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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: enter image description here

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.

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

1 Answer 1

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enter image description here

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 $$\tan\beta=\frac{1}{6}=\frac{h}{10}\\ \implies h=\frac{5}{3}$$

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  • $\begingroup$ Thank you very much!! So answer will be 5/3 * 108= 180 $\endgroup$
    – err or
    Sep 4, 2020 at 17:57
  • $\begingroup$ @error, volume is $\frac{1}{3}Ah$. So answer is $60$, as given in the answer key. $\endgroup$
    – SarGe
    Sep 5, 2020 at 1:35

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