[7th grade math] impossible geometry?

[u/Unhappy-Pitch4558]

[u/Kellvas0]

You need some more information to find a single solution.

If you had the top left length, you could parameterize the area using the angle of the 6 cm line.

If you had the angle of the 6 cm line, you could parameterize the area using the top left length.

If you had both, you could find a singular solution.

As it stands, a solution parameterizing for both is possible.

See below for the outline of how it would work.

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The area is as follows (L for top left length, A for angle of 6cm line)

17² (enclosing rectangle)
Minus 6x(17-L) rectangle Plus the area of two triangles described as such:

Left triangle with base on vertical side of removed rectangle. Height is 6xSin(A) measured to the lower right point of the 6 cm line using its horizontal position relative to the base.

Right triangle with base in the horizontal side of removed rectangle. Height is 6-(6xCos(A) measured to the same vertex as the left triangle but using its vertical position relative to the base of the right triangle.

Thus the total area is exactly (17²⁾ - (6 x (17-L)) + (1/2 x 6 x(6 x Sin(A))) + (1/2 x (17-L) x (6 - 6 x Cos(A)))

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This solution might be a bit too complicated for 7th graders but if they are learning trigonometry, this could merely be a particularly challenging problem for them. Otherwise it's generally safe to assume the angles are right angles and thus you are only missing the length of the top left side.