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Question

In the given figure, ΔABC is right- angled at C. Let BC = a, CA = b, AB = c and CD = p, where CD 1p2 = 1a2 + 1b2

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Solution

(i) Area of ΔABC = 12AB × CD = 12cp.

Also, area of ΔABC = 12BC × AC = 12ab.

12cp = 12ab ⇒ cp = ab

(ii) cp = ab ⇒ p = abc

p2 = a2b2c2

1p2 - c2b2b2 - a2+b2a2b2 [ ∵ c2 = a2 + b2]

1p2 - 1a2 + 1b2


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