# 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:** 20\ **Page:** 1
**Author:** ![Contemplation](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@Contemplation](https://boards.straightdope.com/u/Contemplation)\ **Post date:** [May 19, 2015, 6:48am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/1 "2015-05-19T06:48:10Z")
What is the surface area of a triangular prism if… the height of the triangle is 6 feet. the base of the triangle is 5 feet. the length of the pyramid is 14 feet. ---
**Author:** ![Contemplation](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@Contemplation](https://boards.straightdope.com/u/Contemplation)\ **Post date:** [May 19, 2015, 6:52am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/2 "2015-05-19T06:52:50Z")
[http://postimg.org/image/ro2n53zex/](http://postimg.org/image/ro2n53zex/) [http://s28.postimg.org/ro2n53zf1/triangular\_prism.png](http://s28.postimg.org/ro2n53zf1/triangular_prism.png) images for you to see it ---
**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\ **Post date:** [May 19, 2015, 7:01am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/3 "2015-05-19T07:01:08Z")
Assume the end triangle is isosceles, so the line down from the top to the middle of the base is perpendicular to the base…;Or if the line through the top is perpendicular at the base, then it bisects the base… Then Pythagoras to work out the missing side length (same for both sides because its isosceles.) Don’t want to do your homework any more though. ---
**Author:** ![Robot\_Arm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robot_arm/32/18280_2.png) [@Robot\_Arm](https://boards.straightdope.com/u/Robot_Arm)\ **Post date:** [May 19, 2015, 7:01am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/4 "2015-05-19T07:01:21Z")
I’ve got a number, but first you have to promise that I’m not doing your homework for you. ---
**Author:** ![Contemplation](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@Contemplation](https://boards.straightdope.com/u/Contemplation)\ **Post date:** [May 19, 2015, 7:05am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/5 "2015-05-19T07:05:13Z")
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. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [May 19, 2015, 7:06am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/6 "2015-05-19T07:06:17Z")
> [@Isilder](#): > > Assume the end triangle is isosceles … Without this assumption, the question as posed does not have a definite answer. ---
**Author:** ![Contemplation](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@Contemplation](https://boards.straightdope.com/u/Contemplation)\ **Post date:** [May 19, 2015, 7:06am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/7 "2015-05-19T07:06:44Z")
> [@Isilder](#): > > Assume the end triangle is isosceles, so the line down from the top to the middle of the base is perpendicular to the base…;Or if the line through the top is perpendicular at the base, then it bisects the base… > > Then Pythagoras to work out the missing side length (same for both sides because its isosceles.) > Don’t want to do your homework any more though. that line is the height of the triangle. ---
**Author:** ![Contemplation](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@Contemplation](https://boards.straightdope.com/u/Contemplation)\ **Post date:** [May 19, 2015, 7:10am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/8 "2015-05-19T07:10:30Z")
> [@Giles](#): > > Without this assumption, the question as posed does not have a definite answer. how is that so? This was the homework problem of a 12 year old sibling. I’ve completed 2 semesters of college level math, and cannot remember this simple problem. The length of the prism is 14 feet. height of the triangle is 6 feet. base of the triangle is 5 feet. what is the surface area of the triangular prism? ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [May 19, 2015, 7:18am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/9 "2015-05-19T07:18:05Z")
> [@Contemplation](#): > > how is that so? Suppose the triangle is ABC, the base is BC, and the height is the length of AH where H is a point somewhere on BC. You need to know either BH or CH to determine what the lengths of AB and AC are. (Once you know either BH or CH, you know the other, because BH + CH = 6 feet.) I suspect that the problem-setter made an unwritten assumption that the triangle is isoceles, and BH = CH = 3 feet. ---
**Author:** ![Contemplation](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@Contemplation](https://boards.straightdope.com/u/Contemplation)\ **Post date:** [May 19, 2015, 7:26am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/10 "2015-05-19T07:26:08Z")
> [@Giles](#): > > Suppose the triangle is ABC, the base is BC, and the height is the length of AH where H is a point somewhere on BC. You need to know either BH or CH to determine what the lengths of AB and AC are. (Once you know either BH or CH, you know the other, because BH + CH = 6 feet.) > > I suspect that the problem-setter made an unwritten assumption that the triangle is isoceles, and BH = CH = 3 feet. I doubt it could be 3 feet. just by going off pythagorean theorem, the sides of the triangle should be 6.5 ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [May 19, 2015, 7:32am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/11 "2015-05-19T07:32:56Z")
> [@Giles](#): > > … (Once you know either BH or CH, you know the other, because BH + CH = 6 feet.) > > I suspect that the problem-setter made an unwritten assumption that the triangle is isoceles, and BH = CH = 3 feet. Sorry, since the base is 5 feet, BH + CH = 5 feet, and if the triangle is isoceles, BH = CH = 2.5 feet. The rest of my argument stands. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [May 19, 2015, 7:45am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/12 "2015-05-19T07:45:23Z")
> [@Contemplation](#): > > I doubt it could be 3 feet. just by going off pythagorean theorem, the sides of the triangle should be 6.5 I assume you mean that one or two sides are 6.5. We are already told that the base is 5 feet, and the base is one of the sides of the triangle. Using my notation, if AB = 6.5 feet, then AB^2 = AH^2 + BH^2 (by Pythagoras), i.e. 6.5^2 = 6^2 + BH^2, i.e., BH^2 = 42.25 - 36 = 6.25, i.e., BH = 2.5 feet, so the triangle is isoceles. However, AB can be any length between 6 feet and 7.81 feet (the square root of 61). ---
**Author:** ![Senegoid](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senegoid/32/6606_2.png) [@Senegoid](https://boards.straightdope.com/u/Senegoid)\ **Post date:** [May 19, 2015, 9:22am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/13 "2015-05-19T09:22:59Z")
> [@Giles](#): > > Without this assumption, the question as posed does not have a definite answer. I think this is correct. To solve the original problem, you need to first find both the _area_ and also the _perimeter_ of the end triangle. You have enough information to find the area, but I don’t think you have enough information to find the perimeter. You need to know if the triangle is isosceles, or right, or otherwise. ---
**Author:** ![bob\_2](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bob_2/32/3341_2.png) [@bob\_2](https://boards.straightdope.com/u/bob_2)\ **Post date:** [May 19, 2015, 9:25am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/14 "2015-05-19T09:25:57Z")
> [@Contemplation](#): > > What is the surface area of a triangular prism if… > > the height of the triangle is 6 feet. > the base of the triangle is 5 feet. > the length of the \*\*pyramid \*\*is 14 feet. This is not a pyramid. ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [May 19, 2015, 11:02am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/15 "2015-05-19T11:02:16Z")
True. But he correctly referred to it as a “prism” in the title & the intro verbiage. So I think we can give him that one as an inadvertent word swap between “pyramid” and “prism”. The other’s objection that the problem implicitly assumes an isoscelese triangle is well-founded. ---
**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 19, 2015, 4:56pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/16 "2015-05-19T16:56:20Z")
> [@Contemplation](#): > > I doubt it could be 3 feet. just by going off pythagorean theorem, the sides of the triangle should be 6.5 It sounds like you should have enough information to solve it then. (This is assuming, of course, the triangles are isosceles and both the front and back triangles are the same.) You just have the area of the two triangles + the area of the three rectangles. ---
**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, 3:23am UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/17 "2015-05-20T03:23:54Z")
I’m thinking the problem is misstated or mis-copied, that the length of the side slope is 5 and the triangle is isosceles, thus giving us the classic 3-4-5 Pythagorean, meaning the prism has a cross-section or end of 6,5,5 edges. If we go by 5 and three as the OP suggests, then the hypotenuse (Hipotenii? Hypotenae? 🙂 ) are sqrt(34). Maybe in this age of calculators they don’t mind messy numbers, but my pre-calculator experience was that numbers always came out nice and round. Especially if the alleged sibling is in lower grades… 254 sq ft? ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [May 20, 2015, 2:16pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/18 "2015-05-20T14:16:59Z")
> [@md2000](#): > > _… don’t forget the end pieces._ Because if you do, then you need to double the surface area of the rectangular faces because now the thing’s hollow. And Oh My God - It’s full of stars! 😃 ---
**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, 3:11pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/19 "2015-05-20T15:11:07Z")
Assuming that it’s an isoceles triangle, the area is: (sqrt(5^2 + 3^2) \* 14)\*3 = 244.89 (Pythagorean theorem to get hypotenuse \* length of the side, times 3 sides) plus ((5\*5.83)/2) \* 2 = 29.15 (area of isosceles triangle formula) \* 2 giving us a total of 244.89 + 29.15 , or 274 ---
**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, 3:22pm UTC](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350/20 "2015-05-20T15:22:56Z")
> [@bump](#): > > Assuming that it’s an isoceles triangle, the area is: > > (sqrt(5^2 + 3^2) \* 14)\*3 = 244.89 (Pythagorean theorem to get hypotenuse \* length of the side, times 3 sides) 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. [Next page](https://boards.straightdope.com/t/surface-area-of-a-triangular-prism/720350.md?page=2)