ABC is a triangle right-angled at C. A semicircle is drawn on AB as diameter. Let P be any point on AC produced such that AP = AB = 10 cm. Further, B and P are joined. If BC = 8 cm, then what is BP equal to ?

Explanation

In right triangle ABC, AB=10 and BC=8, so AC = \sqrt{10^2 - 8^2} = 6. Point P lies on AC produced such that AP=10. This implies CP = AP - AC = 10 - 6 = 4. Triangle BCP is right-angled at C, so BP = \sqrt{BC^2 + CP^2} = \sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5} cm.

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