The length of a side of the sqaure is ______ units.
2006-09-07 05:52:46 · 13 answers · asked by Fries 4 in Science & Mathematics ➔ Mathematics
7
7*7=49
7+7+7+7=28
dif = 21
2006-09-07 05:54:33 · answer #1 · answered by Good luck chief 3 · 0⤊ 1⤋
7
2006-09-07 05:58:54 · answer #2 · answered by Carol G 2 · 0⤊ 1⤋
7
2006-09-07 05:57:02 · answer #3 · answered by Anonymous · 0⤊ 1⤋
The length of each side of this square is: 7 units.
Consider the following:
Area of a square is given by (s)(s)=s^2
Perimeter is given by s1+s2+s3+s4= 4s
Since: s1=s2=s3=s4
We're given:
Area of this square is 21 more than its perimeter 4s
i
ts obvious from here. Just plug n play
s^2=21+4s
grouping gives s^2-4s-21=0
factoring gives (s+3)(s-7)=0
so s=--3 or s=7
Therefore s=7
since s=-3 is nonsensical (negative length)
Check this value by plugging it into the given eq.
7^2=21+4(7)
49=49
Magic huh! U got 2 love the graceful simplicity of math
2006-09-07 06:45:08 · answer #4 · answered by mydcmbrgirl 2 · 1⤊ 0⤋
Side of the square = 7
2006-09-07 05:57:43 · answer #5 · answered by Anonymous · 0⤊ 1⤋
The Area of a square is 21 more than its perimeter.
LENGTH*LENGTH=4*LENGTH+21
LENGTH^2-4*LENGTH-21=0
LENGTH^2-7*LENGTH+3*LENGTH -21=0
LENGTH(LENGTH-7)+3(LENGTH-7)=0
SO, LENGTH=(7,-3)
SO, THE LENGTH(SIDE)=7,WHICH WHEN SQUARED WILL BE 21 MORE THAN ITS PERIMETER(28).
2006-09-07 08:20:33 · answer #6 · answered by Anonymous · 0⤊ 0⤋
A=P+21
S=P/4
A=S^2
To Solve:
S^2=S4+21
0=-S^2+S4+21
0=S^2-S4-21
0=(S-7)(S+3)
S=7;S=-3
Since I like positive numbers to express distances, The length of a side is 7 units.
2006-09-07 06:38:44 · answer #7 · answered by Scott S 4 · 0⤊ 0⤋
x (squared)=4x + 21
==> x= 7
2006-09-07 06:07:46 · answer #8 · answered by Antsan 2 · 0⤊ 0⤋
s^2=4s+21
s^2-4s-21=0
(s-7)(s+3)=0
the length of a side will be 7.
2006-09-07 05:55:46 · answer #9 · answered by dan 4 · 1⤊ 0⤋
x^2=4x+21
x^2-4x-21=0
(x-7)(x+3)=0
x-7=0 x+3=0
x=7 x=-3
7 is correct answer
obviously -3 cannot be correct
proof: 7^2=4(7)+21
49=28+21
49=49
2006-09-07 06:06:40 · answer #10 · answered by kjfabre 2 · 0⤊ 2⤋