If a square and a circle have the same perimeter. Which one will have the bigger area?
Mathematics
- 7265 दृश्य
- 36 उत्तर
36 उत्तर
-
-
Rishav
- 12 जुलाई
- 0 टिप्पणी
-
-
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 जुलाई
- 0 टिप्पणी
-
-
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 दिसम्बर
- 2 टिप्पणी
-
- during which year of my bachelor degree should I appear for GRE??
- GRE SYLLABUS FOR 2017
- GRE - About, age, vacancy, salary, Eligible?
- eligibility for GRE ? ,
- GRE - BOOKS, AUTHORS, PLEASE SUGGEST SOME BOOKS?
- GRE exam pattern
- Please suggest a good Study plan for GRE Exam ??
- GRE - Exam Pattern, Syllabus, Question, Answer, time?
- Can someone please let me know about the dates of GRE Exam?
- GRE - Preparation Tips ,Tricks ,Cracks ,Dos & Don's ?
मॉक परीक्षण अभ्यास के लिए
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.