Where did I make a mistake? Calculate b using only surface area and a.

[Ana.stasia]

I tried looking up how to say this pyramid in English and I found some sources calling it the "a regular fringed pyramid" however I am not sure If this is correct. I drew it so that should be enough, however, please inform me of the correct name.

The question: The angle between the side edge (what I marked as s) and the bottom base's edge (what I marked as a) is 60 degrees. Calculate the top base's edge (what I marked as b) only using bottom base's edge and P (P is the the surface of the pyramid).

This is how I did it:



My solution is very similar to the one from the book. Only where I have 2a², they have 3a² and where I have 2P they have 4P. Where did I make a mistake?

Thank you in advance

 

[skeeter]

truncated pyramid, frustrum of a pyramid, or in this case, the frustrum of a regular tetrahedron

total surface area = top area + bottom area + area of 3 trapezoidal sides

[MATH]P = \dfrac{\sqrt{3}}{4}(a^2 + b^2) + \dfrac{3s\sqrt{3}}{4}(a+b) = \dfrac{\sqrt{3}}{4}\bigg[(a^2+b^2) + 3s(a+b)\bigg] [/MATH]
[MATH]b = a-s \implies s = a-b[/MATH]
[MATH]P = \dfrac{\sqrt{3}}{4}\bigg[(a^2+b^2) + 3(a-b)(a+b)\bigg][/MATH]
[MATH]P = \dfrac{\sqrt{3}}{4}\bigg[(a^2+b^2) + 3(a^2-b^2)\bigg][/MATH]
[MATH]P = \dfrac{\sqrt{3}}{2}\bigg[2a^2- b^2 \bigg][/MATH]
[MATH]b = \sqrt{2a^2 - \dfrac{2P}{\sqrt{3}}}[/MATH]
I agree w/your solution

 

[Ana.stasia]

truncated pyramid, frustrum of a pyramid, or in this case, the frustrum of a regular tetrahedron

total surface area = top area + bottom area + area of 3 trapezoidal sides

[MATH]P = \dfrac{\sqrt{3}}{4}(a^2 + b^2) + \dfrac{3s\sqrt{3}}{4}(a+b) = \dfrac{\sqrt{3}}{4}\bigg[(a^2+b^2) + 3s(a+b)\bigg] [/MATH]
[MATH]b = a-s \implies s = a-b[/MATH]
[MATH]P = \dfrac{\sqrt{3}}{4}\bigg[(a^2+b^2) + 3(a-b)(a+b)\bigg][/MATH]
[MATH]P = \dfrac{\sqrt{3}}{4}\bigg[(a^2+b^2) + 3(a^2-b^2)\bigg][/MATH]
[MATH]P = \dfrac{\sqrt{3}}{2}\bigg[2a^2- b^2 \bigg][/MATH]
[MATH]b = \sqrt{2a^2 - \dfrac{2P}{\sqrt{3}}}[/MATH]
I agree w/your solution

Thank you.
One question. I wrote that the angle 60 was between s and a. Now in the text it actually states that s and the base create an angle of 60 degrees. I thought this was the same thing which is why I wrote it like I did, however, due to the solutions not matching I am rethinking this. When It says that an edge that I marked as s creates an angle of 60 degrees with the base, is that angle also true for the angle between a and s because a is the edge of the base?

 

[skeeter]

One question. I wrote that the angle 60 was between s and a. Now in the text it actually states that s and the base create an angle of 60 degrees. I thought this was the same thing which is why I wrote it like I did, however, due to the solutions not matching I am rethinking this. When It says that an edge that I marked as s creates an angle of 60 degrees with the base, is that angle also true for the angle between a and s because a is the edge of the base?

back to the drawing board ... ¯\_(ツ)_/¯

 

[Ana.stasia]

back to the drawing board ... ¯\_(ツ)_/¯

 

[skeeter]

If the angle between [MATH]s[/MATH] and the base is [MATH]60^\circ[/MATH] , the angle between [MATH]s \text{ and } a = \arctan\left(\sqrt{\dfrac{13}{3}} \right) \approx 64.3^\circ[/MATH]
Left diagram is the base equilateral triangle with sides = to [MATH]a[/MATH]
Right diagram is a tetrahedron where edge [MATH]s[/MATH] makes a 60 degree angle to the base. Working with that value makes P a mess.

 

[Ana.stasia]

I need to calculate the slant height using a and b and knowing that s and the base make an angle of 60 degrees.
This is how I tried to do it.
However, I am suspicious of the result.

Is there a better way to do this?

 

[skeeter]