# An area, A, consisting of the sum of two squares is 1000. The side of one square is 10 less than 2/3 of the side of the other square. What are the sides of the squares?

## Let x,y be the side of each of the squares.

x^{2} + y^{2} = 1000

y=(2/3)x-10

substitute (2/3)x-10 for y in the first equation:

x^{2} + ((2/3)x-10)^{2} = 1000

Expand and solve for the quadratic equation to get x=30 or x=-270/13.

Reject second solution to get x=30, y=10

## Why did the square go to therapy? Because it had too many sides and couldn't make up its mind!

Let's solve this riddle and find the sides of the squares.

Let's call the side of one square "x" and the side of the other square "y".

According to the problem, the area A, which is the sum of the squares, is 1000.

So, we have:

x^2 + y^2 = 1000

The problem also states that the side of one square is 10 less than 2/3 of the side of the other square.

So we can set up another equation:

x = (2/3)y - 10

Now we have two equations:

x^2 + y^2 = 1000

x = (2/3)y - 10

We can substitute the value of x from the second equation into the first equation:

((2/3)y - 10)^2 + y^2 = 1000

Expanding and simplifying, we get:

(4/9)y^2 - 8/3 y + 100 + y^2 = 1000

Combining like terms:

(13/9)y^2 - 8/3 y + 100 - 1000 = 0

Simplifying further:

(13/9)y^2 - 8/3 y - 900 = 0

Now we can solve this quadratic equation to find the values of y, and then we can substitute those values back into the equation x = (2/3)y - 10 to find the corresponding x-values.

But hey, quadratic equations can be a bit serious, can't they? Let me tell you another joke while we solve it!

Why don't scientists trust atoms? Because they make up everything!

Okay, now let's solve that quadratic equation and find the sides of the squares!

## Let's assume the side of one square is x, and the side of the other square is (2/3)x - 10.

The area of the first square is x^2, and the area of the second square is [(2/3)x - 10]^2.

According to the problem, the sum of the two squares is 1000:

x^2 + [(2/3)x - 10]^2 = 1000

Expanding and simplifying the equation:

x^2 + [4/9x^2 - 8/3x + 100] = 1000

x^2 + 4/9x^2 - 8/3x + 100 = 1000

13/9x^2 - 8/3x + 100 = 1000

Now, we subtract 100 from both sides:

13/9x^2 - 8/3x = 1000 - 100

13/9x^2 - 8/3x = 900

Multiplying both sides by 9 to eliminate the fractions:

13x^2 - 24x = 900 * 9

13x^2 - 24x - 8100 = 0

Now, we can solve this quadratic equation.

## To find the sides of the squares, we can set up an equation based on the given information and solve for the variables.

Let's assume the side of one square is x, and the side of the other square is y.

We are given that the area, A, consisting of the sum of two squares is equal to 1000:

A = x^2 + y^2 ...(Equation 1)

Now, we are also given that the side of one square is 10 less than 2/3 of the side of the other square:

x = (2/3)y - 10 ...(Equation 2)

To solve for x and y, we can substitute the value of x from Equation 2 into Equation 1:

A = ((2/3)y - 10)^2 + y^2

Expanding this equation, we get:

1000 = (4/9)y^2 - (40/3)y + 100 + y^2

Combining like terms, we have:

1000 = (13/9)y^2 - (40/3)y + 100

Now, let's multiply through by 9 to eliminate the fraction:

9000 = 13y^2 - 120y + 900

Rearranging this equation, we have a quadratic equation:

13y^2 - 120y + 9000 - 900 = 0

Simplifying further:

13y^2 - 120y + 8100 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.

After solving the quadratic equation, we can substitute the value of y back into Equation 2 to find the value of x.

Finally, we will have the sides of the two squares, x and y, which solve the given conditions.