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13. tirzah wants to put a fence around her garden. she has 22 yards of …

Question

  1. tirzah wants to put a fence around her garden. she has 22 yards of fence material. does she have enough to go all the way around the garden? explain why or why not. tirzahs garden 4\frac{2}{3} yards

Explanation:

Response

Assuming the garden is a rectangle (since it's a common garden shape) and from the visible side length of \(4\frac{2}{3}\) yards, we need to know the other side. Wait, maybe it's a rectangle with length and width? Wait, maybe the garden is a rectangle with length \(6\frac{3}{4}\) yards (maybe the other side is partially visible? Wait, maybe the garden is a rectangle with length \(6\frac{3}{4}\) yards and width \(4\frac{2}{3}\) yards? Let's assume that (since it's a common problem setup). So to find the perimeter of a rectangle, the formula is \(P = 2\times (l + w)\), where \(l\) is length and \(w\) is width.

Step1: Identify length and width

Let's assume the length \(l = 6\frac{3}{4}\) yards and width \(w = 4\frac{2}{3}\) yards (common for such problems, as the visible width is \(4\frac{2}{3}\) and maybe length is \(6\frac{3}{4}\) from typical problems). First, convert mixed numbers to improper fractions. \(6\frac{3}{4}=\frac{6\times4 + 3}{4}=\frac{27}{4}\), \(4\frac{2}{3}=\frac{4\times3+2}{3}=\frac{14}{3}\).

Step2: Calculate \(l + w\)

Find a common denominator for \(\frac{27}{4}\) and \(\frac{14}{3}\), which is 12. \(\frac{27}{4}=\frac{27\times3}{4\times3}=\frac{81}{12}\), \(\frac{14}{3}=\frac{14\times4}{3\times4}=\frac{56}{12}\). Then \(l + w=\frac{81}{12}+\frac{56}{12}=\frac{137}{12}\) yards.

Step3: Calculate perimeter \(P\)

Perimeter \(P = 2\times\frac{137}{12}=\frac{137}{6}\approx22.83\) yards (or convert to mixed number: \(\frac{137}{6}=22\frac{5}{6}\) yards).

Step4: Compare with 22 yards

Tirzah has 22 yards of fence. The perimeter of the garden is \(22\frac{5}{6}\) yards, which is more than 22 yards.

Answer:

No, she does not have enough. The perimeter of the garden (assuming length \(6\frac{3}{4}\) yd and width \(4\frac{2}{3}\) yd) is \(2\times(6\frac{3}{4}+4\frac{2}{3}) = 2\times(\frac{27}{4}+\frac{14}{3}) = 2\times(\frac{81 + 56}{12}) = 2\times\frac{137}{12}=\frac{137}{6}=22\frac{5}{6}\) yards, which is greater than 22 yards. So she doesn't have enough fence material.