Two squares have an average perimeter of 36 and an average area of 90. What is the side length of the larger square?
side of first square ---- x
side of 2nd square ----- y
(4x + 4y)/2 = 36
x+y = 18 --- y = 18-x
(x^2 + y^2)/2 = 90
x^2 + y^2 = 180
x^2 + (18-x)^2 = 180
x^2 + 324 - 36x + x^2 = 180
x^2 - 18x + 72 = 0
(x - 12)(x - 6) = 0
x=12 or x=6
if x=12, then y = 6
if x = 6 , then y = 12
we have symmetric solutions
the side of the larger square is 12
The perimeter of two squares is 58in. The side length of the larger square is twice the side length of the smaller square. What is is the length of one side of the smaller square?
To find the side length of the larger square, we need to set up a system of equations using the given information.
Let's denote the side lengths of the two squares as s1 and s2, where s1 is the side length of the larger square.
We know that the average perimeter of the two squares is 36. The perimeter of a square is given by P = 4s, where s is the side length. So, we can write the equation:
(Perimeter of s1 + Perimeter of s2)/2 = (4s1 + 4s2)/2 = 36
Simplifying the equation, we get:
2s1 + 2s2 = 36
We also know that the average area of the two squares is 90. The area of a square is given by A = s^2. So, we can write the equation:
(Area of s1 + Area of s2)/2 = (s1^2 + s2^2)/2 = 90
Simplifying the equation, we get:
s1^2 + s2^2 = 180
Now, we have a system of equations:
2s1 + 2s2 = 36 --(1)
s1^2 + s2^2 = 180 --(2)
To solve this system of equations, we can use substitution or elimination.
Let's use elimination. Multiply equation (1) by 2 and subtract it from equation (2):
4s1 + 4s2 = 72
-(2s1 + 2s2 = 36)
------------------
2s1 = 36
Dividing both sides of the equation by 2, we get:
s1 = 18
Therefore, the side length of the larger square (s1) is 18 units.