The city wants to turn a triangular plot of land into a park. Grass seed will be

Question

The city wants to turn a triangular plot of land into a park. Grass seed will be purchased to cover the triangular plot. It’s estimated that about 87 pounds of seed will be needed for each acre. Fine gravel will be laid down outside the boundary of the plot to create a running path that’s three feet wide and 2 inches deep.

You’re asked to determine approximately how many pounds of seed and how many cubic yards of gravel will be needed. You are given that two sides of the plot of land were measured to be 173 yards and 212 yards and that the angle between them was measured to be 55 .

Round the seed needed up to the nearest ten pounds and the volume of gravel up to the nearest cubic yard.

HINT: Since we just need an approximation here, don’t over complicate the calculation for the volume of the path. A computation of (total length)×(width)×(depth) should do, where ‘total length’ is the perimeter of the park.

Explanation / Answer

given two sides of the plot of land were measured to be 173 yards and 212 yards and that the angle between them was measured to be 55o .

area of the plot,A=(1/2)*173*212*sin55o

=>A=15021.61 square yards

1 acre = 4840 square yards

=>A=15021.61/4840 acre

87 pounds of seed will be needed for each acre

amount of seed needed = (15021.61/4840)*87

amount of seed needed = 270

270 pounds of seed is needed

by law of cosines ,

length of third side = [1732+2122-2*173*212*cos55o]

length of third side =181.1077614

1 feet = (1/3) yard , 1inch =(1/36) yard

volume of the gravel =(181.1077614+173+212)*(3*(1/3))*(2*(1/36))

volume of the gravel =31.45

volume of gravel =31 cubic yard

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