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anonymous
 4 years ago
Find the volume for the regular pyramid.
*drawing included*
anonymous
 4 years ago
Find the volume for the regular pyramid. *drawing included*

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327696893067:dw

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1V=1/3(area of base)(height)

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Find the area of the base and multiply by the height which it looks like might be 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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: \[Vol = \int_{0}^{h}B\left(1\frac{y}{h}\right)^2dy = \frac{B}{h^2}\int_0^h (hy)^2dy\] \[=\frac{B}{h^2} \int_{0}^{h}(h^2  2hy + y^2)dy = \frac{B}{h^2}(h^2y  hy^2 + \frac{y^3}{3}) \vert_0^h\] \[ = \frac{B}{h^2}\left(\frac{h^3}{3}\right) = \frac{1}{3}Bh\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0ah hold on a sec...this is your base triangle isn't it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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
 4 years ago
Best ResponseYou've already chosen the best response.0if 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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