Tuesday, June 20, 2023

figure it out

My answer will be in the comments section.



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Posted by Kevin Kim at 6/20/2023 12:03:00 AM

1 comment:

Kevin Kim said...

OK, so we're supposed to find the area of the blue trapezoid. We're given side lengths of

(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.

June 20, 2023 at 12:19:00 AM GMT+9

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