Asked by Alyssa

Babylonian Problem - circa 1800 BC

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?
Answered by Marth
let x = the side of the 1st square

let 2x/3-10 = the side of the 2nd square ("10 less than 2/3 of the side of the other square")

The area of a square is x^2. Therefore, if the sum of the areas = 1000, than
x^2 + (2x/3-10)^2 = 1000

foil the polynomial:

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

You can simplify it and use the quadriatic equation to solve. The numbers do not seem very friendly though - is this a calculator problem? (If so, just use the graphing function to find solutions.)
Answered by Mike
Side of square 1 is 30

Side of square 2 is 10

Answered by roy
peter is twice a old as paul was when peter was asold as paul now. the combined ages of peter and paul is 56 years. how old are peter and paul now?
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