The sides AB and AD of a square are extended 10 cm and 6 cm. to form sides AE and AF of a rectangle. At most how long is the side of the square if the perimeter of the rectangle is at least twice the perimeter of the square? Please show equation and solve
2006-09-11 15:45:17 · 5 answers · asked by somerandomdude2006 1 in Science & Mathematics ➔ Mathematics
Perimeter of square: 4x
Perim. of rectangle: 2(x+10) + 2(x+6) = 4x + 32
4x+32 >= 2(4x)
32 >=4x
8>=x
x<=8
The side of the square is at most 8 cm long.
2006-09-11 15:52:36 · answer #1 · answered by jenh42002 7 · 0⤊ 0⤋
Perimeter of square (s) = 2(AB+AD) = 2(x)
Perimeter of rectangle (r) = 2(AE+AF) = 2((AB+10)+(AD+6)) = 2(AB+AD+16) = 2(x+16) = 32+2(x)
r > 2s
32 + 2x > 2(2x)
solve for x
2006-09-11 22:55:09 · answer #2 · answered by Anonymous · 0⤊ 0⤋
s = length of any side of square
Perimeter of Rectangle = 2*(s+6)+2*(s+10) = 4s+32
Perimiter of Square = 4s
4s+32 >= 2*4s
4s<=32
s<=8
2006-09-11 22:56:54 · answer #3 · answered by Andy S 6 · 0⤊ 0⤋
I'm against inequality, it's in the Constitution.
2006-09-11 22:49:33 · answer #4 · answered by Anonymous · 1⤊ 0⤋
http://surl.in/HLSCI231793MBXZTQH
2006-09-11 22:47:21 · answer #5 · answered by manoj k 2 · 0⤊ 0⤋