If a square and a circle have the same perimeter. Which one will have the bigger area?
Mathematics
- 7065 Views
- 36 Answers
36 Answers
-
-
Rishav
- 12 Jul
- 0 Comment
-
-
If the perimeter is fixed the circle will have the maximum area possible compared to any other shape including that of square.br /We can check that as follows.br /Let P be perimeter then, br /P = 2*π*radius {For Circle} = 4*side {For Square)br /thereforebr /Circle Area = π*radius^(2) = π*(P/(2π))^(2) = P^2/(4π) = 0.07958*P^(2)br /Square Area = side^(2) = (P/4)^(2) = P/16 = 0.0625*P^(2)br /hencebr /Circle Area > Square Area
-
Mohammad
- 06 Jul
- 0 Comment
-
-
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 area
-
Rajat
- 12 Dec
- 2 Comment
-
- GRE - Important dates, results, exam dates, INTERVIEW ?
- What is a GRE test? What is GRE exam eligibility? Is the GRE test hard?
- GRE - Admit Card, Hall Ticket, DOCUMENTS, Details?
- GRE - BOOKS, AUTHORS, PLEASE SUGGEST SOME BOOKS?
- What business schools accept the GRE ?
- What is a perfect score on a GRE? Is 160 a good GRE score?
- GRE exam question
- What is the best GRE prep book? Why is GRE required? When should I take the GRE?
- What is the difference between the GMAT and GRE?
- GRE - Exam Pattern, Syllabus, Question, Answer, time?
Practice Mock Test
gre
if the perimeter is fixed the circle will have the maximum area possible compared to any other shape including that of square.br /We can check that as follows.br /Let P be perimeter then, br /P = 2*π*radius {For Circle} = 4*side {For Square)br /thereforebr /Circle Area = π*radius^(2) = π*(P/(2π))^(2) = P^2/(4π) = 0.07958*P^(2)br /Square Area = side^(2) = (P/4)^(2) = P/16 = 0.0625*P^(2)br /hencebr /Circle Area > Square Areabr /Let’s now find the area of circle (Ac)Ac) w.r.t its circumference/ perimeter (PP), where rr is the radius of the circlebr /P=2πrP=2πrbr /Therefore, r=P2πr=P2πbr /Ac=πr2Ac=πr2br /Ac=π(P2π)2Ac=π(P2π)2br /Ac=P24π(1)(1)Ac=P24πbr /Let’s now find the area of square (As)As) w.r.t its perimeter (PP), where ll is the side of the squarebr /P=4lP=4lbr /l=P4l=P4br /As=l2As=l2br /As=P216(2)(2)As=P216br /Let’s now find the area of equilateral triangle(Ae)Ae) w.r.t its perimeter (PP), where ss is the side of the trianglebr /P=3sP=3sbr /s=P3s=P3br /Ae=3√4∗s2Ae=34∗s2br /br /Ae=3√4∗P29Ae=34∗P29br /Ae=P3√36Ae=P336br /Ae=P123√(3)(3)Ae=P123br /Comparing the denominators of equations 1, 2 and 3,br /4π<16<123–√4π<16<123br /As the above numbers are in the denominator,br /Ac>As>AeAc>As>Aebr /br /Thus, the circle contains he maximum area and the equilateral triangle contains the minimum area.