# Pyramid's height

• December 3rd 2010, 05:50 AM
regdude
Pyramid's height
Hi!
A pyramid has parallelogram at its bottom, the surface area of it is 360 cm^2, but the area of pyramids surface (without the parallelogram) is 768 cm^2. Edges or parallelogram is 20 and 36. Need to calculate the height of pyramid. Said also that the hight is where both parallelogram's diagonals cross. I only know the answer - 12 cm.
http://img139.imageshack.us/img139/3777/pyr0.jpg
(E on top)
So the area of ABCD = 360 cm^2, Area of AEB + BEC + CED + DEA = 768 cm^2
AB = 20 cm, BC = 36 cm. EF is the height.
Sadly I didn't get far, found a way to get diagonals length (both) and found that it could be really easy if I somehow got EB, then I get 2 right triangles and easy to get the height. Can't figure out what to do with the surface area...
found that the area is 2 times area of both side triangles, but found a problem that I can't get the height of the triangle and there is no connection with the parallelogram.
• December 3rd 2010, 08:54 AM
regdude
If it's a rectangle then why isn't it's area 360 cm^2?
The area is given btw.
Since EB and EC isn't the same size because DB and AC isn't, then the X doesn't cut BC in half.
• December 3rd 2010, 09:38 AM
earboth
Quote:

Originally Posted by regdude
If it's a rectangle then why isn't it's area 360 cm^2?
The area is given btw.
Since EB and EC isn't the same size because DB and AC isn't, then the X doesn't cut BC in half.

You are right: I've made a fatal mistake at the very beginning. Sorry!
• December 3rd 2010, 10:32 AM
Opalg
First, look at the parallelogram that forms the base of the pyramid. Area of ||gram = base times vertical height. If the base AB is 36cm and the area is 360 cm^2 then the height is 10cm. The point F where the diagonals intersect is in the centre of the parallelogram, so the perpendicular distance FX (in the attached figure) from F to AB is 5cm.

Similarly, if we take BC as the base, then the "vertical" height will be 18cm, so the perpendicular distance FY from F to BC is 9cm.

Now look at the diagram of the pyramid. Let T denote the point at the top of the pyramid and let h denote the height FT. The (sloping) height of the triangle BTC is TY. But from Pythagoras in the triangle TFY it follows that $TY = \sqrt{h^2+9^2}$. So the are of triangle BTC (and also of triangle DTA) is $\frac12(20)\sqrt{h^2+9^2}.$ Similarly, the area of triangle ATB and CTD is $\frac12(36)\sqrt{h^2+5^2}$.

Thus the total area of the sloping sides of the pyramid is $20\sqrt{h^2+9^2} + 36\sqrt{h^2+5^2} = 768.$ Solve that equation to get $h=12$.
• December 3rd 2010, 10:49 AM
regdude
No wonder I couldn't figure it out by my self, I thought the "vertical" height didn't go trough F.
The rule that allows us to connect these slopes with ||gram heights is the "3 perpendicular line rule"? (sorry if written wrong, don't know how to translate it right)
Thanks :)