Maths
posted by stressed girl on .
OPQRS is a right pyramid whose base is a square of sides 12cm each. Given that the slant height of the pyramid is 15cm. Find the height of the pyramid. The volume of the pyramid and the total surface of the pyramid.

Since 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~ :)