# solution-verification | Calculate the sine of the angle between two side faces of a trunk

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

Drawing

$$a)$$ 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!

• Enough with the trivial edits, already. Commented Jul 8 at 12:55
• Your answers for the lateral area and the sine of the angle are correct. Good work. (+1) Commented Jul 8 at 14:31
• @mathlove What is the info you are talking about? I used the fact that the trapezoids can be circumscribed Commented Jul 8 at 16:55
• @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. Commented Jul 9 at 11:33
• @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." Commented Jul 9 at 12:06

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$$\sin\angle DXB=2\cdot\frac{3}{4}\cdot \frac{\sqrt7}{4}=\frac{3\sqrt 7}{8}$$

• Thanks for you answer! So the discussion in the comments isn't right, because still cant make the difference between source 1 and 2? Commented Jul 10 at 10:47
• 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….” Commented Jul 10 at 13:04

$$V=\frac{2h}{3}A_B-\frac h3 A_b.$$
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 $$\cos\theta=-\frac18$$, $$\sin\theta=\frac{3\sqrt7}8.$$