The area of a square is 45 more than its perimeter. Find the length of a side.
Please say how you got your answer as well.
2006-11-01 11:09:21 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
A = P +45
S^2 = A
(the area of a square is length times width, which the length and width are the same)
4S = P
(the perimeter of a square is the length of all sides added together)
so... if you plug S^2 into the first equation for A and 4S into the second equation for P:
S^2 = 4S + 45
now solve for S:
S^2 - 4S - 45 = 0
(S - 9) (S + 5) = 0
S = 9, S = -5
since the length of something cant be negative, the only answer is 9
The length of one of the sides is 9
2006-11-01 11:17:43 · answer #1 · answered by Anonymous · 0⤊ 0⤋
A=s^2
P=4s
Therefore s^2= 4s +45
s^2-4s-45 =0
(s-9)(s+5)=0
s=9 or s =-5 S=-5 is rejected because s can not be negative.
Therefore, s=9
That means area = 9^2 = 81 and perimeter = 9*4 =36
A-P 81-36 =45 as requested.
2006-11-01 19:18:19 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
A=s^2
p=4s
A=p+45
s^2=4s+45
s^2-4s-45=0
(s-9)(s+5)=0
s=9
s=-5
Since s>0
s=9
check
9^2=81
4*9=36
81/36=45
2006-11-01 19:12:34 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
Let the length of the side be x.
Then area is x^2 and perimeter is 4x.
The statement is x^2 = 4x + 45,
and I guess you already know how to solve this since you've now gone on to problems using quadratic equations. Only thing is, of your two answers, one is negative. Use just the positive one.
2006-11-01 19:13:11 · answer #4 · answered by Hy 7 · 0⤊ 0⤋
Perimeter = 4s
Area = s^2
s^2 = 4s + 45
s^2 - 4s - 45 = 0
(s-9)(s+5) = 0
s = 9 or -5 (reject, doesn't work for side)
So each side = 9.
2006-11-01 19:13:14 · answer #5 · answered by just♪wondering 7 · 0⤊ 0⤋