Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m then find the sides of the two squares.

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 14 and the other, 8.

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.

two data sets with the same median will not have the same varaibility

step-by-step explanation:

step-by-step explanation: