Question

Length of the altitude on the hypotenuse of a right angled triangle with area $A{\text{cm}}^2$ and one of the sides equal to $b\text{cm}$ is __ cm.

• A. $\frac{2A{b}^2}{\sqrt{b^4+4A^2}}$
• B. $\frac{2b}{\sqrt{b^4+4A^2}}$
• C. $\frac{2b^{2}A}{\sqrt{b^4+A^2}}$
• D. $\frac{2bA}{\sqrt{b^4+4A^2}}$

Answer 1

The correct option is D $\frac{2bA}{\sqrt{b^4+4A^2}}$


Let PQR be a right-angled triangle at Q such that $QR=b$ and $A=\text{Area of}\Delta PQR.$
Now, draw $QN\perp PR.$
$A=\text{Area of}\Delta PQR=\frac{1}{2}(QR\times PQ)=\frac{1}{2}(b\times PQ)\Rightarrow PQ=\frac{2A}{b}...\left(i\right)$
Now, in $\Delta PNQ$ and $\Delta PQR$,
$\angle PNQ=\angle PQR$ [each ${90}^{\circ }$]
and $\angle QPN=\angle QPR$ [common]
$\therefore \Delta PNQ\sim \Delta PQR$ [by AA similarity]
$\Rightarrow \frac{PQ}{PR}=\frac{NQ}{QR}...\left(ii\right)$
Now, in $\Delta PQR$, by Pythagoras' theorem,
$P{Q}^2+Q{R}^2=P{R}^2.\Rightarrow \frac{4A^2}{b^2}+b^2=P{R}^2\Rightarrow PR=\sqrt{\frac{4A^2+b^4}{b^2}}=\frac{\sqrt{4A^2+b^4}}{b}\therefore \frac{2A}{b\times PR}=\frac{NQ}{b}$
[from Eqs. (i) and (ii)]
$\Rightarrow NQ=\frac{2A}{PR}\Rightarrow NQ=\frac{2Ab}{\sqrt{4A^2+b^4}}$