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Sagot :
Let P1 represent the perimeter of the larger square and let P2 represent the perimeter of the smaller square. Since the piece of wire is 32 cm long and both squares are made of this wire, we have the following:
[tex]P_1+P_2=32[/tex]Now let x be the length of the side of the smaller square. Since the larger square has sides 4cm longer than the length of the side of the other square, we have the following:
[tex]\begin{gathered} P_1=4(x+4) \\ P_2=4x \end{gathered}[/tex]using these expressions on the first equation and solving for x, we get:
[tex]\begin{gathered} 4(x+4)+4x=32 \\ \Rightarrow4x+16+4x=32 \\ \Rightarrow8x=32-16=16 \\ \Rightarrow x=\frac{16}{8}=2 \\ x=2 \end{gathered}[/tex]we have that x = 2. Then, the length of the shorter piece of wire will be the perimeter of the smaller square, therefore, the length of the shorter piece of wire is P2 = 4(2) = 8 cm
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