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ahrayia [7]
4 years ago
11

A farmer is enclosing a rectangular area for a pigpen. He wants the length of the pen to be 15 Ft longer then the width. The far

mer can only afford no more than 150Ft of fencing. Want is the pen’s greatest possible length?? (((((Solve the problem and SHOW ALL WORK...EXPLAIN in DETAIL, How you got each answer??)))) Thanks so much
Mathematics
1 answer:
agasfer [191]4 years ago
7 0
Let the width = x.
The length is 15 ft more than the width, so the length is x + 15.
The perimeter is less than or equal to 150 ft.

Perimeter formula:
P = 2L + 2W

The perimeter must be less than or equal to 150 ft.
2L + 2W \le 150

Now we solve for x, the width.

2(x + 15) + 2x \le 150

2x + 30 + 2x \le 150

4x + 30 \le 150

4x \le 120

x \le 30

x is at most 30 ft.
x is the width.
The length is x + 15, so the length is 30 + 15 = 45.
The length is at most 45 ft.
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Using the same pattern we have  

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1)

Break up the irregular shape into two rectangles

12 * 4.5 = 54
2 * 5 = 10
54 + 10 = 64 cm^2

2)

Break up the irregular shape into a triangle and rectangle

24 * 8 = 192

To get the base of the triangle:
24 - 6 - 6 = 12
To get the height of the triangle:
16 - 8 = 8

1/2(12 * 8) = 48

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Separate into triangle and semi circle

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(pi (8)^2)/2 = 100.5

120 + 100.5 = 220.5 cm^2

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Separate half circle from rectangle

(pi (7.5)^2)/2 = 88.4

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(7+9/2)(3.6) = 28.8

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