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15 answers

πr² = area of a circle

(π)(4²) = 16π = 50.27

2007-04-03 01:29:40 · answer #1 · answered by MamaMia © 7 · 0 0

Area of a circle = pi × r(squared)

where pi = 22/7
r - radius which in your case = 4
therefore your answer = 22/7 × 4 × 4 = 50.3 ( with your units suared, you didn't state what your unit of the radius was m or cm)

2007-04-03 01:42:24 · answer #2 · answered by Anonymous · 0 0

for a circle A=radious squared situations pi the section divided with the help of pi equals the radious squared then you definately basically discover the sq. root of the quantity you get pi=3.14 in case you probably did no longer understand :) so for fifty you do 50/pi then you definately take the quantity you get and discover the sq. root

2016-12-15 14:50:30 · answer #3 · answered by Anonymous · 0 0

Formula:
Area=pi x r ^2(squared)
Area equals pi multiplied by radius squared

A= pi x 4^2 (Area equals pi multiplied by 4 squared)
A= pi x 16 (Area eals pi multiplied by 16)
A= 16 pi or 50.26548246
A= 16pi or 50.3

2007-04-03 01:42:08 · answer #4 · answered by sum1 w/ @n @nsw3r 5 · 0 0

Use this formula for finding
the area of a circle = πr²

= (3.1416) (4)²

= 50.26 unit.

2007-04-03 01:38:45 · answer #5 · answered by edison c d 4 · 0 0

Area = pi*r^2
pi = 3,1415926 (approximately)
r = 4 => r^2 = 16
Area = 3,1415926 * 16
Area = 50,2654816

2007-04-03 01:46:46 · answer #6 · answered by dandikyuzir 1 · 0 0

the area of any circle can be given by the formula

πr² sq. units where r is the radius of the circle
so its 3.14*4*4 =50.14 sq.units

2007-04-03 01:34:26 · answer #7 · answered by Not this time 1 · 0 0

Area of circle = (pi) (r^2)

wherein pi = 3.1416
r (radius) = 4

then plug it in the formula...

Area of circle = (3.1416)(4^2)
=(3.1416) (16)
= 50.2656
= 50.27 square units.

hope i had helped you.ü

2007-04-03 01:41:42 · answer #8 · answered by bonjette 2 · 0 0

Pie Are Square

2007-04-03 01:30:22 · answer #9 · answered by oscarsnerd 2 · 0 0

Pi * r^2 = 3.14 * (4)^2 = 50.265

2007-04-03 01:31:35 · answer #10 · answered by Anonymous · 0 0

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