A wire 260 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures have the same area, what are the lengths of the two pieces of wire to the nearest tenth of an inch?
a
211.1 in. and 48.9 in.
b
137.8 in. and 122.2 in.
c
82.7 in. and 177.3 in.
d
142.0 in. and 118.0 in.
I think it might be c. 82.7 in and 177.3 in. Am I right?
2 answers
Can anyone help me with this?
let the radius of the circle be r
perimeter of circle = 2πr
so the piece left over for the square = 260 - 2πr
and each piece will be (65 - πr/2)
area of circle = πr^2
area of square = (65 - πr/2)^2
= 4225 - 65πr + π^2 r^2/4
4225 - 65πr + π^2 r^2/4 = πr^2
16900 - 260πr + π^2 r^2 = 4πR^2
3πr^2 + 260πr - 16900 = 0
solve for r
a rather messy quadratic, I leave it up to you to solve
or
you could try the numbers.
let's try yours:
Since the circle gives the largest area for a given perimeter, the smaller number should be the radius of the circle
r = 82.7, area = π(82.7)^2 = appr 21486
side = 177.3 , area = 177.3^2 = 31435
so , nope , your answer is incorrect
perimeter of circle = 2πr
so the piece left over for the square = 260 - 2πr
and each piece will be (65 - πr/2)
area of circle = πr^2
area of square = (65 - πr/2)^2
= 4225 - 65πr + π^2 r^2/4
4225 - 65πr + π^2 r^2/4 = πr^2
16900 - 260πr + π^2 r^2 = 4πR^2
3πr^2 + 260πr - 16900 = 0
solve for r
a rather messy quadratic, I leave it up to you to solve
or
you could try the numbers.
let's try yours:
Since the circle gives the largest area for a given perimeter, the smaller number should be the radius of the circle
r = 82.7, area = π(82.7)^2 = appr 21486
side = 177.3 , area = 177.3^2 = 31435
so , nope , your answer is incorrect
4 answers