a semicircular region and a square region have a equal perimeters . The area of square region exceeds that of the semicircular reg

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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  1. 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

    Reply
  2. 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 :

    • 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

    [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]

    Reply

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