Answer:
The distance around the lot is 4.05 miles .
Step-by-step explanation:
Formula
[tex]Perimeter\ of\ rectangle = 2 (Length + Breadth)[/tex]
As given
[tex]Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ 1 \frac{1}{8}\ miles\ by\ \frac{9}{10}\ mile.[/tex]
i.e
[tex]Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ \frac{9}{8}\ miles\ by\ \frac{9}{10}\ mile.[/tex]
Here
[tex]Length = \frac{9}{8}\ miles[/tex]
[tex]Breadth = \frac{9}{10}\ miles[/tex]
Put in the formula
[tex]Perimeter\ of\ rectangle = 2(\frac{9}{8}+\frac{9}{10})[/tex]
L.C.M of (8,10) = 40
[tex]Perimeter\ of\ rectangle = \frac{2\times (9\times 5+9\times 4)}{40}[/tex]
[tex]Perimeter\ of\ rectangle = \frac{2\times (45+36)}{40}[/tex]
[tex]Perimeter\ of\ rectangle = \frac{2\times 81}{40}[/tex]
[tex]Perimeter\ of\ rectangle = \frac{81}{20}[/tex]
Perimeter of rectangle = 4.05 miles
Therefore the distance around the lot is 4.05 miles .
Answer:
Step-by-step explanation:
The problem is asking for the perimeter of the rectangular figure which is defines as
[tex]P=2(l+w)[/tex]
Where [tex]l=1\frac{1}{8}=\frac{9}{8}[/tex] and [tex]w=\frac{9}{10}[/tex]
Replacing this values, we have
[tex]P=2(\frac{9}{8}+\frac{9}{10})\\P=2(\frac{90+72}{80})=2(\frac{162}{80})=4.05[/tex]
Therefore, the distance around the lot, the rectangular figure is 4.05 miles approximately.
