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(D) The area $A$ of an isosceles triangle with equal sides $x$ and included angle $	heta$ is given by $A = \frac{1}{2} x^2 \sin	heta$.To find the ra

Vedclass · Accepted Answer

$2^{1/2}\left( {\frac{1}{2} + \frac{\pi }{5}} \right)$

Vedclass · Answer

(D) The area $A$ of an isosceles triangle with equal sides $x$ and included angle $	heta$ is given by $A = \frac{1}{2} x^2 \sin	heta$.To find the rate of change of area with respect to time $t$,we differentiate with respect to $t$:$\frac{dA}{dt} = \frac{1}{2} \left( 2x \frac{dx}{dt} \sin	heta + x^2 \cos	heta \frac{d	heta}{dt} ight) = x \frac{dx}{dt} \sin	heta + \frac{1}{2} x^2 \cos	heta \frac{d	heta}{dt}$.Given values: $x = 12 \ m$,$	heta = \pi/4$,$\frac{dx}{dt} = 1/12 \ m/hr$,and $\frac{d	heta}{dt} = \pi/180 \ 	ext{rad/hr}$.Substituting these values:$\frac{dA}{dt} = (12) \left( \frac{1}{12} ight) \sin(\pi/4) + \frac{1}{2} (12)^2 \cos(\pi/4) \left( \frac{\pi}{180} ight)$$\frac{dA}{dt} = (1) \left( \frac{1}{\sqrt{2}} ight) + \frac{1}{2} (144) \left( \frac{1}{\sqrt{2}} ight) \left( \frac{\pi}{180} ight)$$\frac{dA}{dt} = \frac{1}{\sqrt{2}} + \frac{72}{\sqrt{2}} \left( \frac{\pi}{180} ight) = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \left( 1 + \frac{2\pi}{5} ight) = \frac{\sqrt{2}}{2} \left( 1 + \frac{2\pi}{5} ight)$.Wait,checking the options,the expression is $2^{1/2} (1/2 + \pi/5) = \sqrt{2} (1/2 + \pi/5) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{5} = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}}$. This matches option $D$.

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