Consider a solid whose base is a circle of radius r, and which has all cross-sections perpendicular to a particular diameter as equilateral triangles.

Find an expression for A(x), the cross-sectional area of the slice.

We did this question in class, so I have the answer, but we went through it really fast so I didn't get it. Now, doing it again, I'm not getting the right answer.

What I did: A typical slice has sides 2√(r²-x²)
A(x) = 1/2 * base * height
base = 2√(r²-x²)
height, I used sine rule and got 2√3 √(r²-x²)
Which gives me A(x) = 2√3 (r² - x²)
It should be √3 (r² - x²)
Help please?

Question

Consider a solid whose base is a circle of radius r, and which has all cross-sections perpendicular to a particular diameter as equilateral triangles.

Find an expression for A(x), the cross-sectional area of the slice.

We did this question in class, so I have the answer, but we went through it really fast so I didn't get it. Now, doing it again, I'm not getting the right answer.

What I did: A typical slice has sides 2√(r²-x²)
A(x) = 1/2 * base * height
base = 2√(r²-x²)
height, I used sine rule and got 2√3 √(r²-x²)
Which gives me A(x) = 2√3 (r² - x²) 
It should be √3 (r² - x²)
Help please?

anonymous · Answer

what is x?

watchmath · Answer

The height should be $\sqrt{3}\sqrt{r^2-x^2}$. Remember you only use half of the base when you do the sine rule.