# 3d Geometry- The base of the pyramid has a trapezoid

The base of the pyramid has a trapezoid whose diagonal is perpendicular to the side, and an $$\alpha$$ angle with the base. The height of the trapezoid is equal to $$h$$. Each lateral edge of the pyramid forms a $$\beta$$ angle with the plane of the base. Find the volume of the pyramid if $$h = 6,\quad \tan α =\frac{1}{3},\quad \tan β = \frac{1}{6}$$

My solution:

But the answer is $$60$$. I've tried solving when trapezoid is isosceles, because I couldn't get anything with regular trapezoid.
I would be very pleased if you help me. Thank you.

• Height $6$ is of trapezoid and not of pyramid. Sep 1, 2020 at 16:27
• HINT: The projection of the vertex on the base must be at the same distance from all vertices of the trapezoid. Sep 2, 2020 at 17:21
• I've tried it again but didn't get right answer can you check it? Sep 3, 2020 at 17:59
• @error, it is not correct. Use hint given Intelligenti pauca. Sep 4, 2020 at 2:21
• So OT projection must be at the center of DKLC rectangle? Sep 4, 2020 at 6:21

Let the vertex $$A$$ be origin. So, the equation of diagonal $$AC$$ will be $$y=\frac{x}{3}$$. Given that $$CF=6$$, we get $$C=(18,6)$$.
Now, we can find equation of $$BC$$ which is perpendicular to $$AC$$. From this, we get $$B=(20,0)$$. And from $$A, B, C$$, we get $$D=(2,18)$$.
It is given that all the lateral sides make equal angle with the base. This is possible only when the distance between the vertices and the projection of apex $$(E)$$ are equal i.e. $$AE=BE=CE=DE$$.
So, for $$AE=DE$$, $$E$$ should lie on perpendicular bisector of $$AD$$. Same for $$BC$$. On solving we get $$E=(10,0)$$.
To find the height of the pyramid, we've $$\tan\beta=\frac{1}{6}=\frac{h}{10}\\ \implies h=\frac{5}{3}$$
• @error, volume is $\frac{1}{3}Ah$. So answer is $60$, as given in the answer key. Sep 5, 2020 at 1:35