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Question

(C) Let the triangular park be $\Delta ABC$,where $AB = AC = x$. Let $AT$ be the altitude from $A$ to the river bank $BC$. Let $\angle ABT = 	heta$.

Vedclass · Accepted Answer

$\frac{1}{2} x^2$

Vedclass · Answer

(C) Let the triangular park be $\Delta ABC$,where $AB = AC = x$. Let $AT$ be the altitude from $A$ to the river bank $BC$. Let $\angle ABT = 	heta$. Then $AT = x \sin 	heta$ and $BT = x \cos 	heta$. Since the triangle is isosceles with $AB=AC$,the altitude $AT$ bisects $BC$,so $BC = 2BT = 2x \cos 	heta$. The area $A$ of the triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2x \cos 	heta) 	imes (x \sin 	heta) = x^2 \sin 	heta \cos 	heta = \frac{1}{2} x^2 \sin(2	heta)$. To maximize the area,we need to maximize $\sin(2	heta)$. The maximum value of $\sin(2	heta)$ is $1$,which occurs when $2	heta = 90^\circ$ or $	heta = 45^\circ$. Thus,the maximum area is $\frac{1}{2} x^2 (1) = \frac{1}{2} x^2$.

$A$ triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $x$. The maximum area enclosed by the park is

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