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March 28, 2015

March 28, 2015

Posted by **stressed girl** on Monday, July 8, 2013 at 5:22am.

- Maths -
**Elena**, Monday, July 8, 2013 at 5:57amh=sqrt{H²-(a/2)²} =

=sqrt{15²-6²} = 13.75 cm,

V=ha²/3 =13.75•12²/3=660 cm³,

S= 4•(ha/2) =2•15•12 =360 cm².

- Maths -
**Jai**, Monday, July 8, 2013 at 6:22amSince we cannot draw the figure here, you can draw the figure yourself.

Note that a right pyramid is a pyramid with a square base and the apex (the tip of the pyramid) is aligned directly at the center of the base.

Now, draw an line from the apex to the center of the base, which serves as the height of the pyramid. Then connect the center of the base to one of the midpoints of the side of the square. The slant height is from that midpoint to the apex.

Note that you have formed a right triangle, where the height is unknown, and the hypotenuse (which is also the slant height) is 15 cm. You can solve for the base of the right triangle by getting the half of the side of the square, which is 6 cm.

Solving for height, recall pythagorean theorem. For any right triangle,

a^2 + b^2 = c^2

where

a = base of right triangle

b = height of right triangle

c = hypotenuse

Substituting,

6^2 + b^2 = 15^2

36 + b^2 = 225

b^2 = 189

b = 3*sqrt(21) cm [height]

Recall that the volume of ANY pyramid is just

V = (Area of the base)*(height)/3

*Area of the base = (length of side of square)^2

V = (12^2)*[3*sqrt(21)]/3

V = 144*sqrt(21) cm^3

Recall that the SA of a SQUARE pyramid is just

SA = 2bs + b^2

where

s = slant height

b = length of side of square

SA = 2*12*15 + 12^2

SA = 504 cm^2

Hope this helps~ :)

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