Find the radius of the circle inscribed in triangle ABC, having sides 10 cm,…

Concept: For any triangle with sides a, b, c, the semi-perimeter is s = (a + b + c) / 2, and by Heron's formula the area is Area = √(s(s − a)(s − b)(s − c)). The radius of the circle inscribed in the triangle (the inradius) relates to the area and semi-perimeter by r = Area / s.

Application:

1. The sides are a = 10 cm, b = 10 cm, c = 16 cm, so the semi-perimeter is s = (10 + 10 + 16) / 2 = 18 cm.

2. Compute s − a = 18 − 10 = 8, s − b = 18 − 10 = 8, and s − c = 18 − 16 = 2.

3. By Heron's formula, Area = √(18 × 8 × 8 × 2) = √2304 = 48 cm2.

4. The inradius is r = Area / s = 48 / 18 = 8/3 ≈ 2.667 cm; the offered value of 2.66 cm is this figure truncated to two decimals.

Cross-check:

Since the triangle is isosceles (two sides of 10 cm), drop the altitude from the apex to the 16 cm base; it splits the base into two 8 cm segments. The altitude is h = √(102 − 82) = √(100 − 64) = √36 = 6 cm. So Area = (1/2) × 16 × 6 = 48 cm2, which matches the Heron's-formula result and confirms r = 8/3 ≈ 2.667 cm (the option's 2.66 cm being this value truncated to two decimals).