#Question about visual geometry

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tall dust
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So here is the instruction in english

  1. No pre-formulated answer - In the figure below, ABCD is a sided square 17; the triangles ADE and ABG are isosceles in E and G and their heights from these respective vertices are 20.5. In addition, AEFG is a rhombus. What is its area?

Obviously, since both triangle ADE and BAG have the same basis and height, they are the same triangle. The sides of the rhombus are the same lenght as the isosceles sides of both triangle. Therefore the area of the rhombus should be equal 2 times the area of one of the two isosceles triangle. I did all that and I ended up with 348,5 but according to the correction the answer should be 348. Am I missing something or is the correction wrong?

2C0F8CDB-BA7B-4A65-91FF-CC954374FECE.jpg
storm shaleBOT
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tall dust
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here is the calculus if anyone wonders

IMG_8787.jpg
toxic prairie
tall dust
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wdym

toxic prairie
# tall dust wdym

I agree that the side length of AGFE equals the leg length of AED, but that doesn't mean the one is twice the area of the other.

tall dust
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why not?

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that’s like saying the area of a square of side 4 isn’t the same as 2 times the area of a rectangle of dimensions 2•4

toxic prairie
tall dust
toxic prairie
tall dust
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You haven’t proved me wrong neither

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jk

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ima do this later

mellow ingot
# tall dust So here is the instruction in english 23. No pre-formulated answer - In the fig...

Sam,

In order for you to assert that the area of the rhombus is equal to twice the area of the isosceles triangle, you must prove that half of the rhombus is congruent to the aforementioned.

Since I am aware of the length that it may take for somebody (at least me!) to solve this problem, I will be generous and offer you a hint: half of the rhombus is VERY similar to the triangle, but it is NOT congruent!

Yes. You have missed the fact that it certainly takes more than two lines to solve this problem. Also, the area of a square with 4 units as each measure of the line segments is not the same as the area of a rectangle with the dimensions of 2x4. I would like you to do me a favor, and recalculate those two.

tall dust
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Area of a square of side 4: 4^2=16
2 Times the area of a rectangle of side 2;4 : 2(2•4)=2•8=16

mellow ingot
tall dust
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I think you just didn’t understand what I said

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no shit the area of a rectangle is w*l

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anyway that’s not the point

mellow ingot
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Yes. So what you are you doing multiplying by 2?

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That is not a proper operation, at least not one that I am familiar with.

tall dust
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because my initial statement was saying « 2 times the area of a rectangle of side 2:4 »

mellow ingot
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Okay, I get what you're saying now. But we know for a fact that those two would yield the same answer, because we know the both of their dimensions.

You do not know the dimensions of the rhombus, so you cannot assert this false equivalence.

tall dust
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the isosceles angles of the triangles can be found using arctan (20,5/8,5) after that I got the equation

360° = 2(arctan(20,5/8,5) + 90° + 2x

x being the isosceles angles of the two triangle of the rhombus

after using my calc there was a difference of like 0.02° that could just be caused by the imprecision of my calc

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so I am still pretty sure these are the same triangles

mellow ingot
# tall dust the isosceles angles of the triangles can be found using arctan (20,5/8,5) after...

Yes, as I stated before, there is a very small difference, which is even negligible if you round to the nearest whole number.

However, it is clear that your teacher did not have this in mind, as he most likely assumed that most students would attempt to solve it in two lines, as you did.

So, when you have found the measures of the missing side lengths needed for the area of a rhombus, you must use the exact decimal approximation to get the correct answer, not the rounded answer.

mellow ingot
tall dust
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pls note that the exercice was made to be solvable without a calc

mellow ingot
# tall dust and how exactly are you supposed to find it

You have to do it by hand?! That's one extra hour of work! 😂

Normally, if you teacher didn't enjoy letting students spend about half a day on a problem, in order to find the angle measures, just apply the same Trigonometric ratios that you did before to find the measure of the isosceles' angles.

It seems that you would have to utilize your knowledge of the unit circle to be able to approximate the decimal values in your head, which is quite torturous!!

obsidian ironBOT
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@tall dust

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@mellow ingot @toxic prairie The user still needs help with this help request.

toxic prairie
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Okay, so taking it from the top, what's the length of AE?

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That's the first thing we need to know.

tall dust
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sqrt of (20,5^2+8,5^2)

toxic prairie
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And what is that?

tall dust
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no idea

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22.19 aproximatyvely

toxic prairie
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Well, you should at least take a crack at simplifying it.

toxic prairie
tall dust
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yes I am but I could do it in my head and I would have find 22.2 tbh

mellow ingot
# tall dust yes I am but I could do it in my head and I would have find 22.2 tbh

Do they seriously expect you to do it all in your head? Even the inverse trig ratios? To me, that just seems cruel! There is such a wide range for error doing it all by hand, and it is already painstakingly time-consuming with a calculator!

I do not understand your teacher's rationale; how many decimal places are you expected to round to up until the final answer?

tall dust
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it was the question of a math competition, it’s not from my teacher, basically there are two types of question:

First one are mcq

Second one are « Open question » for which you need to answer with a natural number in the range [0;999]. It’s the case for this question. When the answer has decimals, they precise “round up to…” which isn’t the case here so 348 should be the exact answer, and found by head, in an average time of 3 minutes

mellow ingot
# tall dust it was the question of a math competition, it’s not from my teacher, basically t...

Aha! I had my suspicions! C'est Olympiades de Mathématiques?

Okay, then this is the advice I suggest for you to heed: round to at least three decimal places when doing calculations, and then round to the nearest whole number for the final answer. If you round to only the tenths place, you will still circle back to 348.5 / 349, which is the incorrect answer.

There must be some sort of mathematical genius out there who can solve this in about three minutes, because I am not aware of the method one can undertake in order to scan such information so quickly, and then proceed to do it all by hand!!

I would like you to know that the process of calculating the inverse tangent function is in fact longer than the rest of the entire problem! I'm not sure how you are expected to do that, and the only two methods I can think of are using a series expansion, or estimating it based on the known values on the Unit Circle, but perhaps they will be generous, and offer you a Log Table!

tall dust
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yes that’s Olympiades de Mathématiques from belgium

thank you for your advice but I am pretty sure using the inverse tangent function (even if it’s working) isn’t the way they expected us to take to solve the problem. There must be smthng we don’t know/see here but since it seems like no one knows it ima close this discussion. Thank you for taking the time to answer to me tough

mellow ingot
# tall dust yes that’s Olympiades de Mathématiques from belgium thank you for your advice b...

You might be right about that. However, utilizing the trig functions is the standard way to solve such a problem. I am aware that there must be a difference in the teachings of Mathematical procedure in Belgium as opposed to America, so there is certainly something I'm missing here. What a shame! Now I am quite curious!

Nevertheless, you are very welcome! I am happy that I could be of help!

tall dust
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+close

obsidian ironBOT
# tall dust +close
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