My answer will be in the comments section.
Tuesday, June 20, 2023
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OK, so we're supposed to find the area of the blue trapezoid. We're given side lengths of
ReplyDelete(x + 1) and (2x + 3) for the horizontals, and
(x + 4) for the vertical, and
(2x + 2) for the diagonal.
So we drop a vertical segment from the upper-left corner of the figure straight down to the bottom leg. From the lower-left corner of the figure to the point of intersection with this new vertical, we know the distance is
(2x + 3) - (x + 1), which simplifies to
x + 2.
We can now solve for x because we have a right triangle with legs of (x + 2) and (x + 4), and a hypotenuse of (2x + 2).
Our Pythagorean equation is therefore
(x+2)^2 + (x+4)^2 = (2x+2)^2
This explodes out to
x^2 + 4x + 4 + x^2 + 8x + 16 = 4x^2 + 8x + 4
This collapses to
2x^2 + 12x + 20 = 4x^2 + 8x + 4
or
2x^2 = 4x + 16, or
x^2 - 2x - 8 = 0,
factoring out to
(x - 4)(x + 2) = 0,
leaving us with two solutions for x:
x = -2, x = 4
We can't use x = -2 because for the side that is (x + 1), we'd get a negative answer, and distances can't be negative. Therefore, x = 4 is the only possible solution.
Knowing that x = 4, we can plug x into the relevant sides to find the area of a trapezoid, which is the height times the average of the two horizontal sides.
height = 8
short horizontal side = 5
long horizontal side = 11
[(5 + 11)/2]•8 = 64.
The trap's area is 64.
QED.