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 areas of the teo region.
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Answer:
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,
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,
Answer:
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)}}}
4×side=r(π+2)
4×side=r(π+2)
By Substituting values we get,
\implies{\sf{4a = r\bigg( \dfrac{22}{7} \bigg) + 2}}⟹4a=r(
7
22
)+2
\implies{\sf{4a = r\bigg( \dfrac{22 + 14}{7}\bigg )}}⟹4a=r(
7
22+14
)
\implies{\sf{ 4a = \dfrac{36r}{7} }}⟹4a=
7
36r
\implies{\sf{7 \times 4a = 36r}}⟹7×4a=36r
\implies{\sf{28a = 36r}}⟹28a=36r
By Dividing 4 both sides we get,
\implies{\sf{7a = 9r}}⟹7a=9r
\implies{\sf{a = \dfrac{9r}{7} }}—–(1)⟹a=
7
9r
−−−−−(1)
Now,
Area of the square = Area of the semicircle + 4cm²
\sf{\pmb{\red{ {(a)}^{2} = \dfrac{\pi {r}^{2} }{2} }}}
(a)
2
=
2
πr
2
(a)
2
=
2
πr
2
By putting a value we get,
\implies{\sf{ {\bigg(\dfrac{9r}{7} }\bigg)^{2} = \dfrac{\pi {r}^{2}}{2} +4}}⟹(
7
9r
)
2
=
2
πr
2
+4
\implies{\sf{ {\bigg(\dfrac{9r}{7} }\bigg)^{2} = \dfrac{\pi {r}^{2} + 8}{2} }}⟹(
7
9r
)
2
=
2
πr
2
+8
\implies{\sf{ \dfrac{81 {r}^{2} }{49} = \dfrac{ \dfrac{22 {r}^{2}}{7} + 8}{2} }}⟹
49
81r
2
=
2
7
22r
2
+8
\implies{\sf{ \dfrac{81 {r}^{2} }{49} = \dfrac{ \dfrac{22 {r}^{2} +56}{7} }{2} }}⟹
49
81r
2
=
2
7
22r
2
+56
\implies{\sf{ \dfrac{81 {r}^{2} }{49} = { \dfrac{22 {r}^{2} +56}{7} } \times \dfrac{1}{2} }}⟹
49
81r
2
=
7
22r
2
+56
×
2
1
\implies{\sf{ \dfrac{81 {r}^{2} }{49} = { \dfrac{22 {r}^{2} +56}{14} }}}⟹
49
81r
2
=
14
22r
2
+56
\implies{\sf{81 {r}^{2} \times 14 = 49( {22r}^{2} + 56) }}⟹81r
2
×14=49(22r
2
+56)
\implies{\sf{1134 {r}^{2} = 1078 {r}^{2} + 2744}}⟹1134r
2
=1078r
2
+2744
\implies{\sf{1134 {r}^{2} – 1078 {r}^{2} = 2744 }}⟹1134r
2
−1078r
2
=2744
\implies{\sf{56 {r}^{2} = 2744 }}⟹56r
2
=2744
\implies{\sf{ {r}^{2} = \cancel \dfrac{2744}{56} }}⟹r
2
=
56
2744
\implies{\sf{\purple{ r = 7cm}}}⟹r=7cm
Now, By putting r value in eq- (1),
\implies{\sf{a = \dfrac{9r}{7}}}⟹a=
7
9r
\implies{\sf{a = \dfrac{9×7}{7}}}⟹a=
7
9×7
\implies{\sf{a = \dfrac{82}{7}}}⟹a=
7
82
\implies{\sf{\purple{a = 9cm}}}⟹a=9cm
Now, we can find area and Perimeter :
Area of Square = (a)²
\implies{\sf{(9)²}}⟹(9)²
\implies{\boxed{\sf{81cm²}}}⟹
81cm²
Perimeter of Square = 4a
\implies{\sf{4×9}}⟹4×9
\implies{\boxed{\sf{36cm}}}⟹
36cm
Area of semicircle = {\bf{\dfrac{πr²}{2}}}
2
πr²
\implies{\sf{\dfrac{\dfrac{22}{7}×7²}{2}}}⟹
2
7
22
×7²
\implies{\sf{\dfrac{\dfrac{22}{7}×49}{2}}}⟹
2
7
22
×49
\implies{\sf{\dfrac{22×7}{2}}}⟹
2
22×7
\implies{\sf{\dfrac{154}{2}}}⟹
2
154
\implies{\boxed{\sf{77cm²}}}⟹
77cm²
Perimeter of semicircle = r(π + 2)
\implies{\sf{r\bigg(\dfrac{22}{7}+2\bigg)}}⟹r(
7
22
+2)
\implies{\sf{7\bigg(22+\dfrac{14}{2}\bigg)}}⟹7(22+
2
14
)
\implies{\sf{7×\dfrac{36}{2}}}⟹7×
2
36
\implies{\sf{\dfrac{252}{2}}}⟹
2
252
\implies{\boxed{\sf{126cm}}}⟹
126cm
Correct 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 :
To Find :
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
[tex]\sf{\pmb{\red{4 × side = r(π + 2)}}}[/tex]
By Substituting values we get,
[tex]\implies{\sf{4a = r\bigg( \dfrac{22}{7} \bigg) + 2}}[/tex]
[tex]\implies{\sf{4a = r\bigg( \dfrac{22 + 14}{7}\bigg )}}[/tex]
[tex]\implies{\sf{ 4a = \dfrac{36r}{7} }}[/tex]
[tex]\implies{\sf{7 \times 4a = 36r}}[/tex]
[tex]\implies{\sf{28a = 36r}}[/tex]
By Dividing 4 both sides we get,
[tex]\implies{\sf{7a = 9r}}[/tex]
[tex]\implies{\sf{a = \dfrac{9r}{7} }}—–(1)[/tex]
Now,
Area of the square = Area of the semicircle + 4cm²
[tex]\sf{\pmb{\red{ {(a)}^{2} = \dfrac{\pi {r}^{2} }{2} }}}[/tex]
By putting a value we get,
[tex]\implies{\sf{ {\bigg(\dfrac{9r}{7} }\bigg)^{2} = \dfrac{\pi {r}^{2}}{2} +4}}[/tex]
[tex]\implies{\sf{ {\bigg(\dfrac{9r}{7} }\bigg)^{2} = \dfrac{\pi {r}^{2} + 8}{2} }}[/tex]
[tex]\implies{\sf{ \dfrac{81 {r}^{2} }{49} = \dfrac{ \dfrac{22 {r}^{2}}{7} + 8}{2} }}[/tex]
[tex]\implies{\sf{ \dfrac{81 {r}^{2} }{49} = \dfrac{ \dfrac{22 {r}^{2} +56}{7} }{2} }}[/tex]
[tex]\implies{\sf{ \dfrac{81 {r}^{2} }{49} = { \dfrac{22 {r}^{2} +56}{7} } \times \dfrac{1}{2} }}[/tex]
[tex]\implies{\sf{ \dfrac{81 {r}^{2} }{49} = { \dfrac{22 {r}^{2} +56}{14} }}}[/tex]
[tex]\implies{\sf{81 {r}^{2} \times 14 = 49( {22r}^{2} + 56) }}[/tex]
[tex]\implies{\sf{1134 {r}^{2} = 1078 {r}^{2} + 2744}}[/tex]
[tex]\implies{\sf{1134 {r}^{2} – 1078 {r}^{2} = 2744 }}[/tex]
[tex]\implies{\sf{56 {r}^{2} = 2744 }}[/tex]
[tex]\implies{\sf{ {r}^{2} = \cancel \dfrac{2744}{56} }}[/tex]
[tex]\implies{\sf{\purple{ r = 7cm}}}[/tex]
Now, By putting r value in eq- (1),
[tex]\implies{\sf{a = \dfrac{9r}{7}}}[/tex]
[tex]\implies{\sf{a = \dfrac{9×7}{7}}}[/tex]
[tex]\implies{\sf{a = \dfrac{82}{7}}}[/tex]
[tex]\implies{\sf{\purple{a = 9cm}}}[/tex]
Now, we can find area and Perimeter :
Area of Square = (a)²
[tex]\implies{\sf{(9)²}}[/tex]
[tex]\implies{\boxed{\sf{81cm²}}}[/tex]
Perimeter of Square = 4a
[tex]\implies{\sf{4×9}}[/tex]
[tex]\implies{\boxed{\sf{36cm}}}[/tex]
Area of semicircle = [tex]{\bf{\dfrac{πr²}{2}}}[/tex]
[tex]\implies{\sf{\dfrac{\dfrac{22}{7}×7²}{2}}}[/tex]
[tex]\implies{\sf{\dfrac{\dfrac{22}{7}×49}{2}}}[/tex]
[tex]\implies{\sf{\dfrac{22×7}{2}}}[/tex]
[tex]\implies{\sf{\dfrac{154}{2}}}[/tex]
[tex]\implies{\boxed{\sf{77cm²}}}[/tex]
Perimeter of semicircle = r(π + 2)
[tex]\implies{\sf{r\bigg(\dfrac{22}{7}+2\bigg)}}[/tex]
[tex]\implies{\sf{7\bigg(22+\dfrac{14}{2}\bigg)}}[/tex]
[tex]\implies{\sf{7×\dfrac{36}{2}}}[/tex]
[tex]\implies{\sf{\dfrac{252}{2}}}[/tex]
[tex]\implies{\boxed{\sf{126cm}}}[/tex]