[tex] \bf \: Given \: : [/tex] ↴ [tex] \sf \: Radius \: of \: the \: circle \: : \: 28 \: meter.[/tex] [tex] \bf \: To \: find \: : [/tex] ↴ [tex] \sf \: Area \: and \: circumference \: of \: circle.[/tex] [tex] \bf \: So, \: Let’s \: Start \: : [/tex] ↴ [tex] \sf \: First \: of \: all \: we \: will \: find \: the \: circumference \: of \: the \: circle. [/tex] [tex] \boxed{{ \purple{ \sf \: Circumference \: of \: the \: circle \: : \: 2πr }}}[/tex] [tex] \sf \: Putting \: the \: value \: : [/tex] ↴ [tex] \bf \longrightarrow \: 2 \: \times \: \dfrac{22}{7} \: \times 28[/tex] [tex] \bf \longrightarrow \: \dfrac{1232}{7} [/tex] [tex] \bf \longrightarrow \: 176 \: m ^{2} [/tex] [tex] \blue{\sf \: So, \: the \: Circumference \: of \: the \: circle : 176 \: m ^{2} }[/tex] _____________________________________ [tex] \sf \: Now, \: we \: will \: find \: the \: area \: of \: the \: circle.[/tex] [tex] \boxed{ \purple{ \sf \: Area \: of \: the \: Circle : πr ^{2} }}[/tex] [tex] \sf \: Putting \: the \: value \: : [/tex] ↴ [tex] \bf \longrightarrow \: \: \dfrac{22}{7} \: \times 28 ^{2}[/tex] [tex] \bf \longrightarrow \: \: \dfrac{22}{7} \: \times 28 \: \times 28[/tex] [tex] \bf \longrightarrow \: \dfrac{17248}{7} [/tex] [tex] \bf \longrightarrow \: 2464 \: m ^{2} [/tex] [tex] \blue{\sf \: So, \: the \: area \: of \: the \: circle : 2464 \: m ^{2} }[/tex] Reply
Let r be the radius of the circle. Then according to the problem we’ve r=28m. Then perimeter will be (P)=2πr (P) = 2×(22÷7)×28 (P) = 176m So the perimeter of the circle is 176m. (A) = π(r)^2 (A) = (22÷7)×(28)^2 (A) = 22÷7)×28×28 (A) = (22÷7)×784 (A) = (22×784)÷7 (A) = 17248÷7 (A) = 2464m So the area is 2464m. ———————————————————– Reply
[tex] \bf \: Given \: : [/tex] ↴
[tex] \sf \: Radius \: of \: the \: circle \: : \: 28 \: meter.[/tex]
[tex] \bf \: To \: find \: : [/tex] ↴
[tex] \sf \: Area \: and \: circumference \: of \: circle.[/tex]
[tex] \bf \: So, \: Let’s \: Start \: : [/tex] ↴
[tex] \sf \: First \: of \: all \: we \: will \: find \: the \: circumference \: of \: the \: circle. [/tex]
[tex] \boxed{{ \purple{ \sf \: Circumference \: of \: the \: circle \: : \: 2πr }}}[/tex]
[tex] \sf \: Putting \: the \: value \: : [/tex] ↴
[tex] \bf \longrightarrow \: 2 \: \times \: \dfrac{22}{7} \: \times 28[/tex]
[tex] \bf \longrightarrow \: \dfrac{1232}{7} [/tex]
[tex] \bf \longrightarrow \: 176 \: m ^{2} [/tex]
[tex] \blue{\sf \: So, \: the \: Circumference \: of \: the \: circle : 176 \: m ^{2} }[/tex]
_____________________________________
[tex] \sf \: Now, \: we \: will \: find \: the \: area \: of \: the \: circle.[/tex]
[tex] \boxed{ \purple{ \sf \: Area \: of \: the \: Circle : πr ^{2} }}[/tex]
[tex] \sf \: Putting \: the \: value \: : [/tex] ↴
[tex] \bf \longrightarrow \: \: \dfrac{22}{7} \: \times 28 ^{2}[/tex]
[tex] \bf \longrightarrow \: \: \dfrac{22}{7} \: \times 28 \: \times 28[/tex]
[tex] \bf \longrightarrow \: \dfrac{17248}{7} [/tex]
[tex] \bf \longrightarrow \: 2464 \: m ^{2} [/tex]
[tex] \blue{\sf \: So, \: the \: area \: of \: the \: circle : 2464 \: m ^{2} }[/tex]
Let r be the radius of the circle.
Then according to the problem we’ve r=28m.
Then perimeter will be (P)=2πr
(P) = 2×(22÷7)×28
(P) = 176m
So the perimeter of the circle is 176m.
(A) = π(r)^2
(A) = (22÷7)×(28)^2
(A) = 22÷7)×28×28
(A) = (22÷7)×784
(A) = (22×784)÷7
(A) = 17248÷7
(A) = 2464m
So the area is 2464m.
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