Question

A semi circular region and a square region have equal perimeter. The area of the square region exceed that of the semicircullar region by 4 sq. cm. Find the perimeter and area of two region

Answer 1

$\sf\large\underline{Given:-}$

$\sf{\implies Area\:_{(square)}=Area\:_{(semicircle)}+4}$

$\sf{\implies Perimeter\:_{(square)}=Perimeter\:_{semicircle)}}$

$\sf\large\underline{To\: Find:-}$

$\sf{\implies Perimeter\:_{(square\:and\: semicircle)}=?}$

$\sf{\implies Area\:_{(square\:and\: semicircle)}=?}$

$\sf\large\underline{Solution:-}$

To calculate the perimeter and area of two regions ,at first we have to assume the radius of semicircle be r and the side of square be a then setting up equation as per the given condition in the question then solve the value of a and r after that calculate it's perimeter and area. By applying formula to set up equation here:-

$\sf{\implies Perimeter\:_{(square)}=Perimeter\:_{semicircle)}}$

$\tt{\implies 4×side=r(\pi+2)}$

$\tt{\implies 4a=r(\dfrac{22}{7})+2}$

$\tt{\implies 4a=r(\dfrac{22+14}{7})}$

$\tt{\implies 4a=\dfrac{36r}{7}}$

$\tt{\implies 28a=36r}$

• Dividing by 4 on both sides here:

$\tt{\implies 7a=9r}$

$\tt{\implies a=\dfrac{9r}{7}------(i)}$

$\sf{\implies Area\:_{(square)}=Area\:_{(semicircle)}+4}$

$\tt{\implies a^2=\dfrac{\pi\:r^2}{2}+4}$

Putting the value of a here:

$\tt{\implies \bigg(\dfrac{9r}{7}\bigg)^{2}=\dfrac{\pi\:r^2+8}{2}}$

$\tt{\implies \dfrac{81r^2}{49}=\dfrac{\dfrac{22r^2}{7}+8}{2}}$

$\tt{\implies \dfrac{81r^2}{49}=\dfrac{\dfrac{22r^2+56}{7}}{2}}$

$\tt{\implies \dfrac{81r^2}{49}=\dfrac{22r^2+56}{14}}$

$\tt{\implies 81r^2×14=49(22r^2+56)}$

$\tt{\implies 1134r^2=1078r^2+2744}$

$\tt{\implies 1134r^2-1078r^2=2744}$

$\tt{\implies 56r^2=2744}$

$\tt{\implies r^2=49}$

$\tt{\implies r=\sqrt{49}=7cm}$

Now putting the value of r=7 in eq (i):

$\tt{\implies a=\dfrac{9r}{7}}$

$\tt{\implies a=\dfrac{9×7}{7}}$

$\tt{\implies a=9cm}$

Now calculate area and perimeter here:

$\sf{\implies Area\:_{(square)}=a^2}$

$\sf{\implies Area\:_{(square)}=9^2}$

$\sf{\implies Area\:_{(square)}=81cm^2}$

$\sf{\implies Perimeter\:_{(square)}=4a}$

$\sf{\implies Perimeter\:_{(square)}=4×9}$

$\sf{\implies Perimeter\:_{(square)}=36cm}$

$\sf{\implies Area\:_{(semicircle)}=\dfrac{22/7×7^2}{2}}$

$\sf{\implies Area\:_{(semicircle)}=\dfrac{22×7}{2}}$

$\sf{\implies Area\:_{(semicircle)}=77cm^2}$

$\sf{\implies Perimeter\:_{(semicircle)}=r(\pi+2)}$

$\sf{\implies Perimeter\:_{(semicircle)}=7(22/7+2)}$

$\sf{\implies Perimeter\:_{(semicircle)}=7(22+14/2)}$

$\sf{\implies Perimeter\:_{(semicircle)}=7×36/2}$

$\sf{\implies Perimeter\:_{(semicircle)}=126cm}$