In a triangle ABC, AB=21 cm, BC=20 cm and CA=13 cm. A perpendicular CD is drawn upon the longest side. What is the area of the triangle BCD?

Explanation

The longest side is AB=21. Semi-perimeter s = \frac{21+20+13}{2} = 27. Using Heron's formula, Area of \triangle ABC = \sqrt{27(6)(7)(14)} = 126. The height CD on base AB is \frac{2 \times 126}{21} = 12. In right \triangle BCD, hypotenuse BC=20, so base BD = \sqrt{20^2 - 12^2} = 16. The area of \triangle BCD = \frac{1}{2} \times BD \times CD = \frac{1}{2} \times 16 \times 12 = 96 square cm.

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