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Would like some help with solving for the grey triangle's perimeter. It is assumed that the grey triangle is equilateral.

enter image description here

My attempt:

Let $x =$ side of grey triangle Let $h =$ height of grey triangle Let $y =$ height of rectangle Let $x =$ length of rectangle

Area of triangle $= \sqrt{150^2 - 120^2} \cdot 120 \cdot 2 = 21600$

$90(120) = (h+y) \cdot 120$

$h + y = 90$

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    $\begingroup$ I think something's missing: that grey equilateral triangle can have any height from the big triangles upper vertex and down to the triangle's basis. Unless we're given some data about that white area or something else I don't think this has solution... $\endgroup$ – DonAntonio Mar 9 '13 at 1:45
  • $\begingroup$ I believe the grey triangle shares the same vertex as the bigger triangle. And the white area is a white rectangle. $\endgroup$ – Quaxton Hale Mar 9 '13 at 1:46
  • $\begingroup$ I know that, @Justin...but how are we to know its height?! $\endgroup$ – DonAntonio Mar 9 '13 at 1:47
  • $\begingroup$ Sorry, I was caught up with trying to solve for the area. I need to solve for the perimeter. $\endgroup$ – Quaxton Hale Mar 9 '13 at 1:51
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    $\begingroup$ It's just the same: there's some data missing. $\endgroup$ – DonAntonio Mar 9 '13 at 1:51
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Added: note that your computation of the larger triangle's area is double what it should be. Recall that the area $A$ of a triangle is $$\dfrac 12 \text{base}\times\text{height}$$

It's height, as you calculated, is $H = \sqrt{150^2 - 120^2} = 90\;$ and it's base is $240$. Hence, $$A = 90 \times \frac 12 (240) = 90 \times 120 = 10800$$

Fortunately, you're calculation the large triangle's height of $90 \implies h + y = 90$ is unaffected. But if you do get more information about dimensions of the rectangle, or the area of the red/green regions, you'll want to have the correct total area of the large triangle.


This is what we can say:

$h + y = 90 \implies h = 90 - y$

$$A_{\text{grey triangle}} = \dfrac 12 x \times h = \dfrac 12 x(90 - y) = 45x - \dfrac{xy}{2}$$

That is all we can say. We need to know the height of the rectangle to solve for area of the grey triangle. Otherwise, the grey equilateral triangle sharing the top vertex of the large triangle could be arbitrarily large (or arbitrarily small).


Per comment added: It is also the case that there is insufficient information to determine the perimeter of the grey triangle as well: $\quad P = 3x.\;$ In this case, we'd need to know the length of the rectangle, which is equivalent to $ x.\;$ With no information about the dimensions of the rectangle, there is no solution.


If the solution you heard is correct, and the perimeter $P$ is supposed to compute to $240$, then

$$P = 240 = 3x \implies x = 80,$$ and that would require the length/base of the rectangle/equilateral triangle to be of length $80 = \dfrac 13(240)$, i.e., $1/3$ of the base of the large triangle. This would be consistent with, and imply that the endpoints of the base of the rectangle partition the base into three segments of equal length.

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    $\begingroup$ No problem! My pleasure. +1 for you for all the work you did and added to your post! $\endgroup$ – Namaste Mar 9 '13 at 1:58
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The question does not have a definite solution. Notice that you can make the equilateral triangle any size: the only constraint is that the $h+y = 90$.

For instance, we could set $y=0$ and have an equilateral triangle of height $90$, consistent with what we have been given, or we could set $h = 0$ and have just a dividing line between the green and red triangles.

To solve the equation, we would need an additional constraint, such as a value for the areas of the red and green shaded regions.

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  • $\begingroup$ My bad, sorry! The question was asking for the perimeter of the grey triangle! $\endgroup$ – Quaxton Hale Mar 9 '13 at 1:50
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    $\begingroup$ Justin, doesn't matter --- same objection! $\endgroup$ – Gerry Myerson Mar 9 '13 at 1:51
  • $\begingroup$ Okay, well some people were saying the answer was 240? I need to find the link to where the problem was posted. $\endgroup$ – Quaxton Hale Mar 9 '13 at 1:52

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