What is the length of the altitude of the triangle drawn from vertex A on BC?

Consider the following for the next three (03) items that follow: ABC is a triangle with sides AB=6\text{ cm}, BC=10\text{ cm} and CA=8\text{ cm}. With vertices A, B and C as centres, three circles are drawn each touching the other two externally.

  1. A. 2.4 cm
  2. B. 3 cm
  3. C. 4 cm
  4. D. 4.8 cm

Correct Answer: D. 4.8 cm

Explanation

The triangle has sides 6, 8, 10, satisfying 6^2+8^2=10^2, so it is a right-angled triangle at A. The area is \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2. The area can also be expressed as \frac{1}{2} \times BC \times h = \frac{1}{2} \times 10 \times h. Thus, 5h = 24 \implies h = 4.8\text{ cm}.

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