A) 8 cm
B) 10 cm
C) \[\frac{10}{\sqrt{41}}cm\]cm
D) \[\frac{20}{\sqrt{29}}cm\]
Correct Answer: D
Solution :
Let length of the other perpendicular side be a cm and altitude on the hypotenuse is p cm. \[\therefore \] \[\frac{1}{2}\times 4\times a=20\] or \[a=10\,cms\] \[\therefore \] Hypotenuse \[=\sqrt{{{a}^{2}}+{{4}^{2}}}=\sqrt{{{10}^{2}}+{{4}^{2}}}=2\sqrt{29}\] Area of the triangle \[=\frac{1}{2}\times 2\sqrt{29}\times p=20\,sq.cm\] \[\therefore \] \[p=\frac{20}{\sqrt{29}}\]
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