I'm a bit confused with this one.
Cutting a Wire. A length of wire 16 inches is to be cut into two pieces, and then each piece will be bent to form a square. Find the length of the two pieces if the sum of the areas of the two squares is 10 square inches.
Thank you!.
John F.
Favorite Answer
First, make the first length of wire x and the second one 16 - x. Now convert the words to mathematical expressions. Since the two lengths of wire are bent to make squares, that must mean that they are bent into 4 equal portions. The portion of the first length is x/4 and the portion for the second length is (16 - x)/4).
When the length of one of these portions is squared, that gives you the area of a square. The sum of the areas of the two squares is 10. Set it up like this:
(x/4)^2 + ((16 - x)/4)^2 = 10
Now try to solve for x. First square the things inside the parenthesis,
(x^2/16) + ((256 - 32x + x^2)/16) = 10
You can combine the expressions because they have a common denominator of 16.
(2x^2 - 32x + 256)/16 = 10
Multiply both sides by 16 to cancel it out.
2x^2 - 32x + 256 = 160
Subtract 160 from both sides to get a quadratic equation.
2x^2 - 32x + 96 = 0
Factor out a 2 to make factoring the quadratic easier.
2(x^2 - 16x + 48) = 0
Now factor the quadratic and cancel out the 2 by dividing it.
(x - 12)(x - 4) = 0
x = 12, x = 4
The lengths of the two wires are 12 inches and 4 inches.
Hope this helped!
sv
pieces are 12 inches and 4 inches giving squares of 3 inch side and 1 inch side and areas 9 inch^2 and 1 inch^2 the sum being 10 inch^2.