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 Given Information Clearly 
We are given two squares with different side lengths. 
Let the side of the larger square be x and the side of the smaller square be y, where x > y. 

We are given two conditions: 
1️⃣ The sum of their areas is 640 m². 
Since area of a square = side², \[ x^2 + y^2 = 640 \] 2️⃣ The difference of their perimeters is 64 m. 
Perimeter of square = 4 × side \[ 4x - 4y = 64 \] Factor out 4: \[ 4(x - y) = 64 \] \[ x - y = 16 \] 
Step 2: Express One Variable in Terms of the Other 
From: \[ x - y = 16 \] \[ x = y + 16 \] 
Step 3: Substitute into the Area Equation 
Substitute \( x = y + 16 \) into: \[ x^2 + y^2 = 640 \] \[ (y + 16)^2 + y^2 = 640 \] Expand: \[ y^2 + 32y + 256 + y^2 = 640 \] Combine like terms: \[ 2y^2 + 32y + 256 = 640 \] Move 640 to left: \[ 2y^2 + 32y + 256 - 640 = 0 \] \[ 2y^2 + 32y - 384 = 0 \] 
Step 4: Simplify the Quadratic Equation 
Divide entire equation by 2: \[ y^2 + 16y - 192 = 0 \] Now we factorise. We need two numbers: • Product = −192 • Sum = 16 These numbers are 24 and −8. \[ y^2 + 24y - 8y - 192 = 0 \] \[ y(y + 24) - 8(y + 24) = 0 \] \[ (y + 24)(y - 8) = 0 \] 
Step 5: Solve for y 
\[ y + 24 = 0 \Rightarrow y = -24 \] \[ y - 8 = 0 \Rightarrow y = 8 \] Since side length cannot be negative, \[ y = 8 \] 
Step 6: Find x 
\[ x = y + 16 \] \[ x = 8 + 16 \] \[ x = 24 \] 
Step 7: Final Answer 
\[ \boxed{\text{The sides of the two squares are } 24 \text{ m and } 8 \text{ m}} \] 
Extra Verification (Always Good in Exams) 
Area check: \[ 24^2 + 8^2 = 576 + 64 = 640 \quad ✔ \] Perimeter difference: \[ 4(24) - 4(8) = 96 - 32 = 64 \quad ✔ \] Both conditions satisfied.