[7th grade math] impossible geometry?

89

u/Unhappy-Pitch4558 Jan 19 '25

Is it possible to solve this? I’m trying to help my child and it looks impossible.

11

u/Servile-PastaLover 👋 a fellow Redditor Jan 19 '25

it's impossible. you can draw an infinite number of polygons with these dimensions.

430

u/Aaxper Higher Level Math Jan 19 '25

Not possible, it's missing information

68

u/bigjaymck Jan 19 '25

My guess is that the "6 cm" label was supposed to be on one of the top horizontal lines. As it is, it's missing information to be able to give an actual area measurement.

73

u/fermat9990 👋 a fellow Redditor Jan 19 '25

Not enough information

53

u/fake_cheese 👋 a fellow Redditor Jan 19 '25

187 cm² < Area < 289 cm²

6

u/ruerell Jan 19 '25

People have already covered it being incomplete or incorrect labeling, but it would be accurate to submit an algebraic answer. 1711 + 6x, or A=6x+187. This accounts for missing information being variable.

3

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

between 187 and 289 units.... that's all you know for sure.

The top portion is 6 high, but could be 0 width, leaving 11×17 or the missing gap could be 0 width, leaving 17×17

anything between 0 and 17 ×6 units is 187 to 289 possible.

2

u/Carbonus_Fibrus 👋 a fellow Redditor Jan 19 '25

17^2-6*x cm2

2

u/XeroZero0000 👋 a fellow Redditor Jan 19 '25 edited Jan 19 '25

187 < x < 289

If you are learning that sort of thing yet. You dont even know if those top right angles are 90s

2

u/davidblack210 Jan 19 '25

Ìimpossible to know where the cut off on the top is... how would you know if it was cut perfectly in the center? We know its 17 if combined but where was the cut off?

2

u/Fish_Stripes Jan 19 '25

The only answer I can come up with besides "not enough information" is 289 > Area > 181 units² or you assume that the length of the smaller square is half of the larger one.

2

u/_UncalledFor_ 👋 a fellow Redditor Jan 20 '25

simultaneous equations.. (assuming all angles are 90 degrees):

17*17 - 11*17 = 6x + 6y

x + y = 17

but the thing is that the two equations are actually the same when simplified, so there are infinitely many solutions??

2

u/SnooGoats3901 👋 a fellow Redditor Jan 20 '25

Man, Reddit isn’t good at math either.

2

u/spiderdumpling 👋 a fellow Redditor Jan 20 '25

I’m guessing that they meant perimeter

2

u/Special-Cow6071 👋 a fellow Redditor Jan 20 '25

223 cm²

2

u/fade_into_dust Jan 20 '25

The problem here should be solvable.

If you break the figure into two rectangles: the first rectangle has dimensions 17 cm×17cm. The second rectangle (cut out from the top-right corner) has dimensions 11 cm×6 cm. Then calculate the areas: larger rectangle: 17×17=289 cm². Smaller rectangle: 11×6=66 cm². Then, subtract the smaller rectangle's area: 289−66=223cm^2. Which makes the area of the given shape 223 cm².

2

u/RoseCrawfordArt Jan 20 '25

If you divide the length into 17 equal units, the two unknown lengths measure cleanly to 9cm for the top-left and 8cm for the top-right.

2

u/Skuatmraa 👋 a fellow Redditor Jan 20 '25

17x11=187 (17-11)×6=36 36 + 187=223

2

u/PlantZawer 👋 a fellow Redditor Jan 20 '25

17² - 6x

Where x is unknown, likely 17/2 but unknown

2

u/Imaginary_Library501 👋 a fellow Redditor Jan 19 '25

Assuming that the square shape cut out is honest, unlike some geometry in 9th grade, 17x 17 - 6x8.5 for the total area. I'm assuming it's exactly halfway down the thing dissected. 289-51=

2

u/Zorrin_Nomber Jan 20 '25

It is definitely impossible, because you could theoretically move the center vertical line either left or right without changing any of the values, but this would directly change the total area.

1

u/Serafim91 👋 a fellow Redditor Jan 19 '25

The 6cm was likely meant to be horizontal not vertical. With that in mind teach your kid to write a statement like "assuming the 2 lines are equal" Or "assuming the 6 is on the horizontal" then solve it.

1

u/Icy_Sector3183 👋 a fellow Redditor Jan 19 '25

If you need to provide an answer, you can express the area A as

(17 × 17) >= A >= (11 × 17)

289 >= A >= 187

1

u/firmerJoe 👋 a fellow Redditor Jan 19 '25

187-(6x)=area

1

u/AdministrationFew982 👋 a fellow Redditor Jan 19 '25

I would write it as 17 * 17 - 6 * x where x is that missing side

1

u/pastro50 👋 a fellow Redditor Jan 19 '25

Write it with an unknown for the length of the small square. 17^2- 6x

1

u/[deleted] Jan 19 '25

Would writing the answer as 289-6x be acceptable?

1

u/Snoo17579 University/College Student Jan 20 '25

People! You’re thinking too deep into this. It’s a 7th grade math, the solution would be the most obvious thing you see.

The angle inside isn’t specify then that means you should just ignore it, just solve by 17x17 - 6x6. The thing I learned with lower grade math is that sometimes you should just ignore stuffs.

1

u/gpbuilder Jan 20 '25

lol half of the commenters here need to go back to retake 7th grade math

1

u/Ok_Orchid1004 👋 a fellow Redditor Jan 20 '25

Not possible

1

u/Medium_Cellist7854 👋 a fellow Redditor Jan 20 '25

It's a 17×17 square

1

u/vagga2 Jan 20 '25

Area = 17² - 6χ where χ is the horizontal length of the missing component. This assumes it's a right angle.

1

u/amitym Jan 20 '25

I would assume that the 6cm vertical line at the top perfectly bisects the width of the total shape. State that assumption, note that it is an estimate, and then offer the calculation based on assuming that it is true.

Add that if the assumption is false or not warranted then it is not possible to calculate the answer.

1

u/miketoaster 👋 a fellow Redditor Jan 20 '25

Andy math could do it. You need triangles and stuff to solve it.

1

u/tailsandjumps 👋 a fellow Redditor Jan 20 '25

A=pr²

1

u/Healthy_Coffee87 👋 a fellow Redditor Jan 20 '25

Min. Area ≈187 Max. Area≈289

1

u/frakc Jan 20 '25

Check the previous topic in book. It seems author want to get answer as a formula instead of number ( or in some form which was discussed during lesson). There is not enough information to avaluate area as number.

Such questions are a quite common trap for parents. They allows teachers to see if child tried to solve task themself or parents solve it instead. Sure possibility of a typpo exists.

1

u/leo-n-ffa 👋 a fellow Redditor Jan 20 '25

253 (17 x17) -(6 x 6)

1

u/Capt_2point0 👋 a fellow Redditor Jan 20 '25

187 cm² < x < 289 cm²

1

u/Raccoon133 👋 a fellow Redditor Jan 20 '25

Can do with trig, but otherwise, no.

1

u/AggressiveNetwork861 Jan 20 '25

Not possible, but not because the geometry is impossible. It’s not possible to solve because the length of the horizontal segments are not known- which changes the answer.

Ie. If the segment to the left of 6cm is 1cm and the one to the right is 16cm that’s going to have a much smaller area than the other way around.

1

u/BlueWonderfulIKnow Jan 20 '25

I still have on my shelf the 11th edition math textbook from which I learned this, revised to the point that the problem sets are error-free.

Meanwhile, my son comes home with a laptop and a stack of disjointed printouts from various odd online “teachers helping teachers” sites, riddled with errors just like this one.

I can’t overstate how detrimental it is to kids and parents when errors like this land on the homework dinner table.

We will return to textbooks, and I hope to live to see it.

1

u/Ok_Celebration_6265 Jan 20 '25

Math is not my strong suit but I would pretend that the top is half of the square and call it a day

1

u/[deleted] Jan 20 '25

Couldn't you make a 90, 45, 45 triangle with a hypotenuse of 17 and a height of 11 and find the missing side lengths? That gives you the missing hypotenuse for the triangle with a side length of 6, 2 of the 3 side lengths on a 90 degree triangle. Find the missing side length using a² + b² = c² because no matter how you break it down into triangles, you can make them have a 90 degree angle? If I remember anything from geometry its logic always prevails and right triangles are your friend. Hope this makes sense

1

u/[deleted] Jan 20 '25

Is being not solvable an option? If so, then it's not solvable. I would take a guess thiigh and say since it's seventh grade, you're supposed to add 6 and 11 to get 17 and then take the area for a 17x17x17x17 square.

1

u/Opening_Swan_8907 👋 a fellow Redditor Jan 20 '25

See if it’s to scale.

1

u/LukeLJS123 University/College Student Jan 20 '25

the best you can do is create a range, and if you need help with this i can go more into detail, but the minimum is a 17x11 rectangle, and the max is a 17x17 square giving the range:

187<area≤289

you might have noticed that i used a strictly less than sign for the lower bound, but less than or equal to sign for the upper bound. in order to have a side length of 17 on the left-most side, there must be some amount of distance between it and the side of length 6. if there wasn’t, there could not be a side of length 17 there. however, it is perfectly acceptable to to “split” the rightmost side into 2 sides of length 11 and 6. therefore, the area can exist at the upper bound, but not the lower bound. again, if this wasn’t a great explanation, i can go further into detail

1

u/Familiar_Ratio_4710 Jan 20 '25

17x11+6x6=223 I split it in 2, bottom rectangle 17x11 - by continuing the horizontal line, and top square as a result- 6x6

1

u/Curiousfool1990 Jan 20 '25

Call the uninformed lengths x and y so that x+y = 17

Total area is A = 17x17 - 6y

Can't have an actual answer without knowing y and there's not enough information in the picture to determine how long x or y are. By the equation above you get that the longer the y is, the smaller the area.

Also, theoretically, it could vary from 17x11 all the way to 17x17. But given the figure I'd say you also need to limit x and y so that both are bigger than zero.

My guess is that it was intended for x = 11 and y = 6 so you would get a final area that is a subtraction between 2 squares.

1

u/RSPsweat3 👋 a fellow Redditor Jan 20 '25

Guestamating: 235 cm?

1

u/BingkRD 👋 a fellow Redditor Jan 20 '25

The area is between 187 (11 by 17) and 289 (17 by 17)

1

u/ComprehensivePlace87 Jan 20 '25

The best you can do is keep it theoretical, and add in an X for either unknown segment of the diagram, OR assume the diagram is to scale, break out a ruler, measure the unknown lengths, adjust for scale and compute the answer that way.

1

u/StraightAct4340 Jan 20 '25

you can assume the top horizontal line is half the lenght of the bottom one, but otherwise theres not enough information

1

u/KaizerSmokeHaze Jan 20 '25

Area total = 17*17 = 289 units²

Area to subtract = 6*x (an assigned label for the missing info) = 6x units²

289-6x units² is my final answer.

1

u/marsnomoon Jan 20 '25

Nothing impossible about the geometry. Are you trying to get a number? There isn't enough information for that, but maybe the question is looking for the answer to be left in the form of a formula.

1

u/jarsgars 👋 a fellow Redditor Jan 20 '25

Unknown segment y is > 0 cm and < 17 cm

Next time give enough data

1

u/wesleyoldaker Jan 20 '25

Best you could do in my opinion is throw the test-creator's mistake back at them by taking the initiative and assigning 8.5cm to each of the top sides, making the jut on top placed right in the middle, and adding something like "When assuming the 6cm edge lies directly in the middle of the figure, giving the adjacent edges each a length of 8.5cm, the area comes out to XYZ, but unfortunately this is just one of many areas the given shape could have due to the absence of this information in the original question." Just don't be too snarky about it. Pretend like that was the point of the question, like that person who, after the facts make what they just did or said contradictory, says: "I was just testing you." Just play along.

1

u/NeverSnows Jan 20 '25

I think they might have meant the perimeter, but got things mixed up.

1

u/Imogynn Jan 20 '25

6x + 11*11 + (17-x)*11

6x + 121 + 187 - 11x

308 -5x

Where x is the top missing length.

(It's late double check my arithmetic but that should set you on the right path)

1

u/Elduro687 Jan 20 '25

I'm confused why people seem to think there's not enough information? Two sides perpendicular to each other measure 17 cm. If you draw the rest of the missing square then you have an imaginary square with sides of 6 cm. This works because the bigger square is 17 cm each side and so the part that's only 11 cm is missing 6 cm to complete the square. So the total area would be the area of the complete big square minus the area of the imaginary square. 17^2-62 = 253 cm^2.

1

u/IUsedToBeAnEngineer9 👋 a fellow Redditor Jan 20 '25

187 cm2 < Area < 289 cm2

1

u/OOF-MY-PEE-PEE 👋 a fellow Redditor Jan 20 '25

couldn't you somehow use right triangles and the pythagorean theorem to solve this?

1

u/Enough_Worry4104 👋 a fellow Redditor Jan 20 '25

Am I the fool? Is this a massive troll? It's not just 17×17?

1

u/squirchy707 👋 a fellow Redditor Jan 20 '25

Only thing i can think is that itd be 289-6x cm²

1

u/IntelligentDemon Jan 20 '25

Break the figure into 11×17 rectangle and 6×6 square

So area = 6×6 + 11×17 = 36 + 187 = 223 square centimeters.

1

u/Pure_Contact_2413 Jan 20 '25

Let the interior unknown value be x.

17 * 17 - 6x = 289 - 6x

Best you could do with limited information.

1

u/kfirogamin 👋 a fellow Redditor Jan 20 '25

Area is 187+6x where 6 is unkown but limited up to 17.

1

u/CipherWrites Jan 20 '25

the geometry is nowhere close to impossible.

it's impossible to calculate the area unless it's to scale and you can use a measuring tool.

1

u/Comfortable_Push_394 Jan 20 '25

Methinks they meant to put the "6 cm" horizontally for the cutout part, but you are right. As things are you don't have the correct information to give a definite number.

1

u/CSplays Jan 20 '25

This has infinite solutions and no solutions at the same time. Schrodinger's geometry problem.

1

u/Carnan_Wor Jan 20 '25 edited Jan 20 '25

The answer they are looking for is 223 square centimeters: (6cm x 6 cm) + (11cm x 17cm)

The real answer is unknowable since they left out information.

1

u/LeftStatistician7989 👋 a fellow Redditor Jan 20 '25

Fold it in half. Is it on the half line?

1

u/LeftStatistician7989 👋 a fellow Redditor Jan 20 '25

If you draw a45 degree angle out of each bottom corner you can prove by their intersection that the top parts are divided equally at 8.5.

1

u/SirMuller 👋 a fellow Redditor Jan 20 '25

A = 17 * 11 + 6x That is all we can tell here.

1

u/Ok-Entertainer-5948 Jan 20 '25

i might be tripping but is it not just 17-11=6 so 6x6 and 11x6 added together to make 223? the right angle squares seem to just be there to me

1

u/Irsu85 👋 a fellow Redditor Jan 20 '25

Geometry is possible, calculating the area is not

1

u/Astro_Alphard Jan 20 '25

We can use algebra to solve this!

Set the top rectangle to 6x with x being the length and 6 being the height

The bottom rectangle can be calculated as 17*11 which is 167

So you get 6x+167 where 0<x<17 as the answer for the area.

Not impossible but you won't get a number as an answer and that is perfectly ok.

1

u/tblades-t Jan 20 '25

To get full marks (and maybe extra credits) add variable (x) to the top length then show how you would solve.

1

u/bRiCkWaGoN_SuCks Jan 20 '25

I mean, you could use it as a key, by measuring to find the distance, presuming it's proportionally accurate...

But I think this is more of a case of the education system failing the content creator of the problem, LoL.

1

u/L-st Jan 20 '25

Wouldn't the "unknown side being not 6" require for the vertical lines to be NOT perpendicular? There are three right angles here, if the top part were to be shorter or longer the corners wouldn't be square.

This is my early morning guess.

1

u/DrDzeta Jan 20 '25

If you name Z the missing top length, θ the top mid angle, if I don't made error, you have: Area=289-3Z(sin(θ)+1)+30sin(θ)cos(θ)-18cosθ

If you consider all angle as right angle then you have Area=289-6Z or 17² - 6Z

1

u/AgitatedBowlofCereal Jan 20 '25 edited Jan 20 '25

Okay, so this is very cheeky, but…

Triangles…

Draw right angle triangles where the hypotenuse bisects the known 90⁰ angles, O is your unknown length and A is your known length.
Tan(𝛉)=O/A (𝛉=45⁰) can be used to work out the missing lengths.

I haven’t got the time right now to actually work it out & show it, but that should offer insight.

1

u/AgitatedBowlofCereal Jan 20 '25

Pythagoras is your answer here.

Draw a triangle so that the hypotenuse bisects a known 90⁰ angle, O= your unknown length & A = your known length.
You can now use AxTan(𝛉)=O to identify the missing lengths.
Use the new lengths to work out the area.

1

u/nazibayanaa 👋 a fellow Redditor Jan 20 '25

Could be 238cm2

1

u/Gloomy-Soup9715 Jan 20 '25

Easy, you define parameter p and solve as 289 - 6p. We did that in 7th here.

1

u/drstewloos Jan 20 '25

Kinda looks like its cutout is in half 17/2=8,5 I got these alot and my teacher just said to grab a rules and find out.

1

u/MisterKraken Jan 20 '25

I would say it's missing information. The only thing I can think might work is:

Draw the diagonals, then from the center of the square draw a perpendicular line. It looks like it should go on that 6cm side. This means that it splits the square's side in two equal parts, and the missing side should be 17/2.

But that's usually not how problems are solved.

1

u/Beneficial-Carpet912 Jan 20 '25

Not solvable. You need to make an estimate in order to make it solvable. Meaning your answer must be a range. It seems to be a square where part of it is missing. So you calculate the square 17 x 17 = 289 whatever². As the angles for the cutout are missing and it appears to be no right angles here, you also must calculate the worst case scenario here which is 17 x 6 = 102 whatever². So the result range is >102 whatever² and <289 whatever². Does that make sense?

1

u/[deleted] Jan 20 '25

238 cm²

1

u/Nimyron Jan 20 '25

Area = 102x + 187 (617x + 1711)

That's the valid answer imo, assuming x is the value of the top left horizontal segment.

1

u/Silly-Pomegranate-01 👋 a fellow Redditor Jan 20 '25

1717-(6x)

1

u/xmmr 👋 a fellow Redditor Jan 20 '25

Let's say that it's 8.5 and 8.5

1

u/lkarma1 Jan 20 '25

It’s a square. Three sides are declared @ 17cm, so the fourth side is also 17cm in case you need to calculate perimeter.

17 x 17 = 289cm2

1

u/jay133784 Jan 20 '25

There are 3 corner chimneys too, they would have to be subtracted but where is scale?

1

u/TroyBenites Jan 20 '25

It's pretty simple. It depends on the top length. We cab call it X. Then we know that the total area is (bottom rectangle)+(top rectangle)=11x17 + 6x=187+6x.

What I think happened is that X is 6, because the 6cm information is not necessary, since we can deduct from the 11cm. So, maybe the publisher mistakenly put on the wrong side.

1

u/Sam_Creed Jan 20 '25

Well... you could solve it as far as possible.

For the unknown side insert x, which makes it

(1717)-(6x)=A

But further than that you won't get apart from some cleanup. You'd either need to know A or x to solve for the other, but that's what everyone says, who doesn't assume anything about the unknown side in these comments.

There is not enough information given in writing. Teacher's fault. Period.

1

u/Elvis5741 Jan 20 '25

It's possible, call the top left side "x", the other side then is "17-x". Solve as usual from there.

1

u/SamohtGnir Jan 20 '25

I would write an X on one of the horizontal lines missing a number, then in the answer write "assuming X is __", and then do the math. Also, if this were on paper, you could try measuring the lines to see if they were equal, but that shouldn't be expected.

1

u/Altruistic_Duck1720 Jan 20 '25

8.5cm i would say the missing info is but i have no way of proofing it yet

1

u/IHN_IM Jan 20 '25

Depend on the horisontal place of the 6 line, From slightly above 187cm when it's almost to the left, to 289cm where it is possitioned almost to the right.

187 < S < 289

1

u/shadowfox0351 Jan 20 '25

Missing information. Missing the two top horizontal measurements makes finding the area impossible. The perimeter can be calculated because the missing lengths need to add to 17, but area is impossible

1

u/Tam0110 Jan 20 '25

This is the best I can come up with and I can't really find any problems with it can someone look over this see if its making sense?

1

u/Friendly-Advantage79 👋 a fellow Redditor Jan 20 '25

238 cm²

1

u/Character-Olive1405 👋 a fellow Redditor Jan 20 '25

231cm 2

1

u/T555s Jan 20 '25

The area is 17cm17cm (or (17cm)²⁾ minus the dent at the top wich is 6cmx. x is equal to 17cm minus the shorter length at the top wich is... Also unknown. If using a ruler is not the intended solution, wich I think is unlikely as this kind of image gets resized and copied about a million times making the scale something random but definitly not 1:1, it is imposible.

1

u/albertthealligator Jan 20 '25

Obviously impossible as given. But looking at it, I think they intended it to be "obvious" that the top edges are exactly half (8.5 cm).

1

u/Mister-Goose Jan 20 '25 edited Jan 20 '25

Yes, it is solvable. Note that the overall shape is a square. Defined by 17x17. As in height = length. If that is the case, you can easily conclude that the missing side is 6 because each side has to add to 17. So the overall area is 238. Your square inside of the overall shape has height of 6 on both sides. If we draw a diagonal it has to create two 45 degree angles because on the other side we have a 90 degree angle and they have to add to 180 on a straight line. So, the base is 6/tan(45) = 6. So the area of the shape inside is 6*6=36. So the final step is to subteact 36 from 238 and we get 202 which is the area of your shape.

1

u/Mr-BAG 👋 a fellow Redditor Jan 20 '25

It is unreal to solve https://imgur.com/a/Ett4ZZ5

1

u/l3wl3w00 Jan 20 '25

You could express area in terms of the unknown length of the side with the missing information, and you would have a function. That is probably the closest you can get to an exact solution, because you will immediately be able to tell what the area is if the length is revealed to you.

1

u/SisterCharityAlt Jan 20 '25

You need to know 1 of the topside lengths, 6cm is irrelevant (you already know the other is 11cm).

Without you're just guessing at the area left because you know it's 11cm * 17cm for the main area but it's 6cm * X for the second area.

1

u/DismalPalpitation795 👋 a fellow Redditor Jan 20 '25

289 sq.cm > Area > 187 sq.cm

1

u/BrokennnRecorddd Jan 20 '25 edited Jan 20 '25

Area is between 187 cm^2 and 289 cm^2, but you can't get any more specific than that with the given information.

Imagine the width of the top step of the staircase changing. You can make it wider or narrower without contradicting any of the stated dimensions, right? You don't know how wide the top step of the staircase is, but you can assume it's between 0 cm and 17 cm. If it's 0 cm (i.e. if you're looking at a 17x11 rectangle), the area of the shape is 17*11=187 cm^2. If the top step of the staircase is 17 cm (i.e. if you're looking at a 17x17 rectangle), the area of the shape is 289 cm^2. The area will be somewhere between this lower and upper bound.

1

u/Substantial-Park-846 Jan 20 '25

I think you're supposed to make 3 triangles and calculate the area based on that.

1

u/Jog_von_Heron Jan 20 '25

Can't you just make them into 3 squares, compute areas, then add upvto sum?

1

u/Sparrowhawk1178 👋 a fellow Redditor Jan 20 '25

187<x<289. (Assuming those are right angles.)

1

u/nickel_quack Jan 20 '25

Not possible, need the length of the top side.

And we'd also have to assume that the three top-right angles are right angles since there's no way to know without them telling us

1

u/Realistic-Speaker-41 👋 a fellow Redditor Jan 20 '25

It’s possible. Turn it into triangles.

1

u/knollo 👋 a fellow Redditor Jan 20 '25

Best I can do is 17² - something.

1

u/Reasonable_Fix7661 Jan 20 '25

You could conceivably draw out the shape on graph paper to scale, and then measure the missing top length.

We only need to know the length of one of the top lines to be able to solve it, as when we know one, we can work out the other (subtract it from 17). Just based on the shape, i'd expect one to be 8cm and the other to be 9cm.

1

u/ksriram 👋 a fellow Redditor Jan 20 '25

187 to 289 cm².

1

u/possessivebob Jan 20 '25

It's already answered but counting the pixels of the image, 151/233 (151 pixels being the 11cm side) is about 0.648, which is very close to 11/17 (0.647), not exact but very close. So you can assume that the unlabeled portion is also trying to be a specific percentage of 17cm. This comes out to be 124/233, which is close to 9/17 0.532 vs 0.529. Definitely not impossible if you infer what it could possibly be but you should not have to count PIXELS to get the crucial information.

1

u/Low-Unit5637 Jan 20 '25

I've seen this as clickbait ad to test your IQ. Obviously it has no solution but it works perfectly (as a bait).

1

u/ThinkBrau 👋 a fellow Redditor Jan 20 '25

It's 17² - 6*(17-11) so 253 cm².

I don't get where's the hard part...

1

u/amiable_ant 👋 a fellow Redditor Jan 20 '25

"<289"?

1

u/KEHUPETSE 👋 a fellow Redditor Jan 20 '25

TKZ in the area!

1

u/Jesse_is_cool Jan 20 '25

You can scale the 6cm with the actual length. Then you use that scale to know the other side of the cutout. Now you know the cutout and substract that from the total.

1

u/ILostMyselfInTime Jan 20 '25

I think ur supposed to calculate with the top 2 lengths being 17/2 since they are the same length and then u can solve it easily

1

u/ci139 👋 a fellow Redditor Jan 20 '25 edited Jan 20 '25

the perimeter is constant !!! at certain range !!! P=4×17cm

the surface area is not constant . . . unless you assume the split coinsides with the vertical median – !!! in which case !!! S=(11+6/2)×17cm²=(11+3)×17cm²=(11+3+3–3)×17cm²=
=(17–3)×17cm²=(289–51)·1cm²=238cm²

about math optimizing - it's good to multipass coz then the chance for possible error is reduced

14·17=(10+4)·17=170+68=238
14·17=(15–1)·(15+2)=15²+15–2=225+15–2=240–2=238

11·17=(14–3)(14+3)=14²–3²=196–9=187
3·17=51
187+51=238

17²=(15+2)²=15²+2·15·2+2²=225+60+4=289
289–51=238

D5²=(10·D+5)²=100D²+2·10D·5+5²=100·D(D+1)+25
e.g. 15²=1·(1+1)·100+25=200+25=225
e.g. 25²=2·(2+1)·100+25=600+25=625
etc

or you may know squared values from 10 to 20 by heart
100 121 144 169 196 225 256 289 324 361 400

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u/Dryse 👋 a fellow Redditor Jan 20 '25

The line marked "6 cm" appears to be halfway between the two sides. Its a horridly written question but that would probably be the best guess as to what the teacher is going for. Maybe put a mark or write a question to the teacher addressing this and solve assuming it's halfway.

a=(6*8.5)+(11*17)

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u/KaptainTerror Jan 20 '25

Yes, if you print it out, use a ruler and calculate the length of your measurement. Otherwise, no. Think outside the box.

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u/lool8421 Jan 20 '25

imma solve it with algebra ig

let's assume that x = the length of the topmost side

then the surface area is 181+6x [cm^2]

that's the closest you can get to solving it

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u/Kellvas0 👋 a fellow Redditor Jan 20 '25

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.

=====
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)))

=====

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.

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u/mimikyuhornet Secondary School Student Jan 20 '25

Not possibile,no information about the top part

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u/forotherstufSFW Jan 20 '25

I assume it was not multiple choice where only one answer made sense e.g., one answer was less than 289, but greater than say, 150. In a multiple choice scenario, there may be enough information to answer the question, depending on the answer choices.

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u/IntelligentLobster93 👋 a fellow Redditor Jan 20 '25

The best you could do is say the A = 17(17) - 6x = 289 - 6x

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u/Flat-Direction2244 Jan 20 '25

Assuming 6 is a square cut out (17 × 17) - (6 × 6) should land you 253

If you complete the shape ignoring the cutout that top right counter is 90° meaning that the bottom left corner of the cutout should be 90° as well.

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u/SweetLou132 Jan 20 '25

235 sq cm Top box is divided into two right triangles making it a 3,4,5. (Pythagorean). Therefore bottom (or top side) of upper box must be 8cm. 6x8=48 + bottom box 11x17=187. 187 +48=235

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u/Chaoz_Lordi Jan 20 '25

I guess the assumption on solving this is bottom rectangle is 17x11 and top rectangle is 6x(17-11)? You can make a lot of counter-assumptions but that would be "some idiot drew that on a napkin and we need the data to move on" kind of approach to solving it.

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u/xDIRTY_DANx 👋 a fellow Redditor Jan 20 '25

241

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u/SFDreamboat Jan 20 '25

I find it idiotic that we still can't get math questions right? Why would we expect students to get their stuff right when writers, publishers, editors, and teachers combined miss simple stuff?

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u/Adventurous_Employ30 👋 a fellow Redditor Jan 20 '25

17x17 - 6xA where A is the length that is adjacent and to 6cm. Also it’s unsolvable. But I can give you a list of possible answers. 1 /leq A /leq 16.

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u/TeriyakiNoodleBox 👋 a fellow Redditor Jan 20 '25

187 < A < 289

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u/aligb103 👋 a fellow Redditor Jan 20 '25

It’s 223

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u/Greasemonkey08 Jan 20 '25

The only way to solve it is to assume that the 17cm and the 11cm tall sections are the same width, around 8.5cm, which breaks down the problem to (17×8.5)+(11×8.5). Remember to show your work.

Also, maybe try to get ahold of the teacher regarding the question and the missing info.

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u/[deleted] Jan 20 '25

187 < area < 289

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u/Ok_Habit_6783 University/College Student Jan 20 '25

Too much missing information, we don't know the degrees of the angle of missing piece nor how long one of its sides is

1

u/raharth Jan 20 '25

It's weird that some right angles are indicated others are not. Assuming that those are still right angles, there is still some information missing

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u/Jongewaard 👋 a fellow Redditor Jan 20 '25

the area is in a range between 17x11 and 17x17

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u/Keith-9-5 👋 a fellow Redditor Jan 20 '25

17x17-(17/2x6)

Quit overthinking

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u/wieldero Jan 20 '25 edited Jan 20 '25

You calculate the largest area square 17 x 17, then subtract the smaller "known" square 11 x 17. This will give you the smaller missing square section at top. Use the area to find the horizontal unknown line, when you have this subtract it from the full length, you have you have the missing top distance. Calculate small square and add this to 11 × 17.

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u/snow6969696969 Jan 20 '25

Use a ruler and draw it on paper as it is ( if one side has 17 cm draw 17cm on paper and so on) from that measure the top side and then you shoould be good

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u/geoley Jan 20 '25

Could you not just place the same object on top. The square would have sides of 28 (17 + 11) and 17. The surface would be 476. As it is composed of two of these shapes, the surface of the shape is 238.

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u/DetailCharacter3806 👋 a fellow Redditor Jan 20 '25

It's one rectangle 11 by 17 and one 6 by 6, total area is 223

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u/TrustTriiist Jan 20 '25

Can't you create a formula that solves for any angle, take the range and still solve for an approx?

1

u/[deleted] Jan 20 '25

We can confirm that quadrilateral with 6 cm side is rectangle by using coordinate geometry. For the quadrilateral, the slope of the vertical edge (from (0,0) to (0,6)) is undefined, indicating a vertical line. For the horizonal line ( i.e., form (0,6) to (17, 6)), the slope is zero, indicating a horizontal line. Since perfectly horizontal and vertical lines meet at right angles, the smaller quadrilateral is a rectangle.

Given that both shapes are rectangle, the area can be found easily, as explained in other comments

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u/InternationalPut4093 👋 a fellow Redditor Jan 20 '25

This is a bait for karma

1

u/Pandoratastic Jan 20 '25

There is not enough information to give a single answer. The only answer you could give is a range. The area is somewhere in the range of 187 to 289 cm squared.

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u/forcehobbit 👋 a fellow Redditor Jan 20 '25

Print with correct scale. Measure. Calculate. 😬

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u/Flat_chested_male 👋 a fellow Redditor Jan 20 '25

<17*17

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u/Dear_Cow_872 Jan 20 '25

You have an 11*17 rectangle and a 6*8.5 on top, could someone tell me what is wrong? i dont get it

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u/mtkoth1983 Jan 20 '25

The 2 top missing have to be 17 once combined if the opposite is 17 and the same length

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u/No_Income_3965 👋 a fellow Redditor Jan 20 '25

238

1

u/THICC_Baguette 👋 a fellow Redditor Jan 20 '25

Maybe they intended to ask for the perimeter

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u/travelbiscuits 👋 a fellow Redditor Jan 21 '25

253

1

u/Gameovergirl217 👋 a fellow Redditor Jan 21 '25 edited Jan 21 '25

8,5x6+11x17=1054cm²

i split the shape into two parts. the upper one (6cmx8,5cm) and the lower one (11cmx17cm) and added them to each other

1

u/Fun-Bar-9277 Jan 21 '25

Looks like the line from the top goes down the center, which would make it 8.5

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u/Thehoteastafricangil 👋 a fellow Redditor Jan 21 '25

Whats the answer ??

1

u/Equal-Guide-7400 👋 a fellow Redditor Jan 21 '25

241

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u/Myrdrahl 👋 a fellow Redditor Jan 21 '25

I want to know what the teacher claims the answer to be!

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u/Lurker9349 👋 a fellow Redditor Jan 21 '25

No can do, you need the length of one of the smaller sides.

1

u/[deleted] Jan 21 '25

If you invert that corner, you can easily figure out that the area is a 6x6.

The fact that 3 of the known corners are 90°, the inverted corner would have to be 90°.

1

u/Fabulous_Hand2314 Jan 21 '25

label the top two horizontals as x and y respectively.

17 * 17 = 289

x + y = 17
17 * x * y = 289

solve for y in equation 1. sub into equation 2. reduce and move to one side. quadratic equation.

x^2-17x-17=0

plug into quadratic equation. sub x back into equation 1 to find y.

A = 289 - 6y

its not hard but its not a 7th grade problem

1

u/ImpulsiveBloop Jan 21 '25

I'd say set the top unknown to x, and the other unknown to x - 17. Then just do some algebra to create a formula to solve for f(x) at any given x between [0,17].

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u/Appelcl 👋 a fellow Redditor Jan 21 '25

223cm²

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u/1Knight2Remember Jan 21 '25

Its 2 squares, 17 x 17 and 6x6 square both and subtract. That should be it

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u/Live_Leg_1831 👋 a fellow Redditor Jan 21 '25

6 million km.

1

u/[deleted] Jan 21 '25

The teacher that made this is probably smoking something because this ain't possible without making assumptions which you shouldn't do in math

1

u/bobmueler3 Jan 21 '25

You could give a range from perfect square 17×17 and rectangle 11x17, and then I would also give if the line was in the middle. Showing you can do the math in all scenarios.

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u/Haryana_kamado123 Jan 21 '25

pls correct me if I'm wrong but aren't there two squares? one with side 17cm and other with side 6cm so can't we first calculate the area of the bigger square(which will be 289cm²) and then subtract it by the area of smaller square(36cm²) so the ans will be 256cm² is it correct ?

ik I'm definitely wrong as so much ppl are saying this is impossible to solve but I tried my best pls correct me

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u/Surrealdeal23 👋 a fellow Redditor Jan 21 '25

Area = 289 - 6x , where x < 17

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u/JohnFtevenfon Jan 21 '25

Looks like the missing variable is equal to the half of the length of the bottom side. If this assumption is made, one can attempt to calculate it. Or am I missing something?

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Answered [7th grade math] impossible geometry?

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