Question
The maximum area of a right angled triangle with hypotenuse $$h$$ is :
A.
$$\frac{{{h^2}}}{{2\sqrt 2 }}$$
B.
$$\frac{{{h^2}}}{2}$$
C.
$$\frac{{{h^2}}}{{\sqrt 2 }}$$
D.
$$\frac{{{h^2}}}{4}$$
Answer :
$$\frac{{{h^2}}}{4}$$
Solution :
$$\eqalign{ & {\text{Let base }} = b \cr & {\text{Altitude }}\left( {{\text{or perpendicular}}} \right) = \sqrt {{h^2} - {b^2}} \cr} $$
$$\eqalign{ & {\text{Area, }}A = \frac{1}{2} \times {\text{base}} \times {\text{altitude}} \cr & {\text{ = }}\frac{1}{2} \times b \times \sqrt {{h^2} - {b^2}} \cr & \Rightarrow \frac{{dA}}{{db}}{\text{ = }}\frac{1}{2}\left[ {\sqrt {{h^2} - {b^2}} + b.\frac{{ - 2b}}{{2\sqrt {{h^2} - {b^2}} }}} \right] \cr & \Rightarrow \frac{{dA}}{{db}} = \frac{1}{2}\left[ {\frac{{{h^2} - 2{b^2}}}{{\sqrt {{h^2} - {b^2}} }}} \right] \cr & {\text{Put }}\frac{{dA}}{{db}} = 0,b = \frac{h}{{\sqrt 2 }} \cr & {\text{Maximum area}} = \frac{1}{2} \times \frac{h}{{\sqrt 2 }} \times \sqrt {{h^2} - \frac{{{h^2}}}{2}} = \frac{{{h^2}}}{4} \cr} $$
$$\eqalign{ & {\text{Let base }} = b \cr & {\text{Altitude }}\left( {{\text{or perpendicular}}} \right) = \sqrt {{h^2} - {b^2}} \cr} $$
$$\eqalign{ & {\text{Area, }}A = \frac{1}{2} \times {\text{base}} \times {\text{altitude}} \cr & {\text{ = }}\frac{1}{2} \times b \times \sqrt {{h^2} - {b^2}} \cr & \Rightarrow \frac{{dA}}{{db}}{\text{ = }}\frac{1}{2}\left[ {\sqrt {{h^2} - {b^2}} + b.\frac{{ - 2b}}{{2\sqrt {{h^2} - {b^2}} }}} \right] \cr & \Rightarrow \frac{{dA}}{{db}} = \frac{1}{2}\left[ {\frac{{{h^2} - 2{b^2}}}{{\sqrt {{h^2} - {b^2}} }}} \right] \cr & {\text{Put }}\frac{{dA}}{{db}} = 0,b = \frac{h}{{\sqrt 2 }} \cr & {\text{Maximum area}} = \frac{1}{2} \times \frac{h}{{\sqrt 2 }} \times \sqrt {{h^2} - \frac{{{h^2}}}{2}} = \frac{{{h^2}}}{4} \cr} $$