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geometry [need explanation]

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by snehit » Mon Jul 27, 2009 6:46 am
a pyramid with four equal-sized flat surfaces and a base of 36 feet square has height of 10 feet. What is the total surface area of the pyramid, excluding the base?
_
1. 12 \/31 [take this sign as root]

2. 12 \/109

3. 24\/31

4. 24\/109

5. 48\/31

Source: from some harvard program
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by vinayakdl » Mon Jul 27, 2009 10:20 am
I will go with 2: 12 \/109

(base)side of the pyramid triangle = sqrt(36) = 6

height of pyramid triangle = sqrt( 3^2 + 10^2) = V109

area of triangle surface = 1/2(6V109)= 3V109

total area = 4x3V109 = 12V109

Vinayak
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by prindaroy » Mon Jul 27, 2009 11:17 am
agree
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by snehit » Tue Jul 28, 2009 10:11 am
Not it is bit complicated
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Re: geometry [need explanation]

by maihuna » Tue Jul 28, 2009 11:14 am
snehit wrote:a pyramid with four equal-sized flat surfaces and a base of 36 feet square has height of 10 feet. What is the total surface area of the pyramid, excluding the base?
_
1. 12 \/31 [take this sign as root]

2. 12 \/109

3. 24\/31

4. 24\/109

5. 48\/31

Source: from some harvard program
Total surface area - base area = slant area of pyramid

Slant area of pyramid = (1/2)*Perimeter of base*Slant Ht

36 here is a base of square each side 6

so slant height will come as a triangle with side (6/2) 10 so slant ht = \/3^2+10^2 = \/109

so slant surface = (1/2)*4*6*\/109
72\/109
Charged up again to beat the beast :)
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by snehit » Wed Jul 29, 2009 9:38 am
you cannot find the side in that way its wrong. Hight wud not be similar for slant. And if you consider right triangle from middle of pyramid you create havoc. you wud get not area of slant but area of centre of the slant to centre of pyramid
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by maihuna » Wed Jul 29, 2009 10:02 am
snehit wrote:you cannot find the side in that way its wrong. Hight wud not be similar for slant. And if you consider right triangle from middle of pyramid you create havoc. you wud get not area of slant but area of centre of the slant to centre of pyramid
Boss in case you are referring to my mail, I am founding the slant ht correctly, actually it is perpendicular to one of the base edge so bisecting it, this combined with given ht will give us slant ht using Pythagoras.

Confusion is abt the shape of the base, what it is, is it an square with edge 6, from four sides I have infered it is an square, if you can clarify about the shape of the base more detail will emerge.
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