Surface area of a triangular prism?

[Nava]

@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)]+[heightbase perimeter], so [265/2]+[4imnotbotheringtocalculatethisone].

If it was a regular triangular pyramid you would have an equilateral triangle as the base, plus the sides.

[pulykamell]

@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.

[bump]

@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 + (514), 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.

[Kobal2]

@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*(65/2) + (145) + 2*(6.5*14). Simplified : 30 + 70 + 182 = 282

[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.

[md2000]

@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?

[md2000]

@md2000

106.6?

Oh, no other duplicate side lengths.
162.1?

[septimus]

The area is a whole number. In fact, each face has a whole-number area.

[md2000]

@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.

[Lance_Turbo]

@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

[septimus]

@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.

[Lance_Turbo]

I’ll leave some time for others to ponder this before posting a more detailed solution. I’ll shoot for 48 hours.

[MrFloppy]

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

[Lance_Turbo]

@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 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.

[septimus]

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.

[Indistinguishable]

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)

[Indistinguishable]

[While we’re on the topic, I suppose I may note that the cleanest, most elegant explication of Heron’s formula I ever saw 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!), but now, I finally appreciate its beauty]