A right triangle is drawn in a semicircle of radius $\frac{1}{2}$ with one of its legs along the diameter. The maximum area of the triangle is

given that, radius$=r=\frac{1}{2}$
Since, its one side is along the diameter
Therefore, diameter is the hypotenuse of the triangle      ....[ prop. of semicircle]
let $a$ & $b$ be the other sides of the triangle
Therefore, $a^2+b^2={\left(2r\right)}^2=1$
We know that A.M$\ge$G.M
so $\frac{a^2+b^2}{2}\ge \sqrt{a^2{b}^2}=ab$
$\Rightarrow ab\le \frac{1}{2}$
Area of right angled triangle $\frac{ab}{2}\le \frac{1}{4}$

Ans: A