Question

The sum of the areas of two squares is 640 $m^2$. If the difference in their perimeters is 64 m, find the sides of the two squares.

Hint:

If the difference in perimeters is $P_{diff}$, then the difference in sides is $P_{diff}/4$. This simplifies the problem immediately to $x - y = 16$.

Answer 1

Step 1: Understanding the Concept:
Area of a square $= s^2$ and Perimeter $= 4s$. We are given two conditions involving two unknown side lengths.
Step 2: Key Formula or Approach:
Let the sides of the two squares be $x$ and $y$ (with $x>y$).
1. $x^2 + y^2 = 640$
2. $4x - 4y = 64 \implies x - y = 16$
Step 3: Detailed Explanation:
From the second equation, $x = y + 16$.
Substitute this value of $x$ into the first equation:
\[ (y + 16)^2 + y^2 = 640 \]
\[ y^2 + 32y + 256 + y^2 = 640 \]
\[ 2y^2 + 32y - 384 = 0 \]
Divide by 2:
\[ y^2 + 16y - 192 = 0 \]
Factor the equation (numbers with product 192 and difference 16 are 24 and 8):
\[ y^2 + 24y - 8y - 192 = 0 \]
\[ y(y + 24) - 8(y + 24) = 0 \]
\[ (y + 24)(y - 8) = 0 \]
Since side length cannot be negative, $y = 8$.
Then $x = y + 16 = 8 + 16 = 24$.
Step 4: Final Answer:
The sides of the two squares are 24 m and 8 m.