Question

A Semicircular region and a square region have equal perimeters. The area of the sq
!
region oxeeds that of the semicircular region by 4cm. Find the perimeters and
the two region​

Answer 1

Question :-

• A semicircular region and a square region have a equal perimeters. The area of square region exceeds that of the semicircular region by 4cm². Find the perimeter and area of the two region.

Given :

• Perimeter of square = Perimeter of semicircle
• Area of square = Area of semicircle + 4cm²

To Find :

• Perimeter of square and semicircle
• Area of square and semicircle

Solution :

• Let the radius of semicircle be 'r'

And The side of the square be 'a'

Now, According to the question by using formula we get,

Perimeter of square = Perimeter of semicircle

$\sf{\pmb{\red{4 × side = r(π + 2)}}}$

By Substituting values we get,

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

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

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

$\implies{\sf{7 \times 4a = 36r}}$

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

By Dividing 4 both sides we get,

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

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

Now,

Area of the square = Area of the semicircle + 4cm²

$\sf{\pmb{\red{ {(a)}^{2} = \dfrac{\pi {r}^{2} }{2} }}}$

By putting a value we get,

$\implies{\sf{ {\bigg(\dfrac{9r}{7} }\bigg)^{2} = \dfrac{\pi {r}^{2}}{2} +4}}$

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

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

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

$\implies{\sf{ \dfrac{81 {r}^{2} }{49} = { \dfrac{22 {r}^{2} +56}{7} } \times \dfrac{1}{2} }}$

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

$\implies{\sf{81 {r}^{2} \times 14 = 49( {22r}^{2} + 56) }}$

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

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

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

$\implies{\sf{ {r}^{2} = \cancel \dfrac{2744}{56} }}$

$\implies{\sf{\purple{ r = 7cm}}}$

Now, By putting r value in eq- (1),

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

$\implies{\sf{a = \dfrac{9×7}{7}}}$

$\implies{\sf{a = \dfrac{82}{7}}}$

$\implies{\sf{\purple{a = 9cm}}}$

Now, we can find area and Perimeter :

Area of Square = (a)²

$\implies{\sf{(9)²}}$

$\implies{\boxed{\sf{81cm²}}}$

Perimeter of Square = 4a or 4 × side

$\implies{\sf{4×9}}$

$\implies{\boxed{\sf{36cm}}}$

Area of semicircle =${\bf{\dfrac{πr²}{2}}}$

$\implies{\sf{\dfrac{\dfrac{22}{7}×7²}{2}}}$

$\implies{\sf{\dfrac{\dfrac{22}{7}×49}{2}}}$

$\implies{\sf{\dfrac{22×7}{2}}}$

$\implies{\sf{\dfrac{154}{2}}}$

$\implies{\boxed{\sf{77cm²}}}$

Perimeter of semicircle = r(π + 2)

$\implies{\sf{r\bigg(\dfrac{22}{7}+2\bigg)}}$

$\implies{\sf{7\bigg(22+\dfrac{14}{2}\bigg)}}$

$\implies{\sf{7×\dfrac{36}{2}}}$

$\implies{\sf{\dfrac{252}{2}}}$

$\implies{\boxed{\sf{126cm}}}$