What is the diameter of a circle with area of $18 π$ $18pi$ $f t^{2}$ $ft^2$ ?

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Answer 1 · 2016-04-03T16:43:02

$6 \sqrt{2} π f t$ $6sqrt2pi ft$

Area of a circle $= π r^{2}$ $= pir^2$

$18 π = π r^{2}$ $18pi = pir^2$
$18 = r^{2}$ $18 = r^2$
$r = \sqrt{18} = 3 \sqrt{2}$ $r = sqrt18 = 3sqrt2$

Perimeter of a circle $= 2 π r$ $= 2pir$

$2 π r = 2 π \cdot 3 \sqrt{2}$ $2pir = 2pi * 3sqrt2$
$= 6 \sqrt{2} π$ $= 6sqrt2pi$

What is the diameter of a circle with area of 18π18pi ft2ft^2?