The area of a field shaped like a right triangle is 750yd^2.One leg of the field is fenced with wood costing $5/yd. the remainde
r of the perimeter of the field is fenced with steel costing $10/yd. the perimeter of the field is 150yd. The total cost of the fencing is $1200. A- What is the length of the leg fenced with wood?
B- What is the length of the leg fenced with steel?
Assume that a and b = the two legs of the right triangle. c = the hypotenuse.
The area of the right triangle is 750 yd², therefore (1/2)*a*b = 750 ab = 1500 (1)
The perimeter is 150yd, therefore a + b + c = 150 (2)
Let the side fenced with wood be a, at $5/yd. Sides b and c are fenced with steel at $10/y. The total cost is $1200, therefore 5a + 10b + 10c = 1200 or a + 2b + 2c = 240 (3)
From (2), obtain c = 150 - a - b (4)
Substitute (4) into (3) a + 2b + 2(150 - a - b) = 240 -a + 300 = 240 a = 60
From (1), obtain 60b = 1500 b = 25
From (4), obtain c = 150 - 60 - 25 = 65
Answer: A. The length of the leg fenced with wood is 60 yd. B. The length of the leg fenced with steel is 25 yd.
Marisa had rounded 3.48 to 4 and times it to 7.33 which she rounded to 8
In fact 3.48 is closer to 3 we know this since the .48 of 3.48 s below .50 if it was .50 or above it would round to 4 but it doesn't the same applies to 7.33 so
