Maths problem

[Kusje]

uh.

 

which triangle?

[1fast1]

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 November 15, 2018 by Turboflat4

[Vid]

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.

 

 

But I dunno the answer yet Only know there are a small, medium and large squares.   

[Ender]

 

But I dunno the answer yet Only know there are a small, medium and large squares.

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

[1fast1]

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).

Edited November 15, 2018 by Turboflat4

[mersaylee]

uh.

 

which triangle?

IKR, maciam the one at bermuda...you know it’s there but can’t see.

[Ender]

Another math question

 

NOT DRAWN TO SCALE

 

DCC2231B-C3A2-47B3-A2AB-05675397B45F.png

 

How to solve?

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

[1fast1]

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

 

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.

[Ender]

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.

Edited November 16, 2018 by Ender

[1fast1]

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 November 16, 2018 by Turboflat4

[1fast1]

5 hours ago, Wind30 said:

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

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.

Edited December 13, 2019 by Turboflat4

[1fast1]

 

6 hours ago, Wind30 said:

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

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.

Edited December 13, 2019 by Turboflat4

[1fast1]

2 minutes ago, Wind30 said:

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... 

As mentioned, the trick is to find a way to show that arctan 1/3 + arctan 1/2 = 45 deg, without trigonometry.

At that point, you try to find diagonals of squares or right isosceles triangles, which is how I came up with my solution.

Using coordinate geometry for it is a doddle, but when you're limited to the tools open to Pri sch kids, then you have to get creative.

[Ender]

7 hours ago, Wind30 said:

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

7 hours ago, Wind30 said:

 

 

Serious primary? Maybe gifted school. 

Thanks to turboflat4 solution. I also loss. 

Edited December 13, 2019 by Ender

[1fast1]

1 hour ago, Turboflat4 said:

 

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.

EDIT: meant "corresponding angles", not alternate angles, sorry. @Wind30

Edited December 13, 2019 by Turboflat4

[Ender]

6 minutes ago, Wind30 said:

it primary school math competitions past year paper.

I see. Competition is math Olympiad or new south wale level. 

[1fast1]

36 minutes ago, Wind30 said:

ya lar I never remember the name anyway.

But you are cheating a little if you knew the answer was 45deg. I did not calculate the answer using trigo so I could not solve it last night after thinking for 15 mins. It is rare I see a primary school question which the answer is not obvious in minutes.

Solve it this morning but my answer was much longer. I drew more lines than you.

On that basis, you could argue the exam setters were also "cheating" since I'm sure they know trig and also used it. Then they figured out how to do it using elementary geometry then set the question for the kiddies. 

That's how competition questions often work. They're constructed using more involved math but designed to be creatively solved using more elementary means.

If a pri sch kid knows coordinate geometry and trig and used those, I'm sure they wouldn't deduct marks, by the way.

[Enye]

wah... when i see this type of complicated maths question which i cannot solve, i console myself that 99% of the global population won’t need to use this in their entire working lives or apply the solution in practical day to day living

🤣