Pythagoras 3-D question?

Thanks for an interesting question, but there was one thing you neglected to specify. I'll try both possibilities, because they'll give different answers.

Namely, it could be that

AB, CD, EF, and GH are the 9 cm sides of the bases;

and that BC, DA, FG, and HE are the 12 cm sides;

or the other way around.

OK, there is a right angle in triangle EBH -- EB is the diagonal of a side, and EH is an edge perpendicular to that side, and thus, to EB. So angle BEH is 90º.

In that triangle, angle B is angle EBH, with opposite side EH = 12 cm (given), and hypotenuse BH. To find BH, you can use Pythagoreas' Theorem twice, or you can do essentially the same thing, because BH is a main diagonal of the box, so

BH^2 = BA^2 + AE^2 + EH^2 = 12^2 + 20^2 + 9^2

Easiest way to see through to the result is regroup to form 3-4-5 right triangles:

9^2 + 12^2 = 15^2

15^2 + 20^2 = 25^2

so BH = 25 cm

So

... angle EBH = arcsin(12/25) = arcsin(0.48) = 28.6854º

If the 12 and 9-cm sides are the other way around, then you can still do essentially the same things, with now side EH = 9 cm, and arrive at

... angle EBH = arcsin(9/25) = arcsin(0.36) = 21.1002º

>> base and top are 12cm x 9cm <<

Which are which?

assuming AB = 12 and BC = 9

angle HEF and HEA are 90 degrees

Point B is on the same plane as AEF, so...

Angle HEB = 90 degrees

EH = 9

EB = sqrt(20^2 + 12^2)

BH = sqrt(20^2 + 12^2 + 9^2)

sin(EBH) = ( 9 / sqrt(20^2 + 12^2 + 9^2) )

and

cos(EBH) = ( sqrt(20^2 + 12^2) / sqrt(20^2 + 12^2 + 9^2) )

and

tan(EBH) = ( 9 / sqrt(20^2 + 12^2) )

You can use the Law of Cosines.

BE = √(12^2 + 20^2) = √(144 + 400) = √544

BH = √(12^2 + 20^2 + 9^2) = √(144 + 400 + 81) = √625 = 25

Call the angle BEH x.

EH^2 = BE^2 + BH^2 - 2(BE)(BH) cos x

9^2 = 544 + 625 - 2(√544)(25) cos x

81 = 1169 - 50√544 cos x

-1088 = -50√544 cos x

0.9329523032... = cos x

x ≈ 21° 6'

The quantity of triangles (all assumed to be equilateral) making up the perimeters of the pyramid is principal -- it might be 3, 4, or 5. Assume that it's 4. Then, a diagonal around the base could have period a sqrt(two), and the gap from nook to middle of that line shall be a sqrt(two)/two. You now have a proper triange, with base the part diagonal, hypotenuse is a, and the peak h to be discovered. Pythagoras regulations!

1. angle EAB is Rt angle Triagle.

so use it. & trigonometry.

EB = √(20²+12²) = 23.32

Now, EB = 23.32, EH = √( 20²+12² +9²) = 25.

In triangle EBH, sides are 20, 25, 9

using cos (theta) = ( 23.32²+12² -9²) / (2*23.32*25)

theta = 21.075 deg. Ans