Active forum topics
- The Chain Rule Explained: Don't Just Memorize, Visualize It
- The Intuition Behind Integration by Parts (Proof & Example)
- Statics
- Calculus
- Hydraulics: Rotating Vessel
- Inverse Trigo
- Application of Differential Equation: Newton's Law of Cooling
- Problems in progression
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
New forum topics
- The Chain Rule Explained: Don't Just Memorize, Visualize It
- The Intuition Behind Integration by Parts (Proof & Example)
- Statics
- Calculus
- Hydraulics: Rotating Vessel
- Inverse Trigo
- Problems in progression
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
Recent comments
- I get it now, for long I was…4 days 19 hours ago
- Why is BD Tension?
is it not…4 days 19 hours ago - Bakit po nagmultiply ng 3/4…1 month 4 weeks ago
- Determine the least depth…11 months 3 weeks ago
- Solve mo ang h manually…1 month 4 weeks ago
- Paano kinuha yung height na…1 year ago
- It's the unit conversion…1 year ago
- Refer to the figure below…1 year ago
- where do you get the sqrt411 month 4 weeks ago
- Thank you so much1 month 4 weeks ago


To get the volume of the pile
To get the volume of the pile of ore, which looks like this:
We need to know the formula for the volume of the figure above. The formula applicable would be:
$$V = Bh$$
where $V$ is volume of the figure, $B$ is the area of base and $h$ is the height of the figure.
We notice that we don't know the area of the end bases of this pile of ore. We need to get it by using the formula to get the area of the triangle.
Getting the other sides of the pile of ore:
$$\cos 45^o = \frac{x}{60}$$ $$x = 30\sqrt{2} \space ft$$
Now getting the area of the ends of this pile of ore:
$$Area = \frac{1}{2}bh$$ $$Area = \frac{1}{2}(30\sqrt{2} \space feet)(30\sqrt{2} \space feet)$$ $$Area = 900 \space ft^2$$
Now we can get the volume of this pile of ore:
$$V= Bh$$ $$V = (900 \space ft^2)(500 \space ft)$$ $$V = 450000 \space ft^3$$
Therefore the volume of your pile of ore is $\color{green}{450000 \space ft^3}$
Alternate solutions are encouraged...heheheheh