Solving for the triangle's perimeter

Question

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

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

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

$h + y = 90$

Answer 1

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.

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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).

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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.

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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.

Answer 2

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.