# Help me with this geometry problem **URL:** \ **Category:** Factual Questions\ **Created:** [January 16, 2004, 3:50am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712 "2004-01-16T03:50:12Z")\ **Posts on this page:** 17\ **Page:** 1
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [January 16, 2004, 3:50am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/1 "2004-01-16T03:50:12Z")
… and don’t tell Ms. Flaherty, my ninth grade geometry teacher, that I couldn’t figure this out a mere 20 years later. Imagine a pyramid with a square base and each face is at a 45 degree slope toward the top of the structure. Call the width of each side of the base _a_, and the line connecting the corner of the base to the top of the pyramid _b_. Now each side of the pyramid is an isosceles triangle with a base of _a_ and two sides of _b_. If you look at this triangle, what is the length of _b_ in terms of _a_? Or, alternatively, what is the angle formed at _ab_? I know it has to be between 45 and 90 degrees, and I want to say it’s 67.5, but that doesn’t seem to be right. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [January 16, 2004, 4:14am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/2 "2004-01-16T04:14:45Z")
Let c be a line that connects the center of base a with the vertex b-b. As I work it out: c = a / sqrt(2) The Pythagorean theorm says that b^2 = c^2 + (a / 2)^2 Combining these, we get b = a \* sqrt(3) / 2 Does that look right? ---
**Author:** ![Rodrigo](https://avatars.discourse-cdn.com/v4/letter/r/4af34b/32.png) [@Rodrigo](https://boards.straightdope.com/u/Rodrigo)\ **Post date:** [January 16, 2004, 4:18am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/3 "2004-01-16T04:18:21Z")
a=b, thus the triangles are equilateral thus all angles are 60° * * * This is how. Lets call the centre of the base-square C, one corner A and the top B. Angle CAB is 45°. If base-sides are b, then line CA is b/(sqr rt 2). If CAB is 45°, then CA=CB=b/(sqr rt 2), thus AB=b Ergo All sides are equal. * * * I may be wrong ---
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [January 16, 2004, 4:19am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/4 "2004-01-16T04:19:53Z")
Hmmm. How did you arrive at the length of _c_? ---
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [January 16, 2004, 4:21am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/5 "2004-01-16T04:21:22Z")
Sorry, that question was for **Xema** not **Rodrigo**. ---
**Author:** ![Ringo](https://avatars.discourse-cdn.com/v4/letter/r/779978/32.png) [@Ringo](https://boards.straightdope.com/u/Ringo)\ **Post date:** [January 16, 2004, 4:26am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/6 "2004-01-16T04:26:42Z")
I _think_ **Xema** ’s right. My own quick pass produced: c as defined by **Xema** b[sup]2[/sup]=c[sup]2[/sup] + (½\*a)[sup]2[/sup] ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 16, 2004, 4:36am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/7 "2004-01-16T04:36:02Z")
The triangle consisting of the diagonal of the base has a side of length a\*sqrt(2), two sides of length b, and two 45 degree angles which are not opposite the diagonal. So the angle opposite the diagonal is a right angle, and the Pythagorean theorem applies: b[sup]2[/sup] + b[sup]2[/sup] = 2a[sup]2[/sup] Therefore, a = b. ---
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [January 16, 2004, 4:39am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/8 "2004-01-16T04:39:10Z")
Sorry, I misunderstood **Xema** to mean c connected to the center of the base of a side, when he meant the center of the square base. **Rodrigo’s** logic looks sound to me, though. ---
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [January 16, 2004, 4:41am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/9 "2004-01-16T04:41:06Z")
Thanks, all. When I was 14 I could have done that in my sleep. That’s what a career in Human Resources will do to you. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [January 16, 2004, 5:41am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/10 "2004-01-16T05:41:00Z")
Um… Maybe I’m missing something here, but people seem to have arrived a two different answers by making two different assumptions. If the sloped **faces** make 45-degree angles with the plane of the base, then **Xema** ’s analysis is correct and b = sqrt(3)\*a/2. If the **edges** of the sloped faces make 45-degree angles with the base, then **rodrigo** and **ultrafilter** are right and a = b. Ain’t solid geometry fun? ---
**Author:** ![sturmhauke](https://avatars.discourse-cdn.com/v4/letter/s/e47c2d/32.png) [@sturmhauke](https://boards.straightdope.com/u/sturmhauke)\ **Post date:** [January 16, 2004, 6:58am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/11 "2004-01-16T06:58:53Z")
If I admitted I understood ultrafilter’s sig, would that make me even geekier than I realized? What if I said I was going to answer the angle question with trig because I thought it would be easier? ---
**Author:** ![Urban\_Ranger](https://avatars.discourse-cdn.com/v4/letter/u/e9c0ed/32.png) [@Urban\_Ranger](https://boards.straightdope.com/u/Urban_Ranger)\ **Post date:** [January 16, 2004, 7:27am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/12 "2004-01-16T07:27:35Z")
Since each **face** is sloping up at a 45 degree, that means the height of the pyramid is the same as the distance from the middle of a base to the centre, at 1/2 a. The diagonal of the base is a_sqrt(2), half of that is a_sqrt(2)/2. Length of b is sqrt((1/2a)\*\*2+(a\*sqrt(2)/2)\*\*2) ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [January 16, 2004, 8:19am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/13 "2004-01-16T08:19:08Z")
My answer agrees with **Xema**. The height of the pyramid is a/2 so the length of a face from tip to the center of the base is a/sqrt(2). b is then sqrt(a[sup]2[/sup]/4 + a[sup]2[/sup]/2) = sqrt(3)\*a/2 ---
**Author:** ![Shade](https://avatars.discourse-cdn.com/v4/letter/s/2bfe46/32.png) [@Shade](https://boards.straightdope.com/u/Shade)\ **Post date:** [January 16, 2004, 12:52pm UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/14 "2004-01-16T12:52:38Z")
> [@sturmhauke](#): > > If I admitted I understood ultrafilter’s sig, would that make me even geekier than I realized? What if I said I was going to answer the angle question with trig because I thought it would be easier? I knew I was guilty when I understood his NAME :smack: ---
**Author:** ![Skammer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/skammer/32/143_2.png) [@Skammer](https://boards.straightdope.com/u/Skammer)\ **Post date:** [January 16, 2004, 2:47pm UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/15 "2004-01-16T14:47:46Z")
> [@David Simmons](#): > > My answer agrees with **Xema**. The height of the pyramid is a/2 so the length of a face from tip to the center of the base is a/sqrt(2). > > b is then sqrt(a[sup]2[/sup]/4 + a[sup]2[/sup]/2) = sqrt(3)\*a/2 I was a lot happier when I went to bed thinking that a=b. Now I’m going to have to figure out where sqrt(a[sup]2[/sup]/4 + a[sup]2[/sup]/2) = sqrt(3)\*a/2 falls on my scale ruler. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [January 16, 2004, 8:13pm UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/16 "2004-01-16T20:13:36Z")
> [@Skammer](#): > > Sorry, I misunderstood **Xema** to mean c connected to the center of the base of a side, when he meant the center of the square base. I didn’t say it very well, but c is supposed to be a line that stats at the top and bisects one of the triangular faces. According to my calculation, the angle sought in the OP is 54.7 degrees. ---
**Author:** ![KP](https://avatars.discourse-cdn.com/v4/letter/k/a9adbd/32.png) [@KP](https://boards.straightdope.com/u/KP)\ **Post date:** [January 17, 2004, 2:57am UTC](https://boards.straightdope.com/t/help-me-with-this-geometry-problem/224712/17 "2004-01-17T02:57:23Z")
> [@Skammer](#): > > Imagine a pyramid with a square base and each face is at a 45 degree slope toward the top of the structure. Call the width of each side of the base _a_, and the line connecting the corner of the base to the top of the pyramid _b_. > > Now each side of the pyramid is an isosceles triangle with a base of _a_ and two sides of _b_. If you look at this triangle, what is the length of _b_ in terms of _a_? Or, alternatively, what is the angle formed at _ab_? Here are two ways of visualizing the solution in your mind. 1. Bisect the pyramid with a vertical plane through the middle of a face. The resultant cross-section is a triangle with a base of a and two sides c 2. The pyramid sides slope at 45 degrees, so the apex is a right angle. c^2 + c^2= a^2 so a=sqrt(2)c 3. c bisects each face into right triangles with sides a/2, c, and b (a/2)^2 + c^2 = b^2 [a/2}^2 + [a/sqrt(2)]^2 = b^2 b= sqrt(3)/2 a Theta= arcsin(c/b) = arcsin[a_sqrt(2) / a_sqrt(3)/2] = arcsin[2/sqrt(6)] = 54.74 degrees **If you are more into 3D visualization, here’s a fun exercise:** 1. Make a cube from 6 pyramids (apices at the center, each base is a face) They fit exactly into a cube, since each edge is a, and each side forms a 45 degree angle to its perpendicular. (45+45=90) 2. bisect the cube with the diagonal plane defined by any pyramid face. The resulting cross-section is a rectangle with sides a and sqrt(2)a. (The rectangle’s long side is a diagonal of a pyramid base/cube face.) The pyramid edges of length b connecting the vertices to the center (where the apices meet) can be shown to form diagonals by symmetry From this figure, you can calculate the angles and the ratio between b (the half-diagonal) and a (the basis of both cross-section sides) 1. (2b)^2 = a^2 + (a sqrt(2))^2 = 3 a^2 b = sqrt(3)/2 a theta = arcsin[a_sqrt(2) / 2_sqrt(3)/2] = arcsin(sqrt(2)/sqrt(3)) = 54.74 degrees