The problem

Let the trunk be a regular quadrilateral pyramid $ABCDA'B'C'D'$ with the side of the large base of $8$ cm and the side of the small base of $4$ cm. The lateral faces are isosceles trapezoids that can be circumscribed in a circle.

$a)$ Determine the lateral area and the volume of the pyramid trunk

$b)$ Calculate the sine of the angle of two side faces of the trunk

My solution


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$a)$ As you can see I drew the whole pyramid including its peak $V$.

Because the lateral faces are isosceles trapezoids that can be circumscribed in a circle we get that $AA'=\frac{B+b}{2}=6$ cm

Then he can calculate the height of the trapezoids $h=4\sqrt{2}$

In the right angled trapezoid $A'O'OA$ we can calculate $OO'=2\sqrt {7}$

Now we have everything to calculate the lateral area and the volume of the trunk

$A_l= \frac{(P_B+P_b)*h}{2}=\frac{48*4*\sqrt{2}}{2}= 96\sqrt{2}$

$V= \frac{h}{3}*(A_B+A_b+\sqrt{A_b+A_B})=\frac{2\sqrt{7}*112}{3}=\frac{224\sqrt{7}}{3}$

For point $b)$ we have to find the sine of the angle between $(VBC)$ and $(VDC)$

I let $BX \perp VC$ and by the congruence of the triangles $DXC$ and $BXC$ we get that $DX \perp VC$ so the angle we look for is $\angle DXC$

Triangle is isosceles with $DX=BX=\frac{16\sqrt{2}}{3}$ and $DB=8\sqrt{2}$ so $OX=\frac{4\sqrt{14}}{3}$ so we can express the area of triangle $DXB$ in $2$ ways the find the sin of that angle and I get that $sin= \frac{3\sqrt{7}}{8}$

I put this as a solution verification because I'm not sure if this part is right, Because the lateral faces are isosceles trapezoids that can be circumscribed in a circle we get that $AA'=\frac{B+b}{2}=6$ cm".

Also, I'm not sure if my calculus and idea for point $b)$ are right

I hope one of you can help me! Thank you!

  • 1
    $\begingroup$ Enough with the trivial edits, already. $\endgroup$ Commented Jul 8 at 12:55
  • 1
    $\begingroup$ Your answers for the lateral area and the sine of the angle are correct. Good work. (+1) $\endgroup$
    – Quadrics
    Commented Jul 8 at 14:31
  • $\begingroup$ @mathlove What is the info you are talking about? I used the fact that the trapezoids can be circumscribed $\endgroup$ Commented Jul 8 at 16:55
  • $\begingroup$ @mathlove Oo.... i thought it the wrong way, tho corect me if I'm wrong, but in the 2nd source, it says that the quadrilateral is circumscribed. $\endgroup$ Commented Jul 9 at 11:33
  • $\begingroup$ @mathlove This is what i was reffering to : ,, Since these quadrilaterals can be drawn surrounding or circumscribing their incircles, they have also been called circumscribable quadrilaterals, circumscribing quadrilaterals, and circumscriptible quadrilaterals." $\endgroup$ Commented Jul 9 at 12:06

2 Answers 2


Point $a)$ is correct. Since we are given that a circle is inscribable in the isosceles trapezoid, we know from the tangency that$$AA’=\frac{AB}{2}+\frac{A’B’}{2}=4+2=6$$and from this the lateral area and volume are as OP calculates.

Point $b)$ is also correct: since$$\sin\angle OXB=\frac{BO}{BX}=\frac{4\sqrt 2}{\frac{16\sqrt 2}{3}}=\frac{3}{4}$$and$$\cos\angle OXB=\frac{OX}{BX}=\frac{\frac{4\sqrt {14}}{3}}{\frac{16\sqrt 2}{3}}=\frac{\sqrt 7}{4}$$and$$\angle DXB=2\angle OXB$$and$$\sin2\theta=2\sin\theta\cos\theta$$then$$\sin\angle DXB=2\cdot\frac{3}{4}\cdot \frac{\sqrt7}{4}=\frac{3\sqrt 7}{8}$$

  • $\begingroup$ Thanks for you answer! So the discussion in the comments isn't right, because still cant make the difference between source 1 and 2? $\endgroup$ Commented Jul 10 at 10:47
  • 1
    $\begingroup$ Every isosceles trapezoid has a circumcircle, but here you have one that has an incircle too. This enables you to say (because of tangency) that $AA’=6$. I think your second sentence should be “Because the lateral faces are isosceles trapezoids that circumscribe a circle….” $\endgroup$ Commented Jul 10 at 13:04

Your numerical results are correct.

In point a, the volume formula is

$$V=\frac{2h}{3}A_B-\frac h3 A_b.$$

In point b, you could also find $\sin\theta$, $\theta=\angle DXB$, by using the formula $$\cos\theta=-\cot^2\alpha$$ where $\alpha$ is the base angle of the trapezoids.
$$\cot\alpha=\frac4{8\sqrt2}=\frac1{2\sqrt2}$$ hence $\cos\theta=-\frac18$, $\sin\theta=\frac{3\sqrt7}8.$


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