(a)
Let:
w1 = Original width
l1 = Original length
w2 = New width
l2 = New length
A1 = Original area
A2 = New Area
so:
[tex]\begin{gathered} w2=3w1=3\cdot10=30ft \\ l2=2l1=2\cdot50=100ft \\ A2=w2\cdot l2=3000ft^2 \end{gathered}[/tex]
Answer:
New length: 100ft
New width: 30 ft
New Area: 3000ft²
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(b)
[tex]\frac{A2}{A1}=\frac{3000}{500}=6[/tex]
Answer:
The area of the new walkway will be 6 times the area of the current walkway
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(c)
[tex]\begin{gathered} 8\cdot500=x\cdot50\cdot4\cdot10 \\ 4000=2000x \end{gathered}[/tex]
Solve for x:
[tex]\begin{gathered} x=\frac{4000}{2000} \\ x=2 \end{gathered}[/tex]
Answer:
Make the new length 2 times the current length.
