Question

If a wire is bent into a shape of a square, then the area of the square is \[81c{m^2}\]. When the wire is bent into a semi-circular shape the area of the semi-circle will be?

A.\[22c{m^2}\]

B.\[44c{m^2}\]

C.\[77c{m^2}\]

D.\[154c{m^2}\]

Answer 1

Hint: Use the area of the square to get the sides of the square and from there we can get the perimeter of the square which is ultimately the length of the wire. So that will also be the perimeter of the semi-circle and from there we can find the value of the radius of the semi-circle which will eventually help to get the area of the semi-circle.

Complete step-by-step answer:

So we are given the area of the square as \[81c{m^2}\] and the formula for the area of the square is given by \[{s^2}\] where s is the side of the square.

So from here we are getting that

${s^2} = 81c{m^2}$

$\Rightarrow s = \sqrt {81}$

$\Rightarrow s = 9$

So we have the side of the square as 9 now the perimeter of the square becomes,

$\therefore perimeter = 4s$

$\Rightarrow perimeter = 4 \times 9$

$\Rightarrow perimeter = 36$

So now we know that the perimeter of the square is the length of the wire given to us which means that the length of the wire is 36cm

Now we know that the Perimeter of the semi circle is given by \[\pi r + 2r\] and the perimeter of the semi-circle will be as same as the perimeter of the square which means that

$\pi r + 2r = 36$

$\Rightarrow r(\pi + 2) = 36$

$\Rightarrow r = \dfrac{{36}}{{\pi + 2}}$

$\Rightarrow r = \dfrac{{36}}{{\dfrac{{22}}{7} + 2}}$

$\Rightarrow r = \dfrac{{36}}{{\dfrac{{22 + 14}}{7}}}$

$\Rightarrow r = \dfrac{{36}}{{\dfrac{{36}}{7}}}$

$\Rightarrow r = \dfrac{{36}}{1} \times \dfrac{7}{{36}}$

$\Rightarrow r = 7$

So we have the radius of the semi circle as 7.

Now we know that the area of a semicircle is given by \[\dfrac{{\pi {r^2}}}{2}\]

$A = \dfrac{{\pi {r^2}}}{2}$

$\Rightarrow A = \dfrac{{\pi \times 7 \times 7}}{2}$

$\Rightarrow A = \dfrac{{\dfrac{{22}}{7} \times 7 \times 7}}{2}$

$\Rightarrow A = \dfrac{{22 \times 7}}{2}$

$\Rightarrow A = 11 \times 7$ 

$\Rightarrow A = 77c{m^2}$

Hence, C is the correct option here.

Note: It must be noted that the perimeter of the semi-circle is \[\pi r + 2r\] because the arc is half the perimeter of an actual circle which is \[2\pi r\] and 2r is basically the length of the diameter which is used to make it a closed figure. But we know that the area will only be half of the total area of the circle i.e., \[\dfrac{{\pi {r^2}}}{2}\].