Question

Helen’s house is located on a rectangular lot that is 1 1/8 miles by 9/10 mile.Estimate the distance around the lot

Answer 1

Answer:

The distance around the lot is 4.05 miles .

Step-by-step explanation:

Formula

$Perimeter\ of\ rectangle = 2 (Length + Breadth)$

As given

$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ 1 \frac{1}{8}\ miles\ by\ \frac{9}{10}\ mile.$

i.e

$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ \frac{9}{8}\ miles\ by\ \frac{9}{10}\ mile.$

Here

$Length = \frac{9}{8}\ miles$

$Breadth = \frac{9}{10}\ miles$

Put in the formula

$Perimeter\ of\ rectangle = 2(\frac{9}{8}+\frac{9}{10})$

L.C.M of (8,10) = 40

$Perimeter\ of\ rectangle = \frac{2\times (9\times 5+9\times 4)}{40}$

$Perimeter\ of\ rectangle = \frac{2\times (45+36)}{40}$

$Perimeter\ of\ rectangle = \frac{2\times 81}{40}$

$Perimeter\ of\ rectangle = \frac{81}{20}$

Perimeter of rectangle = 4.05 miles

Therefore the  distance around the lot is 4.05 miles .

Answer 2

Answer:

4.05 miles approximately.

Step-by-step explanation:

The problem is asking for the perimeter of the rectangular figure which is defines as

$P=2(l+w)$

Where $l=1\frac{1}{8}=\frac{9}{8}$ and $w=\frac{9}{10}$

Replacing this values, we have

$P=2(\frac{9}{8}+\frac{9}{10})\\P=2(\frac{90+72}{80})=2(\frac{162}{80})=4.05$

Therefore, the distance around the lot, the rectangular figure is 4.05 miles approximately.