A friend of mine came to me with a problem earlier today and I'm unsure if the answer I arrived at was correct. I simply found a formula on wolfram and used the provided values to work backwards to a solution. I know there has to be an easier way to solve this but I'm stumped on it and any help would be appreciated. The person I was helping has already turned in the assignment so it won't be aiding someone surreptitiously if that is a concern and I'll change the specific values given in the original problem. So anyway the problem goes:
You are given two octagonal pyramids with height 8. The distance from the base point of pyramid A to the midpoint of one of its sides is 6, and the distance between the base point of pyramid B and the midpoint of one of its sides is 15. The surface area of pyramid A is 306.4, what is the surface area of pyramid B?
Edit: sorry I didn't include my attempt at this initially.
The method I used started with a formula I found on wolfram alpha for the surface area of an octagonal pyramid. That formula is
$2 s (\sqrt{4 h^2 + s^2 cot^2 (π/8)} + s cot(π/8))$
Where h is the height and s is the side length of one of the sides of the base octagon. I plugged in the height, and equated the whole formula to the surface area of pyramid A to get the side length s. I then used the fact that the resulting right triangles made from the base octagons of either pyramid should be similar hence we can find the side length of the base octagon of pyramid B by using:
$b/15 = a / 6$
where a is $1/2$ the side length of the base octagon of pyramid A and b is $1/2$ the side length of the base octagon of pyramid B.
I used this second formula to find the value of b:
since
$ a = (383/2)/\sqrt{5*(320 + 383 \sqrt{3 + 2 \sqrt{2}}}) = 2.428$
then
$ b = 15*a/6 $
so plugging $2 * b$ in for s in our original formula gives the surface area of pyramid B as
SA = 1521.78
So by my logic that should be correct but I can't imagine that's how the problem is meant to be solved, nor do I think my answer is correct, though that's simply because I threw together formulas until I got an answer and that leaves me no way to verify my answer. Basically my hope it for some answer that bypasses the use of these formulas and is more straightforward. Any help would be greatly appreciated. Thank you for your time!
$\endgroup$ 82 Answers
$\begingroup$Just to see if your answer is correct I've taken a slightly different tack.
The area of a regular octagon with side length $a$ is given by $A_{base}=2(1+\sqrt2)a^2 $
The area of the sides of a pyramid is given by $A_{sides}=pl/2$ where $p$ is the perimeter of the base and $l$ is the slant height of the pyramid. For an octagonal pyramid this becomes $A_{sides}=4al$, with $a$ being the side length of the octagon.
For pyramid $A$, $l_A=\sqrt{8^2+6^2}=10$
Putting this all together gives $TSA_A=2a_A^2(1+\sqrt 2)+40a_A=306.4$
and solving for $a$ gives $a_A\approx 4.836$ (There is another solution but it is negative and can be ignored). This value more or less agrees with the above ($a$ here is the whole side, not half as above).
For pyramid $B$, $l_B=\sqrt{8^2+15^2}=17$. Also, $a_B=\frac {15} 6 a_A$
Again, putting this all together gives $TSA_B=\frac{25} 2 a_A^2(\sqrt 2+1)+170a_A$
Substituting for $a_A$ gives $TSA_B\approx 1528.084$ which is not that different to above - rounding errors maybe?
This is still a lot of work. If the original question were asking for the surface areas of just the sides of the pyramids, then it would be a more straightforward ratio problem.
$\endgroup$ 2 $\begingroup$I think there is a problem with this question, as the given area for pyramid A is incorrect (assuming a right pyramid, i.e. the apex above the centre of a regular octagon). Nevertheless, I'll solve this problem using that incorrect figure using as few calculations as possible.
In pyramid A, the octagonal base can be split into 8 isosceles triangles, with height 6. Their base is an edge of the pyramid, and it has length L, which I will not need to calculate. The other faces of the pyramid are also eight triangles with base L, but with height $\sqrt{6^2+8^2}=10$.
Therefore the area of the pyramid is split between the base and the other eight faces proportional to the height of the triangles - the area of the octagonal base of A is $\frac{6}{10+6}*306.4$ and the total area of the other eight sides is $\frac{10}{10+6}*306.4$.
In pyramid B, the octagonal base is scaled relative to A by a factor of $\frac{15}{6}$, so the area of the base is scaled by that squared, and must have area $\frac{15^2}{6^2}\frac{6}{10+6}*306.4 = 718.125$. The other eight faces have a base length that is scaled by that factor $\frac{15}{6}$ compared to pyramid A, but their height is now $\sqrt{15^2+8^2}=17$, which is $\frac{17}{10}$ times its previous value. The area of these eight faces is therefore $\frac{15}{6}\frac{17}{10}\frac{10}{10+6}*306.4 = 813.875$. This gives a total area for pyramid B of $1532.0$.
With regards to the incorrect given area:
The value of L is $2*6\tan(22.5) \approx 4.97$. The area of pyramid A should therefore be $8*\frac{6L}{2}+8*\frac{10L}{2} = 64L \approx 318.116$, and not $306.4$. You can adjust the values for B's area accordingly.
