JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 15)
A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then $$\left( {{4 \over \pi } + 1} \right)k$$ is equal to _____________.
Answer
36
Explanation
Let x + y = 36
x is perimeter of square and y is perimeter of circle side of square = x/4
radius of circle = $${y \over {2\pi }}$$
Sum Areas = $${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$$
$$ = {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$$
For min Area :
$$x = {{144} \over {\pi + 4}}$$
$$\Rightarrow$$ Radius = y = 36 $$-$$ $${{144} \over {\pi + 4}}$$
$$\Rightarrow$$ k = $${{36\pi } \over {\pi + 4}}$$
$$\left( {{4 \over \pi } + 1} \right)k$$ = 36
x is perimeter of square and y is perimeter of circle side of square = x/4
radius of circle = $${y \over {2\pi }}$$
Sum Areas = $${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$$
$$ = {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$$
For min Area :
$$x = {{144} \over {\pi + 4}}$$
$$\Rightarrow$$ Radius = y = 36 $$-$$ $${{144} \over {\pi + 4}}$$
$$\Rightarrow$$ k = $${{36\pi } \over {\pi + 4}}$$
$$\left( {{4 \over \pi } + 1} \right)k$$ = 36
Comments (0)
