Question

Please help immediately

Answer 1

Let a = arc distance of the sector
let r = radius of the circle


a + 2r = 8
a = (8-2r)

Find the portion of the circle in the sector (arc/circumference)

a/2r •r

Replace a with ( 8-2r)

(8-2r/2•square root • r)

Simply

(8-r/square root •r)

( 4-r) / square root R


Find the area of the sector
(4-r)/square root •r • square root •r ^2

Cancel pi*r


A = (4-r) * r
A = -r^2 + 4r

Max of this quadratic equation occurs at the axis of symmetry; x=-b/(2a)


R=-4/2•(-1)

R= 2 cm

Find the max area

A = -2^2 + 8(2)
A = -4+ 16
A = 12 sq cm max area