This question is actually impossible

This problem seems like it is solvable, but it is actually an impossible problem. Can you figure out what is wrong with the question?
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[@kseliascryser5259]

In her final math exam at school a friend of mine was given a bunch of info on a hamster cage and then they asked to calculate a bunch of stuff on the bunny cage. She wrote that there was no information given on any bunny cage and therefore she could not calculate anything. Got full marks :D

[@Player-fg4ub]

are u in primary sch

[@Lizkyyy]

@@Player-fg4ub could've been an optimisation question?

[@Mikustan39]

@Nix_UKI feel like mine would have as well.

[@MichaelOnines]

Third "solution": Triangles are similar. Ratio of areas is 1:4, thus ratio of sides is 1:2. Base is broken into two parts 5cm and 10cm long, height is broken into two parts 2cm and 4cm long. Area of requested rectangle is 10cm x 2cm = 20 cm^2. If the bottom dimension were 12 instead of 15 the problem becomes self-consistent.

[@greatbriton8425]

Well done!

[@shaswatadutta4451]

I had thought exactly in your line. The areas of the triangles would be 20 cm^2 and 5 cm^2.

[@jhouck1969]

This was where I went as well.

[@Echinacae]

This

[@justpaulo]

Same way I did

[@NestorAbad]

There's a quick way to see that this setup is impossible: let x and y be the horizontal and vertical legs of the 4 cm² triangle, so xy=8. As the 16 cm² triangle is similar and 16/4=4, then the ratio of similarity is √4=2, so 2x and 2y are the legs of the 16 cm² triangle. But then 3x=15 and 3y=6, meaning that x=5, y=2, so xy=10, contradiction! It was fair to give full marks for all students! As usual, thanks for sharing.

[@L17_8]

Jesus loves you ❤️ please turn to him and repent before it's too late. The end times described in the Bible are already happening in the world.

[@proosee]

That was exactly my solution, you saved me tons of writing 😂

[@deandelvin9924]

Bro how did you post this 1 day ago when the video is 1 hour old?

[@handanyldzhan9232]

Just came here and solved the exact same way.

[@prometheus7387]

I basically use similar triangles and solved for areas and lengths which is similar to what's being done here

[@seinfan9]

"This is impossible to solve. Keep watching to learn how to solve it."

[@stuchly1]

And THAT'S the answer! 😂

[@hakeemcanonio3840]

In engineering, we call this over-constraining the information.

[@LegendaryBea]

In high school language Is it like giving 3 non concurrent linear equations?

[@pierrecurie]

@@LegendaryBea yes, except for the linear part

[@LegendaryBea]

@@pierrecurie why ? How is it wrong

[@pierrecurie]

@@LegendaryBea The equations here aren't linear, but you have the right general idea.

[@LegendaryBea]

@@pierrecurie yes i meant it in that way only as the original comment said "in engineering" so i thought of it in an easier way

[@TrapShooter68]

After scouring my brain to prove there IS a solution, I've come to the conclusion that the answer is 42. The answer to the Ultimate Question of Life, the Universe, and Everything

[@danmerget]

What was the question?

[@mike1024.]

​@@danmerget In this case, the question doesn't matter. The answer is still 42. :-P

[@mitchbandes2245]

What is 6 x 9?

[@charlesnewborn3760]

Cringe.

[@sans1331]

@@mitchbandes2245-4

[@markusf8445]

Consider the lower left rectangle plus the upper right rectangle. Their total area is 40. However, if we visually move the central intersection along the diagonal, the equivalent total approaches 90 at both corners (the full rectangle) and reaches a minimum of 45 in the middle (half of the rectangle). The total of 40 is thus too small.

[@wombat4191]

That's an interesting way to see the error in the numbers!

[@fylosofer]

Yup, that's what I noticed.

[@meekrab9027]

Yes, there's simply no way to construct the initial conditions in Euclidean space.

[@GamezGuru1]

this, much faster than anything in the video, and how I realised it cannot be solved...

[@ericrawson2909]

@meekrab9027 Does it work on a spherical or hyperbolic surface? If so what is the Gaussian curvature of the surface? How about that for an exam question

[@mike1024.]

My gut reaction to this problem was that 4 was awfully small given the dimensions of the rectangle, meaning that 16 was probably too small for the 4. So I used similar triangles to determine what the dimensions would have to be for the smaller triangle, calculated the other dimensions, and sure enough, the corresponding area would have to be bigger than 16. You can actually do a single dimension if you note the height to base ratio is 2/5, so the area of these triangles are (1/5)b^2, where b is the base of the current triangle.

[@dunstvangeet1500]

It the bottom side was 12 cm, instead of 15 cm, then it becomes consistant with the given information, and all the information would have been consistant. Under 12, you can solve it any of the ways. I used similar triangles, based upon the areas, which seemed to most intuitive for me. Under the original problem, the lower triangle had a area that was 1/4th of the area of the large triangle. With what we know of the angles (that they're the same), we know that the sides of the smaller triangle is exactly 1/2th of the length of the larger triangle. So. If it was 12 cm on the bottom, that means that the length of the smaller bottom is 4 cm, and the length of the bigger triangle (the length we're looking for) is 8 cm. Now, on the other side (6 cm), we can divide this into a 4 cm, and 2 cm. Taking the right sides, we get a scenario that is 2 cm x 8 cm, or 16 sq cm. Now, to check, we have 16 + 16 + 4 = 36 sq cm. This is exactly 1/2 the size of 72 sq cm, which is what we get if we multiply 6 cm x 12 cm. So, the bottom length was actually meant to be 12 cm.

[@MathsMadeSimple101]

impossible just stands for I'm possible.

[@ARichli]

@@dunstvangeet1500 your reverse engineering of the problem saved me a while of figuring out what would have been the error. Thank you!

[@mike1024.]

@@ARichli changing the given areas to 5 and 20 also corrects the issue.

[@JordHaj]

If the diagonal is a straight line, the problem is indeed impossible. However, if it is not, then there are two proper solutions. Let the horizontal leg of the lower left blue triangle be x, and that of upper right one be y. Then the vertical legs of each of them will be 8/x and 32/y, respectively. Consider the area of the biggest rectangle, 15 * 6 = (x + y) * (8/x + 32/y), which expands to 40 + 8 * (y/x) + 32 * (x/y). Since the area of the rectangle we want to find is y * 8 / x, substitute t = y/x. Having solved the equation, we get t = (25 ± 3sqrt(41))/8, and the area is thus 8*t = 25 ± 3sqrt(41)

[@barttemolder3405]

Agreed. Clever to directly aim for the area, and your math is correct (even if it uses a different position for x than the video does, but that greatly simplifies the equation).

[@qinqiabc-uo6yq]

you are right !

[@mynameissang]

Yup, I really wish this video would have clearly explained that the problem, as presented here, is an impossible construct; but if we suppose the problem to avoid "over-constraining" in its presentation, then there can, indeed, be a solution. Then we can work out an actual solution, and let this example be an exercise on why the statement "image not drawn to scale" is important, and also point out that, at that point, we've probably stepped outside of the intent of the problem's original thought process.

[@JohnDoe-wg9oh]

It would be nice to have a visual explanation that shows why there is no unique answer.

[@johanneschristopherstahle3395]

It's not about unique answers. Actually the sizes given in the problem are not possible to be true all together at the same time. Since both triangles are similar and the ratios of the sides is equal to 6/15 (or 2/5), you can determine the exact side lengths of each of the given triangles. But if you do so, you will find that both triangles together would not be large enough to touch each other and the corners of the rectangle. One could fix this by either changing the sizes either of at least one of the triangles or the rectangle in a way that the proportions of the sides to the given areas become valid. You can come up with solutions that seem valid, but if you check the sizes of the triangles and the rectangle, you will find, that suddenly at least one of the given sizes will have changed. Or in other words: you can prove that the sizes given can't exist. This prove should be the only valid solution to the problem.

[@barongerhardt]

Take a square of sides length one. Then draw the diagonal and mark as length two. What is the area of the resulting triangle? Some would think that is impossible, but they are assuming it is drawn on a surface with zero curvature.

[@VinyJones2]

​@@johanneschristopherstahle3395showing the contradiction, that prove that the problem data are false, 👍🏼

[@sparshsharma5270]

*When there's a solution, it can be 42, but when there's no solution, then it's definitely 42*

[@peterpan408]

Close, 44.21 or 5.79.

[@TheBrilliantShadow]

I went the funny approach by plotting the two sides. I split the 6 to a and b, once with 6 times x, once with 6 times 1-x. Same with the 15 to c and d. Overlaying a second plot with a(x) times d(y) = 32 you get a nice line that fits the equation. Same with b(x) times c(y) = 8. You get two crossing points for the ratios that fit the setup. Solving for: a=5.0806, b=0.9194, c=8.7015, d=6.2985 (small rounding error included). That makes R = 5.7908 cm^2 or 44.2088 cm^2 for the other.

[@Feltare]

I solved the problem before watching the video in a similar method of finding a, b ,c, and d. I noticed as i was solving id get two sets of values for those variables and you mustve gotten the second set bc i solved for the other set and got the same area of 44.2. All my values satisfy the given statements that are deemed "true" by the values in the image

[@rojnx9]

I did this problem from the thumbnail before clicking, and I had a completely different way: The gradient of the diagonal is 6/15 and so both the upper right and lower left rectangles have a height to width ratio of 6/15. The upper right rectangle is 32cm^2 (double the 16cm^2 triangle) A=L*W W=15/6*L (where L is height, W is width) A=15/6*W^2 32=15/6*W^2 sqrt(12.8)=W This is also the width of the unknown area The lower left rectangle is 8cm^2 (double the 4cm^2 triangle) A=L*W L=6/15*W (where L is height, W is width) A=6/15*L^2 8=6/15*L^2 sqrt(20)=L This is also the length/height of the unknown area Finally: Calculating the unknown area A=L*W A=sqrt(20)*sqrt(12.8) A=sqrt(20*12.8) A=sqrt(256) A=16cm^2 I think it is interesting that (even though the diagram is impossible) there were so many different methods for this one question.

[@emremokoko]

i tried a few solutions and all of them gave a different answer (with one giving "no possible answer".) it's really "interesting".

[@ashleyzinyk399]

I addressed it with similar triangles. You quickly find that the upper right rectangle is 10x4 cm, thus an area of 40 cm^2, but half of 40 isn't 16 cm^2.

[@MrElvis1971]

Exactly, literally took seconds to work out it's not possible.

[@herculesatan4514]

My guess is 25 cm^2. A triangle with height 6 and base 15 would be 45. Subtract 4 from it to get 41. Since the rectangle with a triangle of 16 cm^2 is split perfectly in half by it, one can assume that the remaining area is also 16 cm^2. Subtract that from 41, and you’re left with 25.

[@philipmcnamara4071]

That‘s what I think as well.

[@paullau3835]

I would give full mark to this answer. All the others, even though their proofs are also correct, should get 9/10 as they are not efficient and wasted too much time. There is a saying - work smart and not harder.

[@samueldeandrade8535]

Good guess. But the real answer is "impossible".

[@TonboIV]

The first thing my brain noticed is that 16 is 4 times 4. 4 is a nice simple ratio, and it's also 2^2, which seems like a very geomotry problem-ish piece of information, the kind of thing that isn't put in a problem by accident. Since the triangles are very obviously similar and the upper one has 2^2 the area, it must have 2x the linear dimensions. That means the rectangles containing them have a 2 to 1 ratio, so we can then divide the dimensions of the outer rectangle into 3, and 15 and 6 are both nicely divisible by 3 (obviously intentional), which means the dimensions of the lower right rectangle are 10 and 2, giving an area of 20. Yet a thind different answer!

[@tomdekler9280]

Yeah it's pretty clear they meant to put 12 at the bottom. That fixes everything.

[@Sasha2CZ]

Yes, the assignment is clearly wrong. A rectangle with sides 15 and 6 cannot have triangles constructed in this way with areas of 16 cm^2 and 4 cm^2.

[@tomdekler9280]

@@Sasha2CZ Why was there a need to restate the conclusion that's in the title of the video like that?

[@stigcc]

@@tomdekler9280It was nice to get a different perspective

[@cmilkau]

It's funny how you can work out properties of things that can't even exist

[@markotrieste]

I think I can prove a third value for the xy rectangle. The 4 and 16 cm^2 triangles are similar in shape. Given that the second has four times the area of the first, it means that their sides must be in 1:2 ratio. Which means that the xy rectangle has sides 2/3 and 1/3 of the outer one, i.e. 10x2 cm.

[@AntoineADubai]

I had the exact same train of thoughts.

[@peterpan408]

The triangles are not similar 😮

[@markotrieste]

@@peterpan408 why not? the three angles are equal.

[@ubulom76]

If you have contradiction in your premises you can prove anything. Try finding a solution for an xy area of, like 10000 :)

[@aliensarerealttsa6198]

The smaller triangle wouldn't even fit into the larger one three times. Def not going to work with 4. Also something that is 16cm2 can't fit into something that is 15 x 6 (or about 9.5cm2)

[@Zarunias]

I got 20 on my first quick glance. All the triangles are similar, and one has an area 4 times as large as the other, therefore its sides must be double the length.This means that the sides of the big rectangle must be cut in thirds. Therefore the sides of the small rectangle R must be 10 and 2, giving it an area of 20. Immediately after that I realized that the rectangle on the bottom left has dimensions 5 and 2, giving the 4cm² triangle an area of 5cm². I wasn't surprised that much as during the presentation of the problem I realized that there is one constraint more than needed.

[@hmata3]

I looked at it differently. Area of largest triangle is 6 cm x 15 cm * 0.5 = 45 The larger white triangle is 16, equal to the blue area right next to it. So green rectangle = 45 - 4 - 16 = 25

[@hmata3]

Guess I didn't realize the area of the smaller triangle isn't 4 😂

[@nmklpkjlftmch]

@@hmata3 This was how I saw it after about 5 seconds of looking at the preview and wondered how he could have considered it impossible, or even difficult. It made sense after watching.

[@to2podemosaprender630]

The lengths and the width of the rectangle are wrong it should be 12 by 6 so the triangles are 8×4÷2=16 and 4×2÷2=4 Or the triangles sizes are wrong they should be 20 and 5, 10×4÷2=20 !and 5×2÷2=5 .... I speak geometrical propprtion... thanks! A typo 1 2 3 4 5 6 7 8 9 0 5 instead of 2

[@salerio61]

I think there's a third answer as well :) the ratio of areas of the two given triangles is 4, that means their lengths have a ratio of 2. So 3h=6 and 3w=15, thus the square has sides of 6/3 and 2*15/3 giving 2X10=20cm^2

[@Actually_Zahren]

I went down a slightly different route that still ended up with a set of numbers that made no sense. I called the lengths of the larger triangle A and B, and the lengths of the smaller triangle C and D. A*B=32 and C*D=8. Knowing that (A*B)+(C*D)+(A*D)+(C*B)=90 you end up with (A*D)+(C*B)=50. From there replace A and B in terms of C and D (since A=15-C and B=6-D) and you end up at 6C+15D-2CD=50. There are three positive solutions to this: C=5 D=4, C=7 D=8, & C=10 D=2, however none of these numbers fit to get the area of the smaller triangle (since (C*D)/2 must equal 4). Thus there must be something wrong with the numbers provided

[@DavidDSimon]

I went the route of blue triangles being similar (small is half side lengths of larger) so the short sides have to be 2 and 4 requiring large sides to be 4 and 8 to get those areas which contradicts the measurements given. 15 needs to read 12.

[@jamesforte605]

I assumed that the drawing was not to scale and used four rectangles. 32cm2 and 8cm2 are known. The other two need to add up to 50cm2. After a bunch of number crunching I have the sides of the 8cm2 as 8.7 x 0.92. And the sides of the 32cm2 as 5.08 x 6.3. This gives the area in question as 6.3*0.92= 5.8cm2. This is the correct answer given the areas in the diagram.

[@christianbennett2001]

interesting! I actually had the same assumptions and answer! but during your "number crunching" I encountered another possible solution. its true that you can have the smaller triangle area 4 have sides of 8.7 and 0.92, which yields the shaded area as 5.8 cm^2. there is another solution (equally valid) where the small triangle has the dimensions 3.481 height and 2.297 width, and in this scenario I calculated the mystery area to be a whopping 42.219 cm^2! (please anyone feel free to check my math)

[@jamesforte605]

Your math checks out! Two correct answers.

[@peterpan408]

Bingo! Spotting the Tricks and Red-herrings in math exams are core skills for a student. Trick - not similar triangles. Red-herring - showing triangles at all! Good bonus question though.. Also 44.21 is the other answer 😉

[@deerh2o]

Really great example. Generally, in logic, if you start with a false premise, you can pretty much conclude anything.

[@godowskygodowsky1155]

EDIT: Disregard this reply. I had skipped to the solution and thought the problem was underspecified rather than overspecified. That's not quite what's happening here. The actual analogy is that you have a model with multiple elementary extensions and thus have multiple theorems consistent with the axioms.

[@samueldeandrade8535]

​@@godowskygodowsky1155 are you sure? Because it doesn't look like what you said refutes OP's comment.

[@godowskygodowsky1155]

@@samueldeandrade8535 No, they're different things. I originally didn't fully watch the video and thought that the final setup was the problem, which was underspecified rather than overspecified. OP is right.

[@samueldeandrade8535]

@@godowskygodowsky1155 oh I totally get your previous comment now. Hehehehehe. Ok, everything is clear now.

[@samueldeandrade8535]

@@godowskygodowsky1155 and you chose to edit intead of deleting your comment. I like you a lot because of that!

[@davidsommer325]

Solved the Problem on a third way and actually got the intended solution (A = 25cm²)

[@3057luis]

The only rectangle that is 32 in area is 8,9X3,6 (with aspect ratio of 15/6). So if the other is not 8 in area we have something imposslble.

[@smilerbob]

The moral of the story… Question everything and trust nothing

[@danc8278]

Total area of rectangle is 90, divide by two for the triangle made by diagonal, now you have 45. At this point subtract the area of the two blue triangles and the green area is 25sq cm

[@antoninoincalcaterra5152]

I calculated the exact same way and I am wondering why nobody notices

[@nogurt6273]

Umm….. yes, but did you watch the video, or go straight to the comments? It’s impossible to answer the math question because the initial numbers given don’t mesh… as thoroughly explained in the video.

[@danielcassell4527]

I got 20. The small triangle and larger one are similar and one's area is 4 times the other, so its sides should be twice the other's. This gives x as 10 and y as 2, using the pictures they came up with. Note that you can use this to find the area of the given triangles and you end up with 5 and 20

[@samueldeandrade8535]

Great. But the real answer is "impossible".

[@Inspirator_AG112]

Using similar triangles and inferring the areas of the 2 rectangles and proportional dimensions, we immediately realize that the bottom length would have to be 12.

[@Qwentar]

I would have said that it wasn't necessarily drawn to scale: that the two smaller triangles aren't necessarily similar to each other, nor the larger triangle, that it isn't a straight line connecting the two opposite corners of the large rectangle. This means we can draw two rectangles for which we need to find the areas. That would likely be the first two results you posted.

[@mike1024.]

This wouldn't be solvable in the original problem because you can't calculate the area of the full lower triangle like he did in the video. Or are you proposing the problem also be changed to finding the sum of the two smaller rectangle areas?

[@peterpan408]

Showing triangles is a classic math exam red-herring thing to do. Oh.. I love triangles.. let's go geometry! Nope! It's a quadratic roots sort of problem 😏

[@finris1]

Use algebra and geometry to prove the problem doesn't make sense. Note that the diagonal of the entire rectangle is also the diagonal of the lower left and upper right rectangles. Then the diagonal has a slope of 2/5. Let x and y be the length and height of the lower left rectangle. Then y = 8/x and y = (2/5)x. From here, 8/x = (2/5)x --> 8 = (2/5)x^2 --> 2x^2 = 40 --> x^2 = 20 --> x = sqrt(20) = 2*sqrt(5). Repeat similar calculations for the upper right rectangle, and you get a length of 4*sqrt(5). By these calculations, the combined lengths of the rectangle is 2*sqrt(5) + 4*sqrt(5) = 6*sqrt(5) which is roughly 13.4. Similarly, the calculated heights combined would be 12*sqrt(5)/5 which is roughly 5.4. Both lengths and heights are less than they should be.

[@ricardoguzman5014]

You don't need 2 variables, only one. Recognizing the fact that for every unit length in the horizontal direction, the length of units in the vertical direction is 6/15. Thus, for the triangle with area 16 square cm, it is very simple. Since x is the long side, the short side must be (6/15)x, which very handily becomes (2/5)x. Area of triangle is 1/2 bh, so 1/2 * x(2/5)x = 16. The same can be done for the triangle with the area of 4 sq cm. The rest follows with simple algebra. Using 2 variables complicates the problem unnecessarily. Good thing that sharp students can find the errors in the given values. Either 4 square cm is correct and 16 incorrect, or 16 sq cm is correct and 4 incorrect. If instead of your final solution of keeping the area of the large triangle as 16 and finding the area of the smaller triangle, if you keep the area of the smaller triangle as 4, the larger triangle would have an area of 49 - 12√5, or about 22.167 sq cm, which would make the area of the lower right rectangle about 18.833 sq cm. Also interesting to note that the lower right rectangle and the upper left rectangle have equal areas.

[@mikespence2836]

Do all of the areas add up to 90 square centimetres? Worth checking...

[@Meshamu]

For those lines cutting up the larger rectangle to form a proper rectangle in the lower right, they gotta be perpendicular to the lines they're drawn between, but the vertical one has to be at some kinda angle, rotated clockwise a bit, for the blue triangles the sizes they are, because they just don't add up to the full length of the larger rectangle if that's perfectly perpendicular. Gotta open up the supposed right angles of the 16 and 4 triangles to about 116.6 degrees. The resulting trapezoid in the lower right is 21 square centimeters.

[@WoodyC-fv9hz]

25. :) Equation is: 90/2 - 16 - 4 = 25. I dearly apologize if I am wrong. I am drunk, dealing with a cardiac arrest on the 28th last month. Two thinks I need to say, I love Presh's super effort and the entire community at MYD.

[@r1zynn998]

I can't see any reason why this isn't correct, we'll done and I wish you the best on your recovery

[@ishkanark6725]

I reached this conclusion about 20 seconds before I scrolled upon your comment, and I am not drunk

[@Pengochan]

One could also note that the upper and lower triangles are like triangles. Since the area of the upper triangle is 4 times that of the lower triangle it is also scaled by a factor of 2 (2x2=4). But if the dimensions are 1:2 and the total height is 6, the individual heights are 2 and 4, also the base length must be 5 and 10. I.e. when the upper triangle has 4x the area of the lower one the dimensions must be 5x2 and 10x4 with areas 5 and 20. If it were e.g. 12cm instead of 15cm it'd work.

[@ericlizama8552]

That's interesting. I used the fact that the larger triangle was four times the are and congruent to to smaller triangle to get the area of the rectangle as 20cm^2. This is because the height of the smaller triangle would be 1/3 the height of 6cm, and the length of the larger triangle would be 2/3 the length of 15cm.

[@Olav_Hansen]

0:40 if you were to make the blue parts rectangles they'd take up 40cm^2, while being half of the total area of 15*6=90 (so 45) 45 can't be equal to 40, so indeed the assignment is wrong. Whoever gave the assugnment probably wanted people to flip the 16 180 degrees, so that 16+4+?=45, which would be 25, but that just plain isn't possible with the drawn angles.

[@SerifSansSerif]

ok. paused it. 15x6=90 Upper triangle has an area of 16 and the lower triangle has an area of 4. The diagonal splits it in half. Theupper triangle is the same as the other half of the quadrant that coincidentally is a portion of the half made by th diagonal. Same goes for the smaller triangle. So if you take half of that 90 area we came up with before, you get the area beneath the diagonal, and you just have to subtract the sum ofthe triangles, or the answer is 90/2=45-(4+16)=25. Coincidentally it will also be the area of the top most rectangle...

[@cpsof]

If the triangles were same sized, they would be 7.5*3/2 = 11.25 each. So their total area would be 22.5. If the triangles are different sizes, their total area becomes larger because that's how 2nd powers work (we're dealing with areas here). So it's not possible that the triangles would have areas of 16 and 4, as their total area would be 20 which is less than 22.5.

[@SimonClarkstone]

That was my reasoning too.

[@leppie]

Did it the same way. Easy enough to do in your head without sweat.

[@mastick5106]

I missed the similar triangles bit and approached it from a different angle. If the lower part of the vertical (which you labeled y) is A, then the left part of the horizontal must be (8/A) since the lower-left quadrant area is 8, making the right part of the horizontal (15 - 8/A). Since the upper part of the vertical would be (6 - A), this means the upper-left area is (6 - A)*(8/A) = (48/A - 8) and the lower-right quadrant area is A*(15 - 8/A) = (15A - 8). The sum of the areas of all quadrants is then (48/A - 8) + 32 + 8 + (15A - 8) = 48/A + 24 + 15A = 90. Thus, 15A - 66 + 48/A = 0 (A cannot be zero). We can multiply both sides by A: 15A^2 - 66A + 48 = 0 Or (dividing both sides by 3) 5A^2 - 22A + 16 = 0. Using the quadratic equation to solve for A gives us A = [22 +/- SQRT(484-320)]/10, meaning: A ~= 1.2806 OR A ~= 3.1194, both of which are within the valid range for A. However, if you use both values to work out the other partial side lengths, neither one results in an upper-right quadrant area of 32 (you get about 41.31 and 35.82, respectively), meaning the initial values are inconsistent even if you don't assume the upper right and lower left quadrant diagonals are on the diagonal of the full rectangle.

[@barttemolder3405]

I just looked at the thumbnail, so I missed the diagonal part. I thought the length of 15 and side of 6 meant the diagonal wasn't straight but slightly bent - not a real diagonal after all. I came up with the same formula save using y for your A as in the video. Yet, you made a mistake in the ABC formula. It should be y = (22 +/- SQRT(484-320))/10 => y ~= (22 +/- 12.806)/10 => y ~= 3.4806 or 0.9194. And then these values do lead to the resulting areas having the correct size. Then as x = 15 - 8/y you can with confidence calculate the area xy = 15y - 8 ~= 44.2093737123 or 5.7906272877.

[@mastick5106]

@@barttemolder3405 I'll have to go back over that later. I was using a spreadsheet to run the final calculations so I could have put a typo in a formula.

[@mastick5106]

Yep. I see the typo. Good catch.

[@bryonnoel4254]

The hypotenuse of the large triangle is 16.155 cm. The angles of the triangle are 21.801 and 68.199 degrees. We can find the triangle lengths from the area and the angles using y = sqrt(2 * area * tan(a)) and x = sqrt(2 * area / tan(a)). The small triangle with area of 4cm has width / height of 4.472 cm / 1.789 cm. The larger 16cm triangle has width / height of 8.944 cm / 3.578 cm. The total width of 4.472 cm + 8.944 cm = 13.416 cm NOT 15 cm. The total height of 1.789 cm + 3.578 cm = 5.367 cm NOT 6 cm. Whomever wrote the question did not verify the the areas or the sides.

[@igisanchez265]

this and this is easier.

[@fjorland_norsk3996]

1- The total of all areas is 15cm*6cm = 90cm^2 2- Divide the total area in two like the diagonal line is shown, you now have two halves of 45cm^2 3- To get the area of the green take 45cm^2 - 4cm^2 - 16cm^2 = 25cm^2 Green area is 25cm^2

[@rcengineer]

In any CAD program this program is what it would call "overdefined," where a perfectly defined diagram constrains all elements based on a given set of driving dimensions, but those any one of those dimensions can be modified without changing any other dimension. An underdefined diagram is one where one or more elements can be moved independently of any given driving dimension. An overdefined diagram contains one or more incompatible driving dimensions.

[@kevinmartin7760]

I wonder if the problem with this problem came about because someone took a problem that had dimensions like 2sqrt(5) and rounded them to the nearest integer.

[@kongolandwalker]

The figure is already fully defined with the 6cm, 15cm, 4cm^2. The only problem there can be is that by solving from bottom up you get other than 16 area of the top triangle. But there might be some curved space within which you can get 16, i think. So you have to find the curvature of the space of the universe, where the figure can exist, to continue solving for green area.

[@chemprofdave]

… thus making the problem suitable for graduate students instead of high school students.

[@emadmustafa7294]

I think I know what is wrong with this question. It gives more information than needed to actually solve it. It should be enough to give us the area of only one of the triangles to solve the problem, but it gives us the areas of two triangles which results in no solution.

[@dj_laundry_list]

Someone probably said this already, but the area of the upper right rectangle (32) plus the area of the lower left rectangle (8) needs to be greater than half of the total area, but 32 + 8 is 40 which is less than 15 x 6 = 90

[@dj_laundry_list]

@@pravlex Imagine moving the intersection point from corner to corner. At either extreme the area is going to be predominantly in on of said rectangles. In the center, the area of both will be half of the total area. Moving from the center to the extreme, the area of both needs to increase monotonically.

[@tontonbeber4555]

Ah yes of course ... I solved it easily, but didn't assume that the central point is aligned on the big rectangle diagonal. No such information is present in the question ... so I did ... (a is your x, b is your y) (15-a)b = 15b-ab = 8 (6-b)a = 6a-ab = 32 6a-15b=24 2a-5b=8 a=4+(5/2)b 15b-4b-(5/2)b² = 8 (5/2)b²-11b+8=0 b=(11+/-V(121-80))/5 = (11+/-V41)/5 a=4+(11+/-V41)/2 = (19+/-V41)/2 ab=(209+41+/-30V41)/10 = 25+/-3V41 Both solutions are acceptable. So answers are area = 44.209372712298546 or 5.790627287701454

[@AzureKyle]

So, just saw this again in my recommended, forgot I already commented on it, but this time I decided to be a bit cheeky and plug this in to Wolfram Alpha to see what they said. First, as MYD did, I set the width of the shaded area to x, and the height to y. Then I turned the triangles into squares, with the smaller having a width of 15-x, and a height of y, totaling to 8cm^2 for the area, and the larger having a width of x and a height of 6-y, with the area being 32cm^2. Then I plugged into Wolfram Alpha (15-x)*y=8; x*(6-y)=32; solve x*y. And it actually gave me two answers. First was 25-(3*sqrt(41)), and the second answer being 25+(3*sqrt(41)), which are approx. 5.79 and 44.21 respectively.

[@arnoygayen1984]

The scaling facter is 2. Find the height: 2x + x = 6 therefore x = 2. Find the base 2y + y = 15 therefore y = 5. Area of the small triangle should be 0.5 × 2 × 5 = 5cm² which is a contradiction. Going farther are of the smaller rectangle = x × 2y = 2 × 10 = 20cm

[@NateHays]

Based on similar triangles with area proportion of 4:1, sides are 2:1. (Using x,y as sides of smaller triangle, unlike Presh). Implies 6 - y = 2y => y = 2. Similarly, 15 - x = 2x => x = 5. Then small triangle is (2x5)/2 = 5, not 4. The given dimensions of 6x15 are not compatible with given areas of 4 and 16.

[@aahanism]

Whole rect area = 15*6=90 2 16² triangles = 32 2 4² triangles= 8 Rest area = 90-40 =50 Remaining 2 rectangles would be equal so area =50/2= 25²cm

[@Tulanir1]

There is no answer, watch the video

[@mikicoal]

It's impossible to solve, because the figures don't add up. The upper right rectangle would be 32cm2 and the lower left rectangle would be 8cm2, of 40cm2 together. Double this for the area of the entire box, and you get 80cm2. However, the box dimensions are 6 by 15 which means an area of 90cm2.

[@vibaj16]

My method: Top right triangle and bottom left triangles are similar. Top right triangle has 4 times the area of the bottom left triangle, so its side lengths are double the bottom left triangle's side lengths. So 6 = y + 2y, so y = 2, so bottom left triangle's base is 4 * 2 / 2 = 4 cm, and top right triangle's base is 16 * 2 / 4 = 8 cm. So the base of the entire rectangle must be 4 + 8 = 12 cm, but it's given as 15 cm.

[@amaury4417]

I got another solution: by solving some systems I find x=(19+sqrt41)/2 and y=(11+sqrt41)/5, or x=(19-sqrt41)/2 and y=(11-sqrt41)/5. So the area is 44,2 or 5,79

[@MCAbdo]

I actually solved in and got 20cm² before realizing the question is wrong.. Seeing the triangles are similar, the rectangles have to be too. Since the big is 4 times the small, means that each side of the the big rectangle is 2 times the length of the small one's. Meaning that it splits fhe sides of the whole thing to 4:2 and 10:5 so we get the dimensions of the green area as 10x2=20

[@Anonymous-df8it]

I saw that the problem was over constrained- imagine that you had the rectangle, with the diagonal, and you got to control where the internal vertical and horizontal lines were (with the condition that they must intersect along the diagonal). Then note that there is only one way to place the lines such that the upper-right triangle has an area of 16 cm^2 (in a generalized version, there may be no solutions, but then we are done!), which constrains the area of the lower left triangle Honestly, they should have only given marks to those who figured out that the question is self-contradictory! :)

[@mglenadel]

"Suppose" the triangle has an area of 16, and the other 4. By "Supposing", it's only saying the ratio of the areas is 4:1, therefore the dimensions of the triangles are 2:1, and you just divide the rectangle sides by 3, one triangle is 2/3, the other 1/3 of the dimensions. Height of the small triangle is 6*1/3, or 2. Length of the large triangle is 15*2/3, or 10. The green area is 2*10=20.

[@louishannett356]

I looked at the problem using the similar triangle approach. To make sense of the areas of the triangles and the ratios of the sides I concluded that the base of the rectangle must be 12cm and not 15cm.

[@ultrametric9317]

There is an easier way. The two similar triangles in blue are in an area ratio of 4, so their sides must be in a ratio of 2. The horizontal and vertical lines must then divide the large rectangle in a ratio of 3 (1 + 2 = 3). And that leads immediately to a conflict with the given data.

[@PlasteredDragon]

My attempt at a solution at the pause point. Moving around the large rectangle from the top left corner labelling the vertices, A, B, C, D, E, F, G, H, and call the intersection point X. We are given the length of AC and GE are 15, and the length of CE and AG are 6. So the area of rectangle ACEG is 90. We are also given the area of triangle BCX as 16 and the area of triangle XFG as 4. Because the diagonal XC cuts rectangle BCDX into two equal triangles, if BCX has an area of 16, then triangle CDX also has an area of 16. And because the diagonal CG cuts rectangle ACEG into two equal triangles, if the area of ACEG is 90, then the area of triangle CEG must be 45. Since we know the area of triangle XFG is 4, and the area of triangle DEX is 16, together these consume 20 of the area of triangle CEG, leaving 25 for the area of rectangle XDEF which was what we were asked for. Now, the dimensions given may represent a figure that is in reality impossible to construct, but be that as it may, it seems like there is a simple path of reasoning to arrive at the solution for the area of the green rectangle. 25 cm^2. To answer the question of whether or not this is actually a possible figure, we just need to break it down to the point where we have the lengths of the line segments and then attempt to rebuild it from those segments. Since triangles BCX and CDX are 16 each, then BCDX has an area of 32. And since we computed the area of XDEF as 25, then BCEF has an area of 25+32 = 57. We can divide this by the height of the figure to get the width of BCEF as 57/6=9.5. Which means that EF is 9.5, and GF must be 15-9.5=5.5. Since BCEF has an area of 57, then ABFG has an area of 90-57=33. We can double check the length of GF by dividing the area of ABFG (33) by the height of the figure (6). This does yield 5.5, so far so good. But things break down when we try to get the areas of the inner rectangles. BCDX must be 32, DEFX must be 25, FGHX must be 8, and ABXH must be 25 (90-(32+25+8)). Since XDEF has an area of 25 and a width of 9.5, it must have a height of 2.63. But it must be the same height as FGHX which has an area of 8 and a width of 5.5. But if FGHX has an area of 8 and a width of 5.5 then it's height would have to be 1.45. Meaning HG is 1.45 and DE is 2.63, but HG and DE need to be equal for DEGH to be a rectangle. The figure is broken and cannot be built with all of the characteristics as described in the diagram -- something is wrong with the diagram.

[@wombat4191]

When you give redundant information, you need to make sure it's not conflicting. Either precise side lengths or precise areas are redundant information in this problem, as you should be able to calculate the areas from the side lengths just by having the ratio of the areas (1 to 4), or side lengths from the areas just by having the ratio of the side lengths (2 to 5). With the area ratio of 1 to 4, these particular side lengths should produce areas of 5 and 20, not 4 and 16. Or if we change the side legths to match the areas, they would be 12/sqrt(5) and 6sqrt(5). So this shape is impossible. The intended answer for the size of the green rectangle is either 16 or 20 depending on whether you calculate it based on side legths or areas, but as it is, there is no actual answer. Edit: ok so there was actually the way that Presh first solved it that actually has yet one additional answer, 25. That also seems like an intended way to solve this, as it explains why the author wanted to provide both the side lengths and the areas. As for which information is more redundant, side lengths or areas, that would be the side lengths. You would need at least the ratio of the two areas to solve the problem if it was presented properly, while the ratio of the side legths doesn't actually matter as long as it's not 0 or infinite.

[@globalolympiadsacademy4116]

If the two blue triangles are similar and their areas are in the ratio of 1:4, then their sides will be in the ratio of 1:2. Here y+2y will be 6, and x+x/2 = 15, x = 10. Therefore the areas of the triangles should have been 1/2,*x/2*y= 5 and 1/2*x*2y =20 and the green rectangle will be 20 given that we assume the ratios of areas correct and the side lengths of the large rectangle correct.else if we consider the triangles areas and ratios correct then xy =16 but then the total area will be 36*2 so may be lengths of 12 and 6 will work.

[@guillaumehuguet3243]

The constraint which makes the problem impossible is the fact that the hypothenuses of the triangles of area 4 and 16 coincide with the rectangle diagonal. If you remove this constraint then the problem has actually two solutions. Let X the top left corner of the rectangle which area A is to be computed. Solving for the coordinates of X gives two solutions: X1 = ((11 + sqrt(41))/2,(11-sqrt(41))/5) in which case A = 25-3*sqrt(41) and X2 = ((11 - sqrt(41))/2,(11+sqrt(41))/5) in which case A = 25+3*sqrt(41)

[@bendewhirst4223]

(Edit: the first sentence, 'construct the diagonal of the rectangle makes the following not accurate, but a solution if this wasn't the diagonal non the less') I got an answer of 5.8cm^2 for the area of the rectangle. And on checking my values, all sections remain consistent with the information given. The first thing i did was change the triangles into rectangles so I had an area of 32cm^2 and 8cm^2 for the top right and bottom left sections. Following this I assigned x to the height of the rectangle we are trying to find out, and y to the width of the rectangle bottom left. This left me with 2 equations; xy = 8 (6-x)(15-y)=32 90-6y-15x+xy=32 xy=8 (subtract one from the other and rearrange) 6y+15x=66 (rearrange into y=mx+c) (I did this so i could sketch my results to see if there was possible answers) y=11-2.5x y=8/x (rearranged from xy=8, into y=mx+c too) Once i sketched these i saw 2 possible values where the lines meet so i set up my equation; 8/x = -2.5x + 11 Into the quadratic 5x^2 - 22x + 16 = 0 Solve for x and y: (there were 2 but the one we want is) x = 0.919375 y = 8.7015635 Check for areas of rectangle we already have: (6-0.92) x (15-8.7) = 32 Check. 8.7 x 0.92 = 8 Check. Therefore, using these values that work, 0.92 x (15-8.7) = 5.8cm^2 Doesn't look impossible to me, but I haven't looked into gradient of the triangle etc. Let me know if there is any problems with this 'solution'

[@peterwright5311]

Setting x as the horizontal length of the smaller triangle, the total area of the two shaded triangles is 45 - 6x + (6/15)*x^2 dA/dx = (12/15)x - 6 minima at dA/dX = 0, where x = (6*15)/12 = 7.5 A (at x=7.5) = 22.5, thus the combined area shown on the diagram of 20cm^2 is impossible.

[@MKPoqdPDeamimzpygxDjg7sJY59]

xy-15y-6x+90=32 and xy=8. You can compare the two solutions and get xy-15y-6x+58=xy-8. Simplify to -15y-6x=66 and simplify further to a y=mx+b equation of y=-0.4x+4.4. Also take back the xy=8 and change it to y=x/8. Then simplify the numbers and change into a binomial equation of "x-11-(2/x)=0" and multiply all terms by "x" to get "x^{2}-11x+20=0." Then you have x=(11±√41)/2. Then solve for y using any equation given before to get y = 16/(11±√41). Then remember that the box is in the right corner of the box and this equation was written for the left corner so let z=15-x. Then take the (zy) and ignore the possibilities when one is the positive radical and one is the negative radical. This leaves the two possibilities as "(15-((11+(41^0.5))/2))(16/(11+(41^.5)))" and "(15-((11-(41^0.5))/2))(16/(11-(41^.5)))" which can probably be simplified but I don't really have the head to do that. Rounded to the nearest thousandth these are 5.791 and 44.209. I think these are reasonable ideas for the green box area. Let me know if I made any errors (I haven't watched the video yet).

[@dhavalvyas77]

Still one more answer 20 sq unit (which is also wrong) is possible of the area of the rectangle. For that, we can use the property that... Ratio of the areas of the similar triangles is equal to the ratio of the square of the corresponding sides. Using this property, we can find length of the sides of the required rectangle are 10 & 2 units.

[@PythonPlusPlus]

I used integration on the smaller triangle to find its width, and ended up with the result 12*sqrt(5)-8. It’s crazy how many results you can get from this.

[@cpasr8065]

For any of the first two solutions to be valid, the area of the triangles must atleast add up to a quarter of the area of the rectangle (22.5). You can prove this with partial differentiation and finding the minima.... but a more intuitive way would be to think of the combined area as a quadratic function. Now a quadratic only has one extreme point (minima or maxima), and since the function gives 45 cm^2 as an answer at the corners, and 22.5 cm^2 in the middle (2(1/2*7.5*3)), 22.5 has to be the minima. Hence, the combined areas have to be atleast 22.5, not 20. Addendum: I can only treat this as a quadratic because the angle of the lines is specified (90 degrees), which ensures a linear dependence b/w x and y. If either one of the angle was not fixed, or one of the sides weren't, or even one of the areas wasn't fixed; this problem would be solvable. Like others said, it's a classic problem of having too many differing constraints.

[@hattoff]

Fun problem! My instinct was to use the side ratios to get the equation of a line and integrate [6/15*x dx] from 0 to y and solve 0.2*y^2 = 4cm^2 and 16cm^2 which gives the base of each triangle (4.47 and 8.94 respectively). Their sum should be 15cm but instead I got a total length of about 13.41cm Making the adjustment for the smaller triangle area gives me a base of 6.05cm instead of 4.47cm, nudging the length out to the expected 15cm. Using the base of the larger triangle (8.94) as the width of your unknown rectangle and 6/15*6.05 for the height gives you a rectangle of about 8.94 X 2.42 ~ 21.6

[@MrMousley]

We can work out the area of the big triangle, which is 15 x 6. We can work out the length of the line going from corner to corner because it's the hypotenuse of a right angle triangle .. and the other two sides are 15 and 6. We know the area of two triangles and so we can work out the area of each of the rectangles that they are both half of .. and we know that the smallest one is 1/4 of the larger one. That has to be enough to do it !

[@rainerheller5451]

Given that the problem was announced as 'impossible' suggested that the areas of the lower triangle A and the upper triangle B could not possibly be 4 and 16 simultaneously, for the given rectangle (15 by 6). To prove this I found the x coordinate of the intersection point, both from triangle A and triangle B to show that the two x-values do not agree: Diagonal is y(x)=6/15 x = 2/5x Area A (lower triangle) = 1/2 x y = 1/2 x 2/5x = 1/5 x^2 Setting A=4 gives x=2 sqrt 5 Area B (upper triangle) = 1/2 (15-x)(6-y) = 1/2 (15-x)(6-6/15x) = ... =45-6x+1/5x^2 Setting B=16 gives x=15-4 sqrt 5 There is no x that can produce A=4 and B=16.

[@kestrelwings]

16 = 4 x 4. That means that the linear diminsion of the small triangle is one half of the large triangle. The area of the upper right rectangle is 2 x 16 = 32. The lower right rectangle has the same horizontal distance, but half of the vertical distance so the area is one half of 32 = 16.

[@pos_itronium]

i think this could be considered with less algebra involved or at least with more clear algebra. the main idea is that if we have a ratio of these triangles' areas we can find the areas themselves. okay, it is quite obvious that the triangles are similar. no algebra so far. and that their sides are the sides of the original rectangle divided in a some ratio, the same for every side. say, the similarity coefficient is α:β, so the first triangle has sides (α/α+β)a and (α/α+β)b. the second - (β/α+β)a and (β/α+β)b. their areas are 1/2 (α/α+β)²ab and 1/2 (β/α+β)²ab so! we can see that A1/A2 = 16/4 = 4 = (α/β)², therefore α/β = 2 and we can pick α = 2, β = 1 (only the ratio matters). after finding out the ratio we can find the areas: A1 = 1/2 (2/3)² ab = 20 and A2 = 1/2 (1/3)² ab = 5. contradiction. and here we can see where this contradiction comes from. we can not set the areas independently because, given the ratio, we can find the areas one by one and this fact may make the mechanics of the problem more clear

[@michaelz6555]

The sum of the areas of the upper and lower triangle can be no more than 45cm^2 (a full-area triangle plus a zero-area triangle), nor can it be less than 22.5cm^2 (two triangles of equivalent area). Because the sum of the area is stipulated to be 20cm^2, it is clear on inspection that the problem was incorrectly constrained from the start.

[@sleepib]

Blue triangles are similar, and area is proportional to length squared, so the ratio of the sides must be sqrt(16) / sqrt(4) = 2. This means the triangles must be 5x2 and 10x4, which do not have the given area. If you didn't notice that, you'd get an answer of 20. Changing the horizontal dimension to 12cm would make the problem work, with an answer of 16 cm^2.

[@pierrecurie]

So Presh has shown us that the system is inconsistent, but he hasn't told us what's going on geometrically. Fundamentally, there are 3 equations (1 for the area of each triangle + main diagonal being a line) and 2 vars (x,y). These are the possibilities: 1) If we insist on the areas being what they are, we get 2 solutions. 1 corresponds to the main diagonal kinking upwards, and other to it kinking downwards. Either way, the main diagonal is not a straight line, but the bottom right remains a proper rectangle (with area depending on the 2 scenarios). 2) If we insist on the main diagonal being a line, we can combine it with a triangle area to find the dimensions of that triangle. Defining f = sqrt(5), the smaller triangle is 2f, 4/f and the larger triangle is 4f, 8/f. They correspond to differing values for x,y but the triangles can still be combined in the same diagram. However, the 2 triangles fail to touch which means that the lower right is a rectangle missing its upper left corner. The rectangle has area 106-36f, while the missing corner is 81-36f. Taking the difference, one finds that the area is 25. This is exactly the area Presh found in solution 1! ... except that it's not a rectangle. What is necessary for the system to be consistent? If we let the smaller triangle have dimensions b, 8/b then the larger triangle will be 2b, 16/b. The rectangle will have dimensions 8/b, 2b with an area of 16 (This is coincidentally the same answer Presh found in solution 2, but if you try solving for x,y you get nonsense). The entire thing will thus have an area of 2(4+16+16) = 72. So the length/width of the containing rectangle doesn't even matter, as long as the area is 72. Using the above logic, one finds that in general, given any 2 of {triangle area(s), small rectangle area, big rectangle area}, we can solve for the remainder. Presh's solution 3 starts with big rectangle + big triangle areas and solves for the rest.

[@sandorfalusi3486]

I solved it a 3rd way, looking at the two given triangle ratios. I divided the 16 cm2 triangle graphically into 4 4cm2 triangles, and so I saw that the sides of the larger triangle is double of the smaller. From this I determined that the rectangle is 6*1/3 multiplied by 15*2/3, so the answer I got was 20.

[@nedmerrill5705]

It's really quite simple, no x's, y's, or fancy equations necessary. The triangles adjacent to the 4 and 16 triangles are obviously congruent to their neighbors. Moreover, the two rectangles (NW, SE) _must_ have equal areas, since the main diagonal divides the large rectangle equally, and the small rectangle areas are what's left after you subtract those equal triangle areas from the half large rectangle value. These are reasonable assumptions given the figure. The rectangle in question is "supposed to be" (half big rectangle area) - (small triangle + large triangle) = (1/2 * 15 * 6) - (4 + 16) = 25, but when you take the sum of the constituent parts of the figure: (2 rectangles) + (2 small triangles) + (2 large triangles) you get (2*25) + (2*4) + (2*16) = 90, and compare it to the area of the rectangle computed by L * W = 15 * 6 = 80 you know something's wrong.

[@Rellton_]

If you plug in -64cm^2 for the answer you can get both outer bounds consistent depending on whether you choose the length or height to be negative. If the rectangle is 16x(-4) then the 15 makes perfect sense and if it's -16x4 then the 6 makes perfect sense. So that answer makes everything consistent but never at the same time. It's a solution that's correct as long as you don't explain it.

[@alanclarke4646]

I get 25: ( entire section above small blue triangle) + large blue triangle= 45. Add small blue triangle = 49 and large white triangle (16) and we get 65. Subtract from area of complete rectangle and we get the green area to be 25.

[@tom.1]

Well the surface areas of the 4cm2 and 16cm2 differ by a factor 4. Since these triangles are uniform in shape, this means the side lenghts differ bij factor square-root(4) = 2. The green square is therefore 2 x 10 = 20cm2.

[@jorogan4ev]

There is very simple solution - 25 cm. We have a rectangle with sides: AB 15cm, BC 6cm, two right triangles (determined by rectangle's diagonal AC and two lines), whit known areas. We search for area of small rectangle, determined by same diagonal and lines. Area of big rectangle is 90 cm^2 (15x6=90) Area of triangle ABC is 45 cm^2 (15x6/2=45) Black and blue triangles above green rectangle are congruent, so srea of black triangle is 16 cm^2. Area of green rectangle is 25 cm^2 (45-16-4=25)

[@R3_dacted0]

This showcases the biggest issue I have with math: Even if I know all of the rules and how things relate to each other, if I can't see that I can apply those rules then I will never come up with a solution. So many math tests where me just staring at an image or an equation and trying to see valid steps to take. I'd blankly stare at a question for literally 15 minutes and then suddenly see something and get excited. It's like looking for Waldo or doing a word search puzzle. And it's exceedingly frustrating for me because even if I know how to do the arithmetic, I'd be stuck just seeing what I can do in the first place.

[@gurupartapkhalsa6565]

The green square is 11x2, to get a triangle with an area of 4, the rectangle it resides in has area 8, the rest can be solved by human quantum intuition. it might be impossible to make a geometric theorem, but the question itself is trivial to solve

[@MusicBlik]

The fact that the larger triangle is 4 times the area of the smaller one, and that the triangles are similar, tells me that the ratio of the two triangles' sides is 2:1. Therefore, the horizontal line divides the right side into segments of 2 and 4, while the vertical line divides the bottom side into segments of 5 and 10. The area of the small triangle should be 5, and the area of the larger triangle should be 20. The area of the green rectangle then is also 20. There are so many ways of correcting this problem. Great discussion going on here!

[@CrafterAurora]

20cm squared?? Must be that because the diagonal line is straight, and 4cm squared is a quarter of 16 cm squared, meaning that the 16cm squared triangle is twice as long and tall as the 4cm squared one. This means that the square in question is 10 cm wide and 2 cm tall because the 16cm squared triangle takes up two thirds of both the horizontal and vertical space. 2x10=20, and thus the answer is 20 cm squared

[@mike1024.]

I wasn't satisfied with your updated question forcing the top triangle to stay 16. Clearly the problem was intended to have integers, so I rewrote it better. If the bottom triangle were 5 and the top triangle were 20 instead, everything works out as anticipated (though of course the final answer changes). This even preserves area ratios that students may find meaningful to help solve the problem, as the top triangle is still exactly double the dimensions.

[@boscobabu3264]

Pause video at 3:31... There are 3 sets of equations, if we expand them and add instead of multiplying we get the equation: 6x+15y-2xy= 40, and 6x+15y=90, thereby giving XY=25cm2. How is the same set of equations giving 2 different answers.

[@captainchaos3667]

The big triangle's area is four times that of the small one, so its sides are exactly half that of the small one. This means that the right sides of the triangles divide the sides of the big rectangle in thirds, making the target rectangle 2 by 10 = 20 cm². Of course it would also make the areas of the triangles 5 and 20 cm², rather than 4 and 16...

[@ericschmidt6440]

The puzzle would be a really nice one - if it would not be "overdetermined" - so its actually just overloaded with too many given inputs. I just removed the 6 cm input and solved the new puzzle using the nice 4:1 ratio of the given triangle areas which translates to a 2:1 ratio in their lengths. That immediately gives x = (2/3)*15 = 10 cm and then the hights of the triangles are easily found to be 3.2 cm and 1.6 cm. So the green rectangle is 16 cm² - and there is no "other solution". There is only one solution if you start with a "reasonable" set of inputs at the beginning.

[@leppie]

The minimum combined area of the given rectangles for any crossing point on the diagonal is 22.5cm^2 (in the middle). Does not compute :D

[@tezismith8795]

I did: (15 - x)(6 - y) = 32 90 - 15y - 6x + xy = 32 90 - 15y - 6x + 8 = 32 15y + 6x = 66 5y + 2x = 22 substitute x = 8/y 5y + 16/y = 22 5y^2 + 16 = 22y 5y^2 - 22y + 16 = 0 after using the quadratic formula to find solutions for y, I substituted this into the other equations, and ended up with: area = 25 (+/-) 3sqrt(41)

[@theshed8802]

It's easy. No advanced maths needed. Halve the large rectangle into 2 triangles and you have the total area of each. Subtract the area of each of the small triangles and you get the remaining unknown area of the questioned rectangle, and it's unquestioned mate. 15×6=90. 90÷2=45. 45-16-4=25cm2. Therefore the area of the unshaded rectangle will also be 25cm2 as both large triangles are the same area

[@migBdk]

The figure is over-determined and the numbers do not match. If you only got the area of one triangle to 4 cm^2, you could calculate the area of the rectangle to one value. But the information that the other triangle is 16 cm^2 suggest a different area. I calculated the position of the vertical lines with integrals btw.

[@juergenilse3259]

Let the sides of the green rectangle be a and b, then we get the equations (15-a)*b/2=4 (the area of the lower triangle) (6-b)*a/2=16 (the area of the upper triangle) We search for the value of a*b. From the first equation, we get: 15b-ab=8 ab=15b-8 From the second equation, we get: 6a-ab=32 ab=6a-32 So we know: 6a-32=15b-8 6a=15b+24 a=2.5b+4 If we insert this in the equation 6a-ab=32,we get: 15b-(2.5b+4)*b-32=0 11b-2.5b^2-32=0 b^2-(22/5)b+32/5=0 pq-formula gives: 11/5+-sqrt(121/25-32/5)=11/5+-sqrt((121-160)/25)=11/5+-sqrt(-39/25) This is not a real number, because the square root is imaginary..There can't be such a figure with that given dimensions .....

[@cmuller1441]

The big rectangle is 6*15=90cm² The first small one is 4*2=8 so the intersect is located at sqrt(8/90) (*15cm) from the left. The second one at sqrt(32/90) (*15cm) from the right For this to work you need the total to be 1 But: sqrt(8/90)+sqrt(32/90)=(2sqrt(2)+4sqrt(2))/3sqrt(2*5)=2/sqrt(5) != 1 (Sqrt(4)=2 so Sqrt(5)>2)

[@isaac_aren]

This is now a brilliant question to give where the correct answer is to find the mistake and give the actual correct answer