question_answer

The area of a right angled triangle is \[20\,c{{m}^{2}}\]and one of the sides containing the right angle is 4 cm. The altitude on the hypotenuse is

A) \[\frac{41}{\sqrt{34}}\,cm\]

B) \[\sqrt{\frac{41}{40}}\,cm\]

C) \[\frac{29}{\sqrt{20}}\,cm\]

D) \[\frac{20}{\sqrt{29}}\,cm\]

Correct Answer: D

Solution :

[d] Let the altitude be x cm. Then, \[\frac{1}{2}\times 4\times x=20\] \[\Rightarrow \]   \[x=10\,cm\] BC = Hypotenuse\[=\sqrt{{{(10)}^{2}}+{{(4)}^{2}}}\] \[=\sqrt{116}=\sqrt{4\times 29}=2\sqrt{29}\] Let \[AD\bot BC.\]Then,             \[\frac{1}{2}\times BC\times AD=\] area of \[\Delta ABC\]             \[\Rightarrow \]   \[\frac{1}{2}\times 2\sqrt{29}\times AD=20\]             \[\therefore \]      \[AD=\frac{20}{\sqrt{29}}\,cm\]