The area of a right-angled triangle is 20 sq.cm and one of the sides containing the right angle is 4 cm. The altitude on the hypotenuse is :
Options:
A . $$\frac{{41}}{{\sqrt {34} }}{\text{ cm}}$$
B . $$\sqrt {\frac{{41}}{{40}}} {\text{ cm}}$$
C . $$\frac{{29}}{{\sqrt {20} }}{\text{ cm}}$$
D . $$\frac{{20}}{{\sqrt {29} }}{\text{ cm}}$$
Answer: Option D Let the length of the other side containing the right angle be x cm Then, $$\eqalign{ & \frac{1}{2} \times 4 \times x = 20 \cr & \Rightarrow x = 10{\text{ cm}} \cr} $$ Hypotenuse : $$\eqalign{ & = \sqrt {{{10}^2} + {4^2}} {\text{ cm}} \cr & = \sqrt {116} {\text{ cm}} \cr & = 2\sqrt {29} \,{\text{cm}} \cr} $$ Let the altitude on the hypotenuse be h cm Then,$$\eqalign{ & \frac{1}{2} \times 2\sqrt {29} \times h = 20 \cr & \Rightarrow h = \frac{{20}}{{\sqrt {29} }}{\text{ cm}} \cr} $$
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