In a right-angled triangle ABC, AB=15 cm, BC=20 cm and AC=25 cm. Further, BP is the perpendicular on AC. What is the difference in the area of triangles PAB and PCB?

Explanation

In right \Delta ABC, BP = \frac{AB \times BC}{AC} = \frac{15 \times 20}{25} = 12 cm. Also, AP = \frac{AB^2}{AC} = 9 cm and PC = \frac{BC^2}{AC} = 16 cm. The difference in their bases is 16 - 9 = 7 cm. The difference in area is \frac{1}{2} \times \text{base\_diff} \times \text{height} = \frac{1}{2} \times 7 \times 12 = 42 sq cm.

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