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Question 149
ABC is a triangle in which $$\angle$$ A = 90$$^\circ$$. Let P be any point on side AC. If BC = 10 cm, AC = 8 cm and BP = 9 cm, then AP =
Solution
In $$\triangle\ $$ABC,
$$AB^2+AC^2=BC^2$$
$$=$$> Â $$AB^2+8^2=10^2$$
$$=$$> Â $$AB^2+64=100$$
$$=$$> Â $$AB^2=100-64$$
$$=$$> Â $$AB^2=36$$
$$=$$> Â $$AB=6$$ cm
In $$\triangle\ $$APB,
$$AP^2+AB^2=BP^2$$
$$=$$> $$AP^2+6^2=9^2$$
$$=$$> $$AP^2+36=81$$
$$=$$> $$AP^2=81-36$$
$$=$$> $$AP^2=45$$
$$=$$> $$AP=3\sqrt{5}$$ cm
Hence, the correct answer is Option D
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