ABC is a right angled triangle , with C = 90∘. If P is the perpendicular from C to AB and a,b,c have the usual meaning, then find the value of 1a2+1b2.

ABC is a right angled triangle , with C = ${90}^{\circ }$. If P is the perpendicular from C to AB and a,b,c have the usual meaning, then find the value of $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}$.

1. A
${p}^{2}$
2. B
$\frac{1}{{p}^{2}}$
3. C
$p$
4. D
$\frac{1}{p}$

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Solution:

Given ABC is a right-angled triangle with the right angle at C.
Let P is the perpendicular from C to AB and it intersects AB at D . Given, the length of the side AB = c, BC = a, CA = b and CD = p .
Area of triangle ABC = $\frac{1}{2}×\mathit{BC}×\mathit{AC}=\frac{1}{2}\mathit{ab}$
Also, Area of triangle ABC = $\frac{1}{2}×\mathit{AB}×\mathit{CD}=\frac{1}{2}\mathit{cp}$
$⇒\mathit{ab}=\mathit{cp}$
Since ${a}^{2}+{b}^{2}={c}^{2}={\left(\frac{\mathit{ab}}{p}\right)}^{2}=\frac{{a}^{2}{b}^{2}}{{p}^{2}}$
$⇒\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{{p}^{2}}$

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