AC is 17 cm.

Step 3: Use the Formula for Inradius of a Right Triangle

For a right-angled triangle with legs of lengths 'a' and 'b' and hypotenuse of length 'c', the radius of the inscribed circle (inradius, denoted by 'r') can be calculated using a specific formula:

\(r = \frac{\text{Sum of the two legs} - \text{Hypotenuse}}{2}\)

In terms of the side lengths of ΔABC, where the right angle is at B, the legs are AB and BC, and the hypotenuse is AC, the formula is:

\(r = \frac{AB + BC - AC}{2}\)

Step 4: Calculate the Inradius

Substitute the values we know into the formula:

\(r = \frac{8 \, \text{cm} + 15 \, \text{cm} - 17 \, \text{cm}}{2}\)

\(r = \frac{23 \, \text{cm} - 17 \, \text{cm}}{2}\)

\(r = \frac{6 \, \text{cm}}{2}\)

\(r = 3 \, \text{cm}\)

Alternatively, the inradius can be calculated using the formula \(r = \frac{\text{Area}}{\text{Semi-perimeter}}\).

Area of ΔABC = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 15 \, \text{cm} \times 8 \, \text{cm} = \frac{1}{2} \times 120 \, \text{cm}^2 = 60 \, \text{cm}^2\)

Semi-perimeter (s) = \(\frac{AB + BC + AC}{2} = \frac{8 \, \text{cm} + 15 \, \text{cm} + 17 \, \text{cm}}{2} = \frac{40 \, \text{cm}}{2} = 20 \, \text{cm}\)

Inradius \(r = \frac{\text{Area}}{s} = \frac{60 \, \text{cm}^2}{20 \, \text{cm}} = 3 \, \text{cm}\)

Both methods yield the same result. The radius of the inscribed circle in ΔABC is 3 cm.

Revision Table: Key Formulas

Additional Information: Properties of Inscribed Circles and Incenters

• The incenter is equidistant from all three sides of the triangle. This distance is the inradius (r).
• The incenter is the point where the angle bisectors of the triangle intersect.
• Every triangle has exactly one incircle and one incenter.
• The formula \(r = \frac{\text{Area}}{s}\) is applicable to all types of triangles, not just right triangles. The formula \(r = \frac{a+b-c}{2}\) is a special case applicable only to right-angled triangles where a and b are the lengths of the legs and c is the length of the hypotenuse.