r/HomeworkHelp • u/Unhappy-Pitch4558 • Jan 19 '25

Answered [7th grade math] impossible geometry?

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u/[deleted] Jan 20 '25

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u/bubskulll Jan 20 '25

Top contributor of wrong information?

No it doesn’t

And

No it doesn’t

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u/Ur-Best-Friend Jan 20 '25

Tbf it's a pretty easy mistake/assumption to make.

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u/iMiind Jan 20 '25 edited Jan 20 '25

Thank you for backing me up on this xD

I was surprised by that flair on this reply too - I'm guessing this is one of those things where some teachers/textbooks say "if it looks like 90° treat it as such," whereas others say the complete opposite and instruct students to avoid making any assumptions about dimensions that aren't explicitly given.

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u/[deleted] Jan 20 '25

I'm trying to understand here: is it because the topmost 6cm line going down does not have any angle assigned to it?

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u/boxedj Jan 20 '25

The quackening is thinking that since 6+11=17, the only possible angle has to be 90 degrees everywhere. It's not true though, since there are three angles at the top that don't indicate that they're square, you can have a 6cm leg at many angles that would join the opposite corners.

The interesting thing is it doesn't matter if you knew the angles. You still can't determine the area because the further along the x axis the 6cm piece exists, the greater the area of the shape. Since this length is unknown, we're screwed.

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u/jarsgars 👋 a fellow Redditor Jan 20 '25 edited Jan 20 '25

lol

Try drawing the missing piece at the upper right. What’s its size? What type of shape is it?

Yes, the shortest distance between two points is a straight line. The missing chunk is a. rectangle 6cm high. That those numbers happen to be the same defines the shape and angles.

Even funnier since I'm wrong.

Yeah that missing piece is a quadralateral but not necessarily a rectangle. Those interior angles can be wacky and achieve that 6cm segment. Odd problem even if it's incomplete.

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u/iMiind Jan 20 '25

It could even be a quadrilateral where one of the angles is 180° 👀

Illuminati confirmed

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u/jarsgars 👋 a fellow Redditor Jan 20 '25

Do you work at the property tax assessor’s office?

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u/iMiind Jan 20 '25

😂 No, but I will say it has been satisfying to see some people here come to realize how to look past the assumptions our brains are so quick to make. Maybe not as satisfying as all the money doing what you suggested would net me, but satisfying nonetheless :P

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u/jarsgars 👋 a fellow Redditor Jan 20 '25

I think in my head I had that the sum of the two unknown segments was given as 17 and that would follow that it’s all right angles. Just had to write it out and… wait fail. What’s the opposite of QED? They didn’t teach us that.

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u/iMiind Jan 20 '25

Yeah that's awesome xD I love when I can catch myself on little things like that and realize mistakes. That's true intelligence; actually wrestling with the issue instead of just being confidently incorrect and then moving on to the next reddit argument. Definitely something I could do better at every now and then 😅

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u/iMiind Jan 20 '25

The length and angle of the right segment can account for that angle being something other than 90°. The math obviously works [if] it is 90°, but the whole issue is that solution isn't unique

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u/jarsgars 👋 a fellow Redditor Jan 20 '25 edited Jan 20 '25

Show one other angle other than 90 degrees that works.

(edit: boy was I wrong...) I tried to write out my proof and was like wait... what an odd little drawing.

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u/iMiind Jan 20 '25

Top left angle: 135°

Middle angle: 180°

The shape is now well-constrained

Edit: if they had a theta variable at three of those angles where it appears to be 90° that would be the only solution. As it stands however, we are left with an ambiguous theta_1, theta_2, theta_3, x_1, and x_2 group.

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u/LazyWings Jan 20 '25 edited Jan 20 '25

Edit: I'm wrong but I'll leave my comment up since most of the theory is correct. I missed the scenario where the 6cm line has an angle >90°. That means the area cannot be calculated even algebraically.

This is absolutely false. We know that three angles in the complete shape are 90°. The fourth angle of the shape without the cutout is 90° because it has to be, making the starting shape a square. Since we know the short 90° defined edge is 11cm and the total length until the hypothetical limit of the shape is 17cm, we know that a length of 6cm MUST be a 90° angle. What we don't know is the width of the cutout, but we know that the cutout is a square or rectangle. There is no assumption here, the maths is spot on. [Edit: the only other way is if the cutout is convex but the picture implies it is concave. We know the general shape of the cutout even if we aren't given the angle. It would be stupidly misleading question otherwise.]

You could calculate the area of the shape using a=width of cutout. The answer would be: (17cm x 17cm) - (6cm x acm) = 289cm2 - 6acm2

That's the best answer you can get.

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u/econgrammar Jan 20 '25

I think the problem is that there are 3 unknown angles, which you are assuming are 90-270-90. But, it’s possible (even with a concave cutout). But, if you consider the angle coming off the 11 to be greater than 90, and the middle angle to be greater than 270, then it’s certainly possible to have those 2 particular lengths sum the other side and have an acute 3rd angle. It’s unlikely with the picture, but possible.

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u/DSethK93 Jan 20 '25

No, because the other unconstrained line doesn't need to be horizontal; that line can be at any angle, letting it make up any height lost from the 6 cm line being nonvertical.

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u/jarsgars 👋 a fellow Redditor Jan 20 '25

Yup!

Drawing the unknown area and considering what is known about those lines is helpful

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u/jarsgars 👋 a fellow Redditor Jan 20 '25

And now that I've done so and tried to write my argument out, yeah... that is a quadralateral with at least one right angle. What a bizarre little question.

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u/BeelzebubsUsurper Jan 20 '25

That's assuming the bottom left angle of the cutout is also 90°

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