ABC is a triangle with sides AB=41 cm, BC=28 cm and CA=15 cm. If D, E and F are the mid-points of AB, BC and CA respectively, then what is the area of the triangle DEF?

Explanation

For \triangle ABC, semi-perimeter s = \frac{41+28+15}{2} = 42. Using Heron's formula, Area = \sqrt{42(42-41)(42-28)(42-15)} = \sqrt{42 \times 1 \times 14 \times 27} = 126 sq cm. The area of the triangle formed by the midpoints is \frac{1}{4} of the original triangle, which is \frac{126}{4} = 31.5 sq cm.

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