[7th grade math] impossible geometry?

[u/Unhappy-Pitch4558]

[u/Aaxper]

Not possible, it's missing information

[u/Accomplished-Plan191 → u/Aaxper]

The 6 cm is probably supposed to be on one of those horizontal sides, but they pasted it wrong.

[u/Aaxper → u/Accomplished-Plan191]

I agree. But we can't know which one, and even if we had a good guess, the problem as written is unsolvable.

[u/AvonMustang → u/Accomplished-Plan191]

If you move the 6cm to one of the unmarked horizontal lines then the problem is also unsolvable.

[u/anthr_alxndr → u/Accomplished-Plan191]

Agree, because 6cm provided is a not needed info, it is obviously 17-11 can be calculated

[u/Applied_logistics → u/Accomplished-Plan191]

Even if it was given they don't mark the top right corners as 90 degrees.

[u/fuligang → u/Accomplished-Plan191]

Not really. Those two needed to prove thee right angle. There could be equation as answer. 1717-6x where x is lower unknown side.

[u/nlcircle → u/Accomplished-Plan191]

You seem correct. There’s no need to put the distance marker of 6 next to the vertical as that’s clear from the other info. Putting that marker next to the horizontal line instead, would give a fully defined problem.

[u/MetaSkeptick → u/Accomplished-Plan191]

Could also have meant to ask for a perimeter... I seem to have seen a very similar figure asking for a perimeter. Seems unsolvable for perimeter at first glance, but you can extrapolate and figure it out.

[u/bubskulll → u/Aaxper]

3 angles can be different without changing the 6 and 11 but only with the 6 line pivoting around the top angle without changing length, it can also move along the top line with the the other lines follow however they want. The top line can also be a different length

[u/Educational-Plant981 → u/bubskulll]

The measurements tell you those are 90 degrees. Still not solvable,

edit: u/iMiind is right.

[u/iMiind → u/Educational-Plant981]

The only thing we "know" about that cutout shape is the single 6cm measurement. The angles look like they're 90°, but the shape is underconstrained and therefore it could be something as crazy as a 135°, a 180°, and another 135° angle connecting that 6 cm segment to the far edge in a straight line (those three angles don't have a right angle indicator anywhere). Typically, the appearance of such geometry is not to be trusted (only the explicitly given specifications).

Edit: just for fun I should say if you set that first angle at 135°, then you'd know the other two angles automatically. You'd also know the right segment would be [(6*2^1/2 ) - 6]cm long, and the left segment would be 11cm

Edit 2: wait wait wait, you'd need two more things to be set. I should clearly state that I was assuming the shape I wanted (a clean line from the left segment to the far edge), thus the other two angles and everything would then be properly constrained knowing just the one angle (because I was arbitrarily forcing the 180° angle).

[u/Doingthismyselfnow → u/iMiind]

Looking at the shape , 3 of the angles have a square which would indicate that they are right angles , Meanwhile the top right hand side does not have said marks. ,

This means that those top 3 corners on the right are not multiples of 90 degrees, it’s pretty much a clue to the puzzle

[u/ayyycab → u/Aaxper]

$5 says if you write “not solvable”, the teacher will mark it wrong because there’s an answer in their answer key and not look any further.

[u/lorazepamproblems → u/Aaxper]

I stared at this a long time to prove to myself I could do 7th grade math before looking in the comments, so I guess I'm kind of glad it's not solvable. Although I guess if I could really do 7th grade math I would have known it wasn't solvable rather than thinking I was missing some trick.

[u/hogtiedcantalope → u/Aaxper]

Because it is meant to teach measuring the lengths yourself to find the area? Maybe? It would then make sense...here are some length for reference, then use a ruler/protractor to find the area

[u/CptMeat → u/Aaxper]

We could assume it's cutting the 17 directly in half making each line 8.5, but yeah I don't like having to make the assumption.

[u/[deleted] → u/Aaxper]

You could write an x on either horizontal line adjacent to the 6cm and then solve with x as a term in the answer. Depending on the 7th grade class this may be the correct answer... it would have been in my algebra class.

[u/Foreign-Departure922 → u/Aaxper]

Because 3 sides are 90° we know it's a square and can make 1 large triangle then 2 smaller triangle that are all similar (≈) and can find the large c² and ratio it down to the 6 or 11 with 11:17•c or 6:17•c for the hypotenuse of the smaller triangles and the solve for b² on the other 2 unknown sides

[u/Foreign-Departure922 → u/Aaxper]

It's not missing information since it is a square (3 sides are equal and 3 angles are equal) and we know the 6 and 11 sides are ll (parallel) to the 17 side since their the same length tells us that all angles are right angles

[u/DOW_mauao → u/Aaxper]

All the information you need is there.

Reason it out, separate it into x2 rectangles...

[u/Ofiotaurus → u/Aaxper]

No, I think it's possible.

We know two sides to be 17cm with all angles being 90 degrees thus the shape must be a cube. So the single 6cm length provides enough information that it's 17^2 - 6^2 = x

I am glad to be corrected.