SadakoMoose
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I have an area of 150 square feet and a perimeter of 62.
How do I determine the dimensions?
How do I determine the dimensions?
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Determining Dimension from Area and Perimeter
- Thread starter SadakoMoose
- Start date
SadakoMoose said:I have an area of 150 square feet and a perimeter of 62.
How do I determine the dimensions?
You don't say what kind of figure this is! Is it a triangle? Is it a rectangle? Is it something else?
Without that information, I don't think we can help you.
Hello, SadakoMoose!
I will assume that the figure is a rectangle.
Let \(\displaystyle L\) = length, \(\displaystyle W\) = width.
\(\displaystyle \text{We have: }\;\begin{array}{cccc}LW &=& 150 & [1] \\ 2L + 2W &=& 62 & [2] \end{array}\)
\(\displaystyle \text{From [2], we have: }\:W \:=\:31-L\)
\(\displaystyle \text{Substitute into [1]: }\;L(31-L) \:=\:150 \quad\Rightarrow\quad L^2 - 31L + 150 \:=\:0\)
\(\displaystyle \text{Factor: }\;(L-6)(L-25) \:=\:0 \quad\Rightarrow\quad L \:=\:6,\:25\)
\(\displaystyle \text{Assuming that: }\,L\,>\,W\text{, we have: }\;L = 25,\;W = 6\)
I have an area of 150 square feet and a perimeter of 62.
How do I determine the dimensions?
I will assume that the figure is a rectangle.
Let \(\displaystyle L\) = length, \(\displaystyle W\) = width.
\(\displaystyle \text{We have: }\;\begin{array}{cccc}LW &=& 150 & [1] \\ 2L + 2W &=& 62 & [2] \end{array}\)
\(\displaystyle \text{From [2], we have: }\:W \:=\:31-L\)
\(\displaystyle \text{Substitute into [1]: }\;L(31-L) \:=\:150 \quad\Rightarrow\quad L^2 - 31L + 150 \:=\:0\)
\(\displaystyle \text{Factor: }\;(L-6)(L-25) \:=\:0 \quad\Rightarrow\quad L \:=\:6,\:25\)
\(\displaystyle \text{Assuming that: }\,L\,>\,W\text{, we have: }\;L = 25,\;W = 6\)
SadakoMoose
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- Oct 9, 2009
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And those are the dimensions of the rectangle?soroban said:Hello, SadakoMoose!
I have an area of 150 square feet and a perimeter of 62.
How do I determine the dimensions?
I will assume that the figure is a rectangle.
Let \(\displaystyle L\) = length, \(\displaystyle W\) = width.
\(\displaystyle \text{We have: }\;\begin{array}{cccc}LW &=& 150 & [1] \\ 2L + 2W &=& 62 & [2] \end{array}\)
\(\displaystyle \text{From [2], we have: }\:W \:=\:31-L\)
\(\displaystyle \text{Substitute into [1]: }\;L(31-L) \:=\:150 \quad\Rightarrow\quad L^2 - 31L + 150 \:=\:0\)
\(\displaystyle \text{Factor: }\;(L-6)(L-25) \:=\:0 \quad\Rightarrow\quad L \:=\:6,\:25\)
\(\displaystyle \text{Assuming that: }\,L\,>\,W\text{, we have: }\;L = 25,\;W = 6\)
yep,
25 feet long and 6 feet high,
perimeter = combined length of all four sides = 25+6+25+6=31+31=62 feet long,
which is the distance you would travel if you walked around the touchline.
The area is the number of squares that fit inside the rectangle.
These squares have sides of length 1 foot.
25 of these will line up along the base,
and you have 6 rows of these 25.
6 multiplied by 25 is 150 squares,
if the shape is rectangular. Is it ?
25 feet long and 6 feet high,
perimeter = combined length of all four sides = 25+6+25+6=31+31=62 feet long,
which is the distance you would travel if you walked around the touchline.
The area is the number of squares that fit inside the rectangle.
These squares have sides of length 1 foot.
25 of these will line up along the base,
and you have 6 rows of these 25.
6 multiplied by 25 is 150 squares,
if the shape is rectangular. Is it ?