Question : a and b are two sides adjacent to the right angle of a right-angled triangle and p is the perpendicular drawn to the hypotenuse from the opposite vertex. Then p2 is equal to:
Option 1: $a^{2}+b^{2}$
Option 2: $\frac{1}{a^{2}}+\frac{1}{b^{2}}$
Option 3: $\frac{a^{2}b^{2}}{a^{2}+b^{2}}$
Option 4: $a^{2}-b^{2}$
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Correct Answer: $\frac{a^{2}b^{2}}{a^{2}+b^{2}}$
Solution :
In triangle ABC,
The hypotenuse of the triangle,
$AC=\sqrt{a^2+b^2}$
The area of triangle ABC,
$\frac{1}{2}\times AB \times BC=\frac{1}{2}\times AC\times BD$
$⇒\frac{1}{2}ab=\frac{1}{2}p\sqrt{a^2+b^2}$
$⇒ab=p\sqrt{a^2+b^2}$
Squaring both sides,
$⇒a^2b^2=p^2(a^2+b^2)$
$⇒p^2=\frac{a^{2}b^{2}}{a^{2}+b^{2}}$
Hence, the correct answer is $\frac{a^{2}b^{2}}{a^{2}+b^{2}}$.
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