The base of a pyramid is a right triangle with the longest side = 12cm. All non base edges are also equal to 12cm. And how can I find the height of the pyramid?
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You have an equilateral triangle, so all angles are 60 and $\sin(60) = \text{???}$. – Christopher K Jun 24 '14 at 19:26
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I don't think this has enough information to solve. If it's a 45-45-90 right triangle, then the height will be smaller than if it is a say 1-89-90 right triangle. – Peter Woolfitt Jun 24 '14 at 19:43
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@Peter Woolfitt - there is enough information - for the arbitrary points of a right triangle for a given hypotonusa lie on a circle of which the hypothenusa is the diameter... – johannesvalks Jun 24 '14 at 20:04
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1@johannesvalks Ah yes, I see now. Thanks for your excellent answer. – Peter Woolfitt Jun 24 '14 at 20:14
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Use Thales' theorem.
You can select the points
$$ A = (-6,0,0), B = (+6,0,0), $$
to form the longest side of the base triangle. The other point $C$ lies on a circle of which $AB$ is the diameter, as
$$ y^2 + \big[ 6 - x \big]^2 + y^2 + \big[ 6 + x \big]^2 = 144, $$
so
$$ x^2 + y^2 = 36 $$
The top point $T$ is given by $(0,0,z)$ and it is clear that
$$ |AT| = |BT| = |CT| $$
as $A$, $B$ and $C$ lie on the same circle, as $|AT| = |BT| = |AB|$, we obtain
$$ z = 6 \sqrt{3} $$
johannesvalks
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2The realization that the top point is over the origin is key. Well done. – Ross Millikan Jun 24 '14 at 20:07
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@Ross Millikan Thanks - the image above is also the top view where $O$ and the top $T$ are the same as seen from above... – johannesvalks Jun 24 '14 at 20:14