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You have an isosceles Triangle on top of a square. The perimeter of the whole thing

Asked Sep 1, 2011, 04:54 AM — 15 Answers
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?

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Stratmando's Avatar
Stratmando Posts: 10,418, Reputation: 2515
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#2

Sep 1, 2011, 05:20 AM
I would think the same as the triangle?
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mhefter's Avatar
mhefter Posts: 2, Reputation: 10
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#3

Sep 1, 2011, 05:35 AM
Ok so what would the triangle be? And the equation to figure it out?
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jcaron2's Avatar
jcaron2 Posts: 983, Reputation: 1034
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#4

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.
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Unknown008's Avatar
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#5

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!
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Stratmando's Avatar
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#6

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?
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Unknown008's Avatar
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#7

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.
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ebaines's Avatar
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#8

Sep 2, 2011, 06:07 AM
Quote:
Originally Posted by Unknown008 View Post
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?
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Unknown008's Avatar
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#9

Sep 2, 2011, 09:35 AM
Quote:
Originally Posted by ebaines View Post
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.
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ebaines Posts: 10,033, Reputation: 5529
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#10

Sep 2, 2011, 10:02 AM
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Originally Posted by Unknown008 View Post
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!
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