Q.If the lengths of the perpendicular sides of a right-angled triangle are \(a\) and \(b\), and the length of the perpendicular drawn from the right-angled vertex to the hypotenuse is \(p\), then – (a) \( \cfrac{1}{p^2} =\cfrac{1}{a^2} +\cfrac{1}{b^2} \) (b) \( \cfrac{1}{p^2} =\cfrac{1}{a^2} -\cfrac{1}{b^2} \) (c) \(p^2=a^2+b^2\) (d) \(p^2=a^2-b^2\)
Answer: A
∴ Hypotenuse = \(\sqrt{a^2 + b^2}\)
∴ Area, \(\cfrac{1}{2} ab = \cfrac{1}{2} \sqrt{a^2 + b^2} \times p\)
or, \(ab = p \sqrt{a^2 + b^2}\)
or, \(a^2 b^2 = p^2 (a^2 + b^2)\)
or, \(\cfrac{1}{p^2} = \cfrac{a^2 + b^2}{a^2 b^2} = \cfrac{1}{b^2} + \cfrac{1}{a^2}\).
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