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(A) Let the two sides of the triangle be $a$ and $b$,and the angle between them be $	heta$.The area of the triangle is given by $A = \frac{1}{2}ab \s

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$\frac{1}{2}\sqrt{a^2 + b^2}$

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(A) Let the two sides of the triangle be $a$ and $b$,and the angle between them be $	heta$.The area of the triangle is given by $A = \frac{1}{2}ab \sin	heta$.For the area to be maximum,$\sin	heta$ must be maximum,which occurs when $	heta = 90^\circ$.Thus,the triangle is a right-angled triangle with sides $a$ and $b$ forming the right angle.Let the vertex containing sides $a$ and $b$ be at the origin $(0,0)$ in the Cartesian plane.The coordinates of the vertices are $(0,0)$,$(a,0)$,and $(0,b)$.The midpoint of the hypotenuse (the side opposite to the right angle) is $M = (\frac{a}{2}, \frac{b}{2})$.The median from the vertex $(0,0)$ to the hypotenuse is the distance from $(0,0)$ to $M$.Length of the median $m = \sqrt{(\frac{a}{2} - 0)^2 + (\frac{b}{2} - 0)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{1}{2}\sqrt{a^2 + b^2}$.

Two sides of a triangle have lengths $a \text{ cm}$ and $b \text{ cm}$. If the triangle is to have the maximum area,then the length of the median from the vertex containing the sides $a$ and $b$ is

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