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(D) Let the two given sides of the triangle be $a$ and $b$,and let the angle between them be $C$.The area $A$ of the triangle is given by the formula:

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${\pi \over 2}$

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(D) Let the two given sides of the triangle be $a$ and $b$,and let the angle between them be $C$.The area $A$ of the triangle is given by the formula:$A = \frac{1}{2}ab \sin C$To find the maximum area,we differentiate $A$ with respect to $C$:$\frac{dA}{dC} = \frac{1}{2}ab \cos C$For the area to be maximum,we set the first derivative to zero:$\frac{dA}{dC} = 0$$\frac{1}{2}ab \cos C = 0$Since $a$ and $b$ are sides of a triangle,$a, b 
eq 0$,so we must have:$\cos C = 0$$C = \frac{\pi}{2}$ or $90^\circ$Thus,the area is maximum when the angle between the given sides is $\frac{\pi}{2}$.

If two sides of a triangle are given,then the area of the triangle will be maximum if the angle between the given sides is

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