QUESTION IMAGE
Question
- how many cubic yards of aggregate must be ordered for a road base 16 in. thick and 2 miles long with a top width of 30 ft, if the side slopes are one on one, or 45°?
Step1: Convert units
First, convert all units to yards.
- Thickness \(h = 16\) inches. Since \(1\) yard \(= 36\) inches, \(h=\frac{16}{36}=\frac{4}{9}\) yards.
- Length \(L = 2\) miles. Since \(1\) mile \(= 1760\) yards, \(L = 2\times1760=3520\) yards.
- Top - width \(b_1=30\) feet. Since \(1\) yard \(= 3\) feet, \(b_1 = 10\) yards.
Because the side - slopes are \(1:1\) (or \(45^{\circ}\)), the bottom - width \(b_2=b_1 + 2h\). Substituting \(h=\frac{4}{9}\) yards, we get \(b_2=10+\frac{8}{9}=\frac{90 + 8}{9}=\frac{98}{9}\) yards.
Step2: Calculate the cross - sectional area
The cross - section of the road base is a trapezoid. The area formula for a trapezoid is \(A=\frac{(b_1 + b_2)}{2}\times h\).
Substitute \(b_1 = 10\) yards, \(b_2=\frac{98}{9}\) yards and \(h=\frac{4}{9}\) yards into the formula:
Step3: Calculate the volume
The volume \(V\) of the road base (a prism) is \(V = A\times L\).
Substitute \(A=\frac{752}{162}\) square yards and \(L = 3520\) yards into the formula:
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Approximately \(16340\) cubic yards of aggregate must be ordered.