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Diagram

I'm looking to solve for the length of side X in the diagram above. Trying to design this enclosure but having difficulty with the math. I know it's simple triangle math that I'm forgetting, but any help would be appreciated! The depth of the side panels isn't decided yet, but if it is important to solving, let me know.

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    The sizes don't add up. $4'3''+2''+5''\ne 5'$ – 2017-02-20
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    Right, faulty labeling on my part. That top 5" should be 7". Will update diagram. – 2017-02-20

2 Answers 2

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A little trig does it: on the side view, from the top of slanted section draw a vertical line down, and from the bottom of slant draw a horizontal to the right. You get a triangle with the top angle $\theta = 180-135 = 45$ degrees. The height of the slanted section is then $z=10 \cos\theta = \frac{10}{\sqrt{2}} \approx 7.071''$.

Then $x\approx 36.25-1.125-7.071 \approx 2' 4.054''$.

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    Assuming we're looking at front view: The width of the top and bottom are the same, 18". It bulges closer to the viewer, not any "wider." – 2017-02-20
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    I thought it was an orthogonal projection. In any case, a little trig does it: on the side view, from the top of slated section draw a vertical line down, and from the bottom of slant draw a horizontal to the right. You get a triangle with the top angle $\theta = 180-135 = 45$ degrees. The height of the slanted section is then $z=10 \cos\theta = \frac{10}{\sqrt{2}} \approx 7.071''$ – 2017-02-20
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    Thanks for your help! Marked as answer, but hopefully you can edit the correct one into the original post. – 2017-02-20
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    @Chris Manning I edited my answer. Thanks – 2017-02-20
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If we assume, that in the front view picture the angle between sloped $10''$ side and side $x$ is $135^{\circ}$, as on the side view, then $$x=3'0.25''-1.125''-5\sqrt{2}''\approx2'4.054'' $$

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    Can't edit: But I believe you meant 2' 4.8" ? – 2017-02-20
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    @ChrisManning Yes – 2017-02-20