Question

A semicircular region and a square region have equal perimeters. The area of the square region exceeds that of the semicircular region by 4 square cm, find the perimeters and the areas of the two region.​
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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:-]

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

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

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

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

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

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

$\sf{\implies Area\:_{(semicircle)}=\dfrac{\pi\:r^2}{2}}$

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

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

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

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

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

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

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

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

Answer 2

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