Answer: x=14 and y=8
*Note: x and y are used as variables to solve the problem. Just know that the sides for one square is <u>14</u> and the other, <u>8</u>.
Step-by-step explanation:
For this problem, we can use system of equations to solve. Let's use x for one side of a square and y for the other.
Equation 1:
x²+y²=260
We get this equation from the sum of the areas. The x² and y² are from the area. Since all sides of a square are equal lengths, we can directly square them.
Equation 2:
4x-4y=24
This equation comes from the difference of the perimeters. 4x and 4y are perimeters because perimeter is adding all sides of the square together. There are 4 sides to a square, therefore we get 4x and 4y.
We can use substitution method to solve.
4x-4y=24 [add both sides by 4y]
4x=24+4y [divide both sides by 4]
x=6+y
Now that we have x, we can plug it into any equation and solve.
(6+y)²+y²=260 [expand]
y²+12y+36+y²=260 [combine like terms]
2y²+12y+36=260 [subtract both sides by 36]
2y²+12y=224 [factor out 2]
2(y²+6y)=224 [divide both sides by 2]
y²+6y=112 [subtract both sides by 112]
y²+6y-112=0 [factor equation]
(y-8)(y+14)=0 [set each factor equal to 0 to solve]
y=8 and y=-14
We know that y=8 because (y+14)=0 gives you y=-14, but a length or area can NEVER be negative, only positive.
Now that we have y, we can plug it into any equation to find x.
4x-4(8)=24 [combine like terms]
4x-32=24 [add both sides by 32]
4x=56 [divide both sides by 4]
x=14
Now, we have x and y. x=14 and y=8.
, so that we rewrite the above as
of the line chosen, which means the limit is path-dependent and thus does not exist.