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Mathematics 14 Online
OpenStudy (anonymous):

URGENT: VOLUME OF A PYRAMID!!!!

OpenStudy (anonymous):

http://lmgtfy.com/?q=volume+of+a+pyramid

OpenStudy (anonymous):

its a complicated problem

Directrix (directrix):

@haley321 Please post the problem.

OpenStudy (anonymous):

A square pyramid has a lateral area of 2,684 cm squared. If the base has a side length of 22cm, what is the volume o the pyramid? (Hint:Find the slant height first)

Directrix (directrix):

Let me find a diagram.

OpenStudy (anonymous):

ok

Directrix (directrix):

Directrix (directrix):

If the lateral area is 2684, then each of the four congruent triangles forming the lateral area has individual area 2684/4 = 671. Do you agree?

OpenStudy (anonymous):

yes i got that

Directrix (directrix):

Directrix (directrix):

The slant height l (l as in "ell") of the pyramid is the altitude of an individual triangular face.

Directrix (directrix):

So, solve this equation for L to get the slant height of the regular pyramid 671 = (1/2)*22*L

Directrix (directrix):

We need the slant height to get the height of the pyramid for its volume.

Directrix (directrix):

@haley321 L = ?

OpenStudy (anonymous):

i dont know

Directrix (directrix):

(1/2)*22*L = 671 11L = 671 What number times 11 equals 671?

OpenStudy (anonymous):

61

OpenStudy (anonymous):

how did you get that eqaution

Directrix (directrix):

The area of a triangle is one-half its base times the altitude drawn to that base.

Directrix (directrix):

Look at this diagram above: regularsquarebasepyramid1.jpg

OpenStudy (anonymous):

ok i understand, how do i find the volume

Directrix (directrix):

Now, to get the altitude of the pyramid. h^2 + 11^2 = 61^2 --> Pythagorean Theorem Solve that for h.

Directrix (directrix):

Super cool. The sides are a Pythagorean Triple. (11, 60, 61)

OpenStudy (anonymous):

h=60

OpenStudy (anonymous):

61

Directrix (directrix):

The last equation to solve: Volume of Pyramid = (1/3) * B * h where B is the area of the base of the pyramid and h is the altitude of the pyramid.

Directrix (directrix):

V = (1/3) * 22 * 22 * 60 --> Note: 22*22 is the area of the square base. V = ?

OpenStudy (anonymous):

is this right ? v=14,762 cm cubed

OpenStudy (anonymous):

i mean 9,841.3

Directrix (directrix):

I did not get that answer. Would you post your calculations? Thanks.

Directrix (directrix):

I did not get that. Maybe I am the one who should post the calculation.

Directrix (directrix):

(1/3) * 22 * 22 * 60 = (1/3) * 60 * 22 * 22 = 20 * 22 * 22 = ?

Directrix (directrix):

@haley321 20 * 22 * 22 = ?

OpenStudy (anonymous):

22*22=484 which is the base, multiply by 61 and divide by two.

Directrix (directrix):

And the final answer is ?

Directrix (directrix):

The volume of the pyramid is ?

OpenStudy (anonymous):

9,841.3

Directrix (directrix):

@hayley321 Why are you multiplying by 61? 61 is the slant height, not the height of the pyramid. Look at the diagram.

Directrix (directrix):

Volume = (1/3) * 22 * 22 * 60 NOT (1/3) * 22 * 22 * 61 --> NO

OpenStudy (anonymous):

ok thanks

Directrix (directrix):

And, the answer is ?

Directrix (directrix):

And, the answer is ?

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