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anonymous

  • 4 years ago

Find the volume for the regular pyramid. *drawing included*

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  1. anonymous
    • 4 years ago
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    |dw:1327696893067:dw|

  2. Mertsj
    • 4 years ago
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    V=1/3(area of base)(height)

  3. Mertsj
    • 4 years ago
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    Find the area of the base and multiply by the height which it looks like might be 3

  4. anonymous
    • 4 years ago
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    |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.

  5. anonymous
    • 4 years ago
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    ah hold on a sec...this is your base triangle isn't it?

  6. anonymous
    • 4 years ago
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    |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.

  7. anonymous
    • 4 years ago
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    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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