# Surface area of a triangular prism? **URL:** \ **Category:** Factual Questions\ **Created:** [May 19, 2015, 6:48am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350 "2015-05-19T06:48:10Z")\ **Posts on this page:** 17\ **Page:** 2
**Author:** ![Nava](https://avatars.discourse-cdn.com/v4/letter/n/da6949/32.png) [@Nava](https://boards.straightdope.com/u/Nava)\ **Post date:** [May 20, 2015, 4:00pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/21 "2015-05-20T16:00:37Z")
> [@Contemplation](#): > > this isnt my homework. I was trying to help a younger sibling with her homework a day ago, and this thought acured to me. > > For some reason, I just cant remember how to do this. It eventually boils down to calculating the area of all sides and adding them. If it’s a prism with isosceles triangles as the bases, then it’s [2\*(base area)]+[height_base perimeter], so [2_6_5/2]+[4_imnotbotheringtocalculatethisone]. If it was a regular triangular pyramid you would have an equilateral triangle as the base, plus the sides. ---
**Author:** ![pulykamell](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/pulykamell/32/3166_2.png) [@pulykamell](https://boards.straightdope.com/u/pulykamell)\ **Post date:** [May 20, 2015, 4:04pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/22 "2015-05-20T16:04:54Z")
> [@Nava](#): > > Your OP assumes but does not actually say that the pyramid is a regular one, but even if it wasn’t, it eventually boils down to calculating the areas of the four sides and adding them. > > For a regular pyramid like yours, there are three isosceles triangles and an equilteral one (the base). He actually means a prism, with two triangular and three rectangular sides. See his illustration. ETA: I see you caught it in the edit. ---
**Author:** ![bump](https://avatars.discourse-cdn.com/v4/letter/b/7c8e57/32.png) [@bump](https://boards.straightdope.com/u/bump)\ **Post date:** [May 20, 2015, 4:11pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/23 "2015-05-20T16:11:25Z")
> [@pulykamell](#): > > No, not quite. Two details: > > The right triangle created by the height has 6 for the vertical side, 2.5 for the bottom side, which gives us a hypoteneuse of 6.5. > > So we have a triangle, assuming it is isosceles, of two sides 6.5, and one side 5. So the are of the rectangular bits are (6.5\*14)_2 + (5_14), or 252 > > Then we have the area of the triangle, which is just a simple base times height divided by two (times two to account for both triangles.) So, 6\*5 (after simplifying) = 30 > > Total area: 282. I think I got my 6 side and 5 sides backward. ---
**Author:** ![Kobal2](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kobal2/32/20_2.png) [@Kobal2](https://boards.straightdope.com/u/Kobal2)\ **Post date:** [May 20, 2015, 4:34pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/24 "2015-05-20T16:34:12Z")
> [@Contemplation](#): > > this isnt my homework. I was trying to help a younger sibling with her homework a day ago, and this thought acured to me. > > For some reason, I just cant remember how to do this. Area of the triangles\*2 + area of the sides So that’s 2\*(6_5/2) + (14_5) + 2\*(6.5\*14). Simplified : 30 + 70 + 182 = 282 ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [May 20, 2015, 4:40pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/25 "2015-05-20T16:40:25Z")
Since OP’s question is {un,}answered, may I please ask for the surface area of my tetrahedron? Unfortunately all I remember is that the sides and area are all whole numbers of meters or square meters; two of the sides are 13 meters; the other four sides are all different from each other and each is less than 100 meters. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 20, 2015, 5:23pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/26 "2015-05-20T17:23:18Z")
> [@septimus](#): > > Since OP’s question is {un,}answered, may I please ask for the surface area of my tetrahedron? > > Unfortunately all I remember is that the sides and area are all whole numbers of meters or square meters; two of the sides are 13 meters; the other four sides are all different from each other and each is less than 100 meters. 106.6? ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 20, 2015, 5:46pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/27 "2015-05-20T17:46:07Z")
> [@md2000](#): > > 106.6? Oh, no other duplicate side lengths. 162.1? ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [May 20, 2015, 5:55pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/28 "2015-05-20T17:55:03Z")
The area is a whole number. In fact, each face has a whole-number area. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 20, 2015, 7:45pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/29 "2015-05-20T19:45:39Z")
> [@septimus](#): > > The area is a whole number. In fact, each face has a whole-number area. Huh? Two sides are 13. Thus, one triangle is 13-12-5 If the other is also 13-12-5 that violates the “two other sides the same” rule (common side 12 or 5 to get two 13’s). I’m trying to see how 13-84-85 paired with a 13-12-5 works into this… no common sides, certainly not with a whole-number triangle. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [May 20, 2015, 9:00pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/30 "2015-05-20T21:00:18Z")
> [@septimus](#): > > Since OP’s question is {un,}answered, may I please ask for the surface area of my tetrahedron? > > Unfortunately all I remember is that the sides and area are all whole numbers of meters or square meters; two of the sides are 13 meters; the other four sides are all different from each other and each is less than 100 meters. 1008 ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [May 20, 2015, 9:28pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/31 "2015-05-20T21:28:15Z")
> [@Lance\_Turbo](#): > > 1008 Thank you **Lance** for jogging my memory. I really didn’t want to get out the tape measure on that huge tetrahedron again. ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [May 20, 2015, 9:40pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/32 "2015-05-20T21:40:37Z")
I’ll leave some time for others to ponder this before posting a more detailed solution. I’ll shoot for 48 hours. ---
**Author:** ![MrFloppy](https://avatars.discourse-cdn.com/v4/letter/m/43a26b/32.png) [@MrFloppy](https://boards.straightdope.com/u/MrFloppy)\ **Post date:** [May 21, 2015, 7:59pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/33 "2015-05-21T19:59:28Z")
It’s 282 as other have said. Area of end triangle is 1/2bh so 2.5\*6=15 So two ends is 30. The hypotenuse of each end triangle is 6.5 because it is a 5,12,13 (2.5, 6, 6.5) right angled triangle. So two of the square sides are 6.5\*14=91 So two sides like that is 182 The bottom side is 5\*14=70. 30+70+182=282 QED ---
**Author:** ![Lance\_Turbo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lance_turbo/32/6156_2.png) [@Lance\_Turbo](https://boards.straightdope.com/u/Lance_Turbo)\ **Post date:** [May 22, 2015, 7:18pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/34 "2015-05-22T19:18:41Z")
> [@septimus](#): > > Since OP’s question is {un,}answered, may I please ask for the surface area of my tetrahedron? > > Unfortunately all I remember is that the sides and area are all whole numbers of meters or square meters; two of the sides are 13 meters; the other four sides are all different from each other and each is less than 100 meters. OK, close enough to 48 hours… assuming anyone cares. What we’re looking for here are four [Heronian triangles](http://en.wikipedia.org/wiki/Heronian_triangle) that can be assembled into a tetrahedron and satisfy other conditions. It turns out the the solution is unique up to rotation and reflection. The triangles are as follows: 13, 13, 24 -\> area 60 13, 40, 45 -\> area 252 13, 40, 51 -\> area 156 24, 45, 51 -\> area 540 This gives a total surface area of 1008. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [May 23, 2015, 7:46pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/35 "2015-05-23T19:46:56Z")
Sorry the tetrahedron puzzle attracted little interest. As an historical note: Heron’s Formula was discovered by other ancient mathematicians including apparently … Archimedes of Syracuse. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2015, 4:30am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/36 "2015-05-24T04:30:14Z")
For what it’s worth, I found the tetrahedron puzzle interesting, and appreciated your posing it. (However, I came to this thread late, so by the time I read it, **Lance Turbo** had already explained the solution in a way I could see no improvement on; hence, I had nothing to say till now) ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2015, 4:38am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/37 "2015-05-24T04:38:21Z")
[While we’re on the topic, I suppose I may note that [the cleanest, most elegant explication of Heron’s formula I ever saw](https://mathgarage.files.wordpress.com/2013/01/heron_by_complex_number.gif) was given by a high school student. Until I saw this, I used to say Heron’s formula was one of the few standard trigonometric formulas I would be hard-pressed to derive under time pressure (I even said it [here](http://boards.straightdope.com/sdmb/showpost.php?p=13943388&postcount=149)!), but now, I finally appreciate its beauty] [Previous page](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350.md?page=1)