ABC is a triangle right angled at C. Let p be the length of the perpendicular drawn from C on AB. If BC=6~\text{cm} and CA=8~\text{cm}, then what is the value of p?

Explanation

Using Pythagoras theorem, AB = \sqrt{6^2+8^2} = 10~\text{cm}. The area of \Delta ABC = \frac{1}{2} \times BC \times CA = \frac{1}{2} \times 6 \times 8 = 24. Alternatively, Area = \frac{1}{2} \times AB \times p = \frac{1}{2} \times 10 \times p = 5p. Equating both areas: 5p = 24 \implies p = 4.8~\text{cm}.

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