If a square and a circle have the same perimeter. Which one will have the bigger area?
Mathematics
- Bernard
- 12 जुलाई
- 36 उत्तर
36 उत्तर
36 उत्तर
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if the perimeter is fixed the circle will have the maximum area possible compared to any other shape including that of square.
We can check that as follows.
Let P be perimeter then,
P = 2*π*radius {For Circle} = 4*side {For Square)
therefore
Circle Area = π*radius^(2) = π*(P/(2π))^(2) = P^2/(4π) = 0.07958*P^(2)
Square Area = side^(2) = (P/4)^(2) = P/16 = 0.0625*P^(2)
hence
Circle Area > Square Area
Let’s now find the area of circle (Ac)Ac) w.r.t its circumference/ perimeter (PP), where rr is the radius of the circle
P=2πrP=2πr
Therefore, r=P2πr=P2π
Ac=πr2Ac=πr2
Ac=π(P2π)2Ac=π(P2π)2
Ac=P24π(1)(1)Ac=P24π
Let’s now find the area of square (As)As) w.r.t its perimeter (PP), where ll is the side of the square
P=4lP=4l
l=P4l=P4
As=l2As=l2
As=P216(2)(2)As=P216
Let’s now find the area of equilateral triangle(Ae)Ae) w.r.t its perimeter (PP), where ss is the side of the triangle
P=3sP=3s
s=P3s=P3
Ae=3√4∗s2Ae=34∗s2
Ae=3√4∗P29Ae=34∗P29
Ae=P3√36Ae=P336
Ae=P123√(3)(3)Ae=P123
Comparing the denominators of equations 1, 2 and 3,
4π<16<123–√4π<16<123
As the above numbers are in the denominator,
Ac>As>AeAc>As>Ae
Thus, the circle contains he maximum area and the equilateral triangle contains the minimum area.0Modal content
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If the perimeter is fixed the circle will have the maximum area possible compared to any other shape including that of square.
We can check that as follows.
Let P be perimeter then,
P = 2*π*radius {For Circle} = 4*side {For Square)
therefore
Circle Area = π*radius^(2) = π*(P/(2π))^(2) = P^2/(4π) = 0.07958*P^(2)
Square Area = side^(2) = (P/4)^(2) = P/16 = 0.0625*P^(2)
hence
Circle Area > Square Area0Modal content
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Circle Let's say that the the perimeter is p. For a square, the side would be, (p/4) and the area would be (p^2/16). For the circle, the radius would be (p/2p) and area would be (p^2/4p). 4p is approx 12.56. Thus, circle had bigger area0Modal content
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circle is p.r^2 n sq only r^2 hence circle0Modal content
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circle0Modal content
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Circle (3.1416:2.42)0Modal content
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circle0Modal content
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circle0Modal content
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circle0Modal content
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Circle0Modal content
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Square0Modal content
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circle1Modal content
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Circle0Modal content
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circle0Modal content
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circle1Modal content
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