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Maths problem


Sosaria
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  On 10/16/2018 at 6:30 AM, Sosaria said:

Why you no take Further Math? Instead of Bio. That would've made JC even more relaxing for you. I did that, but of course, the doors to med course is well and truly closed once take that option, no matter how good the results.

Doctor father said must become doctor so I did lor. :D

 

Now I'm a don't see patient kind of specialist so there's a cautionary tale in there somewhere. [laugh]

 

Because of my interest, I knew the entire F math syllabus (and more) anyway.

Edited by Turboflat4
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Better show a method. I assume the figure is just rather poorly drawn and that lower diagonal is supposed to intersect the "northeast" vertex of the medium sized square. I also have to assume that you intended the "easternmost" quadrilateral to be a square too, otherwise the problem doesn't have a unique solution.

 

For simplicity's sake, let the side length of the smallest square be one unit. Note that the lower diagonal line has a gradient of 1/3, which basically means for a three unit horizontal increment, it undergoes a rise of one increment. Now if we let the largest square have a side of "A", we can easily derive an equation: A/3+2 = A, giving A = 3, so we've shown that the largest square has a side that's equal to three times the smallest square (or equivalently, the sum of the sides of the small and medium squares).

 

We can subdivide the triangle into two portions using the upper ("northern") side of the medium square. Note that the triangular portion below this has base 2 units and height 1 unit, so its area will be equal to the area of the small square, which is 5. The upper portion shares the same base and has equal height, so it also has an area of 5. Total area of the large triangle = 10. Done.

Edited by Turboflat4
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  On 11/15/2018 at 1:12 AM, Turboflat4 said:

Better show a method. I assume the figure is just rather poorly drawn and that lower diagonal is supposed to intersect the "northeast" vertex of the medium sized square. I also have to assume that you intended the "easternmost" quadrilateral to be a square too, otherwise the problem doesn't have a unique solution.

 

For simplicity's sake, let the side length of the smallest square be one unit. Note that the lower diagonal line has a gradient of 1/3, which basically means for a three unit horizontal increment, it undergoes a rise of one increment. Now if we let the largest square have a side of "A", we can easily derive an equation: A/3+2 = A, giving A = 3, so we've shown that the largest square has a side that's equal to three times the smallest square (or equivalently, the sum of the sides of the small and medium squares).

 

We can subdivide the triangle into two portions using the upper ("northern") side of the medium square. Note that the triangular portion below this has base 2 units and height 1 unit, so its area will be equal to the area of the small square, which is 5. The upper portion shares the same base and has equal height, so it also has an area of 5. Total area of the large triangle = 10. Done.

 

[thumbsup]

 

But I dunno the answer yet [:p] Only know there are a small, medium and large squares.  :XD: 

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  On 11/15/2018 at 1:26 AM, Vid said:

[thumbsup]

 

But I dunno the answer yet [:p] Only know there are a small, medium and large squares. :XD:

Is biggest square a 3 x 3 of the smallest square of area 5? Then I also get 10.
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  On 11/15/2018 at 4:34 AM, Ender said:

Is biggest square a 3 x 3 of the smallest square of area 5? Then I also get 10.

 

It is, but you have to show it.

 

More precisely: it is a square (has to be assumed as a given). It is of side 3 times that of the smallest square (needs to be shown).

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  On 11/15/2018 at 2:59 PM, Ender said:

Keep these questions coming. Since I doing the solution for my daughter, might as well share the ppt here too.

post-52704-0-77233600-1542319726.jpg

 

Sorry, it's not clear to me that you've shown the square demarcated in red has side 3 units.

 

A scale drawing is not adequate. A geometric argument is needed.

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  On 11/15/2018 at 10:10 PM, Turboflat4 said:

post-23584-0-59152100-1542293971_thumb.jpg

 

Sorry, it's not clear to me that you've shown the square demarcated in red has side 3 units.

 

A scale drawing is not adequate. A geometric argument is needed.

The longest lines is a 3x1 diagonal, that makes the biggest square a 3x3.

It's a primary school question, unlikely a argument is needed for an answer, I expect the question to have stated the biggest square is a 3x3 or at least drawn to scale to show the line to be a 3x1 diagonal line.

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  On 11/16/2018 at 2:53 AM, Ender said:

The longest lines is a 3x1 diagonal, that makes the biggest square a 3x3.

It's a primary school question, unlikely a argument is needed for an answer, I expect the question to have stated the biggest square is a 3x3 or at least drawn to scale to show the line to be a 3x1 diagonal line.

Yes, that's basically a gradient argument like I made (but simpler and better suited to a primary school question, so that's good).

 

If we produce the line to the same diagonal length, we can enclose it in the same 3 x 1 rectangle. There are two units below the rectangle, so we've made a 3 x 3 square. Nice and simple.

 

But it does need to be shown. Can't be assumed. Got to be rigorous even from a young age.

 

(sorry for the late reply, just saw this, notifications for this thread not working).

Edited by Turboflat4
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  On 12/13/2019 at 12:12 AM, Wind30 said:

Can u guys solve this? No usage of sin cosine as it is primary question.

D1E14F90-F8B3-4698-BCA1-797F4EEA1AD8.jpeg

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See attached, which is just a rotated version of your figure with extraneous info removed. You need to find angle DGB. By alternate angles, this is equal to angle EAB Since AB is the diagonal of square AEBF, angle EAB is 45 degrees, and that's the value of x.

The problem is essentially to show arctan 1/3 + arctan 1/2 = 45 degrees without trig. That means finding right angled isosceles triangles and/or squares with their diagonals.

geometry.png

Edited by Turboflat4
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  On 12/13/2019 at 12:12 AM, Wind30 said:

Can u guys solve this? No usage of sin cosine as it is primary question.

D1E14F90-F8B3-4698-BCA1-797F4EEA1AD8.jpeg

Expand  

I was rushing for time before, so I did a better diagram that corresponds to your original question. Hopefully, it's clearer.

You want angle KQI. Since KQ is coincident with KP, which is parallel to LA, angle KQI = angle LAI (alternate angles) = 45 degrees (angle made by diagonal AI of square ALIN with side LA). Done.

geometry3.png

Edited by Turboflat4
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thanks. Your solution is more elegant than mine. Mine was a bit convoluted. I thought most of such questions have easy solution once you get the trick. 

I think the main trick is to shift the shorter line KP to LA. From then on, it is kind of obvious. 

My daughter is still trying to solve this... 

Edited by Wind30
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