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Post by tickingmask on Apr 3, 2023 19:19:36 GMT
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Post by Flying Monkeys on Apr 3, 2023 19:49:48 GMT
Label half of a side as x.
Area = 2x * 2x = 4x2 = 68 + ?
Therefore ? = 4(x2 - 17)
Therefore x2 > 17, i.e. x > than 4
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Looking at the section which is 32cm2, its area is > x2
Therefore 32 > x2
Therefore x < 6
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Assuming we are dealing with whole numbers, x = 5
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Post by tickingmask on Apr 3, 2023 20:23:21 GMT
Label half of a side as x. Area = 2x * 2x = 4x 2 = 68 + ? Therefore ? = 4(x 2 - 17) Therefore x 2 > 17, i.e. x > than 4 ---------------------------------------------------------------- Looking at the section which is 32cm 2, its area is > x 2Therefore 32 > x 2Therefore x < 6 ----------------------------------------------------------------- Assuming we are dealing with whole numbers, x = 5 Ok, your reasoning that half the side of the square (x) is between 4 and 6 is sound, but x is not necessarily a whole number.
Indeed, if you were going to argue that because x = 5, the area of the square = 100 and therefore the area of the missing bit is 100 - 16 - 20 - 32 = 32 then that's not the right answer I'm afraid :-(
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Post by Deleted on Apr 4, 2023 13:54:58 GMT
Ill attempt this tomorrow
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Post by tickingmask on Apr 4, 2023 19:19:01 GMT
Ill attempt this tomorrow If you need a hint then:
{Spoiler}
The area of a triangle = ½ x base x height :-) |
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Post by Deleted on Apr 5, 2023 11:09:29 GMT
Ill attempt this tomorrow If you need a hint then:
{Spoiler}
The area of a triangle = ½ x base x height :-) I got nothing lol That didnt really help. I was thinking of an isosceles triangle and trig but you dont have any cm, just area You have Monkeys already trying algebra Obviously your clue is meant to help but Im just not seeing it |
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Post by tickingmask on Apr 6, 2023 7:51:33 GMT
I got nothing lol Obviously your clue is meant to help but Im just not seeing it No worries! I'll give one further hint that might give my original clue some context, but I won't give the answer yet, in case anybody else wants to give it a go.
{Further hint}
Draw lines that connect the central point where all four areas meet, to each of the four corners of the square. This turns the four quadrilaterals into eight triangles.
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Post by Jep Gambardella on Apr 7, 2023 12:22:30 GMT
I got nothing lol Obviously your clue is meant to help but Im just not seeing it No worries! I'll give one further hint that might give my original clue some context, but I won't give the answer yet, in case anybody else wants to give it a go.
{Further hint}
Draw lines that connect the central point where all four areas meet, to each of the four corners of the square. This turns the four quadrilaterals into eight triangles.
I was trying to break up each of the sections into triangles but I didn't think of doing it that way. I don't have pen and paper now but I will figure it out later. |
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Post by bomtombadil on Apr 7, 2023 13:47:09 GMT
Well based on the inverse ratio of Avogadro's number to Euler's number and subsequently having the quotient algebraed by calculus, the solution to this can only be 7 areas.
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Post by Jep Gambardella on Apr 12, 2023 0:09:24 GMT
The answer is 36 sq cm Each side of the square measures 2 x sqrt(26) |
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Post by tickingmask on Apr 12, 2023 7:37:51 GMT
The answer is 36 sq cm Each side of the square measures 2 x sqrt(26) I'm sorry, I really wanted this answer to be correct... but it isn't. ![[frown]](http://storage.proboards.com/6870873/images/FySwlyxIVZHczdwVyYqH.gif)
I think I understand how you tried to do it - use the known areas to work out the distance from the edge of the square to the mid-point from each side, thence the side of the square itself, and you could probably do it that way (with 3 areas and 3 unknowns). Indeed I initially tried to do the same myself but got horribly bogged down in the number-crunching and decided to give up and look for an easier way! Which I found, {Hint}
one that doesn't require you to work out the length of the side of the square in advance and comes out with a consistent answer regardless whether it's a rectangle or a square (nasty bit of sneakiness, there, making it a square and tempting people down this rather complicated alley). |
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Post by Jep Gambardella on Apr 12, 2023 13:22:51 GMT
I hope I am right this time. On my previous solution, I had falsely assumed that breaking up each of the areas into two triangles, the two triangles would have the same area. My new solution is:
Missing area is 28. Side of the square is 2xsqrt(24)
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Post by Jep Gambardella on Apr 12, 2023 14:01:20 GMT
Just to clarify, it seems that you know the correct answer but not how to calculate it. Is that so?
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Post by tickingmask on Apr 12, 2023 15:23:04 GMT
Missing area is 28. Side of the square is 2xsqrt(24) Correct! (Although I'd have simplified it to 4sqrt(6)) In answer to the question in your other post, yes, I do know how to calculate it, all I was trying to say was that you don't have to work out the length of a side, or indeed the area of the square in order to do so. Here's my method: {Spoiler}Ok, let O be the point at which the four different areas all meet. As I said in the hint, construct four straight lines, going from O to each of the four corners of the square. This splits the four quadrilaterals into eight triangles.
Now take the two triangles at the bottom - i.e. the ones where one side occupies half the bottom of the square. Both triangle bases are the same length - half the length of the side of the square - and they both have their apex at point O, so they both have the same height as well. Therefore they both have the same area - half base times height. Let us call this area a.
Now the triangles on the left. By the same argument - equal base lengths, common apex - they both have the same area as each other so let's call this area b.
Same goes for the triangles at the top - area c, and the ones on the right - area d.
Now if we look at the areas we are told about, we get:
a+b = 16 (bottom left quadrant)
b+c = 20 (top left)
c+d = 32 (top right)
d+a = ? (bottom right, the area we want to find)
And we can see that d+a = a+d = (a+b) - (b+c) + (c+d) = 16 - 20 + 32 = 28. So that's our answer! Another way of looking at it is that each set of diagonally opposite quadrants has a total area a+b+c+d so add up to the same value. i.e. 16 + 32 = 20 + 28.
As mentioned in my last post, this method is just as valid for a rectangle as for a square, so long as all the corners are 90 degrees. If you liked the problem I recently found another which I thought was harder. I'm happy to put that one here as well, if anybody's interested. |
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Post by Jep Gambardella on Apr 13, 2023 16:05:48 GMT
Missing area is 28. Side of the square is 2xsqrt(24) Correct! (Although I'd have simplified it to 4sqrt(6)) In answer to the question in your other post, yes, I do know how to calculate it, all I was trying to say was that you don't have to work out the length of a side, or indeed the area of the square in order to do so. Here's my method: {Spoiler}Ok, let O be the point at which the four different areas all meet. As I said in the hint, construct four straight lines, going from O to each of the four corners of the square. This splits the four quadrilaterals into eight triangles.
Now take the two triangles at the bottom - i.e. the ones where one side occupies half the bottom of the square. Both triangle bases are the same length - half the length of the side of the square - and they both have their apex at point O, so they both have the same height as well. Therefore they both have the same area - half base times height. Let us call this area a.
Now the triangles on the left. By the same argument - equal base lengths, common apex - they both have the same area as each other so let's call this area b.
Same goes for the triangles at the top - area c, and the ones on the right - area d.
Now if we look at the areas we are told about, we get:
a+b = 16 (bottom left quadrant)
b+c = 20 (top left)
c+d = 32 (top right)
d+a = ? (bottom right, the area we want to find)
And we can see that d+a = a+d = (a+b) - (b+c) + (c+d) = 16 - 20 + 32 = 28. So that's our answer! Another way of looking at it is that each set of diagonally opposite quadrants has a total area a+b+c+d so add up to the same value. i.e. 16 + 32 = 20 + 28.
As mentioned in my last post, this method is just as valid for a rectangle as for a square, so long as all the corners are 90 degrees. If you liked the problem I recently found another which I thought was harder. I'm happy to put that one here as well, if anybody's interested. Sure, yes, let's have it. |
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