The rectangular part of a field is 110 yards long, and the diameter of each semicircle is 42 yards. Find the cost of fertilizing the field at $0.70 per square yard.
2007-06-12 18:10:30 · 4 answers · asked by vanessandavid 1 in Science & Mathematics ➔ Mathematics
area of rectangle (L times W) = 110•42 = 4260
area of circle (πr²) = (3.1416)(21)² = 1385.44
total area 6005.44 yd²
times $0.70 / yd² = $4203.81
2007-06-12 18:15:50 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
First, visualize the field. It is composed of a rectangle with two semicircles on each end. The length of the rectangle is 110 yards, and since the diameter of the semicircle is the same as the width of the rectangle, we can say that the area of the rectangle is 4,620 square yards (110 x 42). Next, calculate the area of the circle (two semi-circles of the same diameter make one circle). The area of a circle is pi times the radius squared. The radius of a circle is half the diameter, so in this case, it is 21 yards. 21 yards squared is 441 times pi is 1385.44236 square yards. Add this to the area of the rectangle (4,620) and we get approximately 6.005.44 square yards. Multiply this by the price per square yard, which is $0.70 and we get $4,203.81.
2007-06-13 01:35:23 · answer #2 · answered by an_ugly_ugly_girl 1 · 0⤊ 0⤋
Hi - I can't see a drawing (if there is one) but if the semicircle is inside the field and touches three sides of the field then the width of the field is 110 x 42 yards or 4,620.00 sq. yards. The fertilizing cost is thus 4620 x .7 = $3,234.00.
2007-06-13 01:24:43 · answer #3 · answered by Helpful 2 · 0⤊ 0⤋
The area to be fertilised can be broken up into a circle ( two semi-circles put together) and a rectangle. The are of the rectangle is 110x42=4620 sq. yards PLUS area of a circle= Pi*r^2. Are of the circle is Pi*21*21= 1385.4 sq yards. Total area is 4620+1385=6005. Cost to fertilise is 6005 * $0.70= $4203.50.
2007-06-13 01:16:53 · answer #4 · answered by Anonymous · 1⤊ 0⤋