So, I was trying to solve this problem from a textbook. The Problem is: "A piece of wire 60 inches long is cut into two pieces and then each piece is bent into the shape of a square. If the sum of the areas of the two squares is 117 square inches, find the length of each piece of wire."
And this is the solution.
"A piece of wire 60 inches is cut into two pieces and then each piece was bent into the shape of a square"
$4x + 4y = 60$
$x + y = 15$
$y = (15-x)$:
"the sum of the areas of the two square is 117 square inches"
$x^2 + (15-x)^2 = 117$
$(15-x)(15-x)+x^2=
x^2 + 225 – 15x – 15x + x^2 = 117$
combine like terms
$2x^2 – 30x + 225 – 117 = 0$
$2x^2 – 30x + 108 = 0$
simplify divide by 2
$x^2 – 15x + 54 = 0$.
Factors to
$x = 6$ and
$x = 9$
therefore
4(6) = 24" is one piece of wire
and
4(9)= 36" is the other
But something doesn't make sense here, If we bend the original 60-inch wire into a square. Then it's area would be 400 square inches.
So why do the two smaller squares area equal, 117 square inches and not 400?
Best Answer
If you bend the $60$-inch wire into a square, the area should be $\left( \frac{60}4\right)^2=225$
In general, we have
$$\left( p+q\right)^2 \ne p^2+q^2$$
Hence $p=\frac{24}4$ and $q=\frac{36}{4}$
We have $(p+q)^2 = p^2+2pq+2^2$. Hence there should be a difference of $2pq=2\left( \frac{24}{4}\right)\left(\frac{36}{4} \right)=108.$
Geometrically: Don't forget the rectangle region.