anonymous
  • anonymous
Find the volume for the regular pyramid. *drawing included*
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1327696893067:dw|
Mertsj
  • Mertsj
V=1/3(area of base)(height)
Mertsj
  • Mertsj
Find the area of the base and multiply by the height which it looks like might be 3

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anonymous
  • anonymous
|dw:1327697482033:dw| |dw:1327697605284:dw| The area of the cross section at height y is given by \[A_y = B\left(1-\frac{y}{h}\right)^2\] To fid the volume of the pyramid, you integrate between the y values 0 and h: \[Vol = \int_{0}^{h}B\left(1-\frac{y}{h}\right)^2dy = \frac{B}{h^2}\int_0^h (h-y)^2dy\] \[=\frac{B}{h^2} \int_{0}^{h}(h^2 - 2hy + y^2)dy = \frac{B}{h^2}(h^2y - hy^2 + \frac{y^3}{3}) \vert_0^h\] \[ = \frac{B}{h^2}\left(\frac{h^3}{3}\right) = \frac{1}{3}Bh\] So just sub in your values for the area of the base, B = 2x2/2 = 4/2=2 and for h which is 3 to find: Vol = 6/3 = 2 cubic units.
anonymous
  • anonymous
ah hold on a sec...this is your base triangle isn't it?
anonymous
  • anonymous
|dw:1327698598539:dw| but we cannot find the area unless we know that it is an equilateral, as we are forced to solve one equation in two unknowns if it is an isosceles.
anonymous
  • anonymous
if it is an equilateral, then \[x=2\] and \[y=\sqrt{3}\] and the area of the base is \[\frac{1}{2}2\sqrt{3} = \sqrt{3}\] Then the volume of the pyramid is \[\frac{1}{3}(3)\sqrt{3} = \sqrt{3}\]

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