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(B) 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 r

Vedclass · Accepted Answer

$\sqrt{2}\left(\frac{\pi}{5}+\frac{1}{2}ight) 	ext{ m}^2/	ext{hour}$

Vedclass · Answer

(B) 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 the area,we differentiate $A$ with respect to time $t$ using the product rule:$\frac{dA}{dt} = \frac{1}{2} \left( 2x \frac{dx}{dt} \sin 	heta + x^2 \cos 	heta \frac{d	heta}{dt} ight) = x \frac{dx}{dt} \sin 	heta + \frac{1}{2} x^2 \cos 	heta \frac{d	heta}{dt}$.Given values: $x = 12 	ext{ m}$,$	heta = \frac{\pi}{4}$,$\frac{dx}{dt} = \frac{1}{12} 	ext{ m/h}$,and $\frac{d	heta}{dt} = \frac{\pi}{180} 	ext{ rad/h}$.Substituting these values into the derivative:$\frac{dA}{dt} = (12) \left( \frac{1}{12} ight) \sin \left( \frac{\pi}{4} ight) + \frac{1}{2} (12)^2 \cos \left( \frac{\pi}{4} ight) \left( \frac{\pi}{180} ight)$.$\frac{dA}{dt} = (1) \left( \frac{1}{\sqrt{2}} ight) + \frac{1}{2} (144) \left( \frac{1}{\sqrt{2}} ight) \left( \frac{\pi}{180} ight)$.$\frac{dA}{dt} = \frac{1}{\sqrt{2}} + \frac{72}{\sqrt{2}} \left( \frac{\pi}{180} ight) = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \left( 1 + \frac{2\pi}{5} ight)$.Note: The provided options seem to contain a typo in the coefficient. Re-evaluating the expression: $\frac{1}{\sqrt{2}} (1 + \frac{2\pi}{5}) = \frac{\sqrt{2}}{2} (1 + \frac{2\pi}{5}) = \sqrt{2} (\frac{1}{2} + \frac{\pi}{5})$.Thus,the correct option is $B$.

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