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mhefter · Sep 1, 2011, 04:54 AM

You have an isosceles triangle on top of a square. The perimeter of the whole thing is 36 and the triangles perimeter is 24. What are the 3 sides of the square?

Stratmando · Sep 1, 2011, 05:20 AM

I would think the same as the triangle?

mhefter · Sep 1, 2011, 05:35 AM

Ok so what would the triangle be? And the equation to figure it out?

jcaron2 · Sep 1, 2011, 07:10 AM

If the sides of the square are length x, that means the base of the triangle is also x. Meanwhile, the other two sides of the triangle are length y.

If the triangle's perimeter is 24, then x + 2y = 24.

Meanwhile, if the perimeter of the whole thing is 36, then 3x + 2y = 36.

There you go. Two equations and two unknowns.

Unknown008 · Sep 1, 2011, 10:58 AM

Well, in the case that the base of the triangle is not equal to the side of the square, you can get an answer only if the base of the triangle is larger than the side of the square. Making drawings help you a lot! :)

Stratmando · Sep 1, 2011, 03:08 PM

I figured if the triangle is 24, 1 side must be 8 (1/3 of 24).
If the base of the triangle is the same size a 1 side of the square, then 3 X 8 = 24.
Only problem with this is, is That total would be 40(8 X 5 = 40?),
Which to me says the triangle should be smaller?

Unknown008 · Sep 2, 2011, 03:19 AM

No no no, you have something assumed that is not permissible.

An isosceles triangle has one side, then two sides equal. You made the assumption that the triangle is an equilateral triangle (3 equal sides) which is what is not permissible.

ebaines · Sep 2, 2011, 06:07 AM

Originally Posted by Unknown008

Well, in the case that the base of the triangle is not equal to the side of the square, you can get an answer only if the base of the triangle is larger than the side of the square. Making drawings help you a lot! :)

First, I think jcaron2 has the answer as intended by the problem. But thinking about whether there is a solution for an isoceles triangle whose base is less than length of the square: this leads to 2 equations in 3 unknowns, and hence infinite number of solutions. For example, the length of the square could be 4, and the isoceles triangle could have base length of 2 and the other two sides each 11. Am I misunderstanding what you meant?

Unknown008 · Sep 2, 2011, 09:35 AM

Originally Posted by ebaines

First, I think jcaron2 has the answer as intended by the problem. But thinking about whether there is a solution for an isoceles triangle whose base is less than length of the square: this leads to 2 equations in 3 unknowns, and hence infinite number of solutions. For example, the length of the square could be 4, and the isoceles triangle could have base length of 2 and the other two sides each 11. Am I misunderstanding what you meant?

Exactly, there are those two little sides of the square besides the base of the triangle that cannot be found in that case. :)

ebaines · Sep 2, 2011, 10:02 AM

Originally Posted by Unknown008

Exactly, there are those two little sides of the square besides the base of the triangle that cannot be found in that case. :)

I thought what you meant was that it was impossible to have the case where the base of the triangle is less than the square. But I guess what you meant to say was that if the base is less then the length of the square then the length of the square can't be determined. In that case - we agree!

Stratmando · Sep 2, 2011, 04:28 PM

A question could be for a right angle triangle, then you would only have 2 answers.

Unknown008 · Sep 3, 2011, 07:10 AM

I'm not sure I understand :confused:

Stratmando · Sep 3, 2011, 07:15 AM

A right angle triangle, right side up, and on its side on top of the square.

Unknown008 · Sep 3, 2011, 07:20 AM

Sorry, I still don't understand... If I did though, I get only one answer given the lengths given :(

jcaron2 · Sep 3, 2011, 07:52 AM

If you picture a non-isosceles right triangle on top of the square such that the hypotenuse is congruent with the top of the square, you end up with three sides of unknown length. Let's use x for the side of the square (and the hypotenuse of the triangle), and y and z for the two legs of the triangle. You then end up with:

3x + y + z = 36
x + y + z = 24
y^2 + z^2 = x^2

Given the quadratic nature of the third equation, this gives two potential answers for y and z (hence Stratmando's statement), but still only one answer for x (which becomes obvious if you subtract the second equation from the first; you simply end up with 2x = 12). And since the original question asked about the length of the sides of the square (there are four of them on a square, by the way, not three :)), Jerry's right. There's only one answer.

By the way, those first two equations should hold no matter what sort of constraints you put on the triangle. Whether the triangle is isosceles, right, or totally unconstrained, if it's perimeter is 24 and it shares a side with the square (resulting in a total perimeter of 36), the side of the square is always 6.

jcaron2 · Sep 3, 2011, 09:22 AM

By the way, to be even more general still, it doesn't even matter if the polygon on top is a triangle or not. It could be a rectangle, an octagon, an irregular 277-sided polygon, a semi-circle, or any random blob with a flat bottom. As long as it has a perimeter of 24 and sits atop a square the same width as the shape's flat bottom, if the perimeter of the result is 36, the width of the base must have been 6.