Missing area of a square with no given dimensions
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Dr.Peterson
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Well, for a start you can choose some variables; maybe z = half a side of the square, and (x,y) = coordinates of the meeting point relative to whatever you want (maybe the center). Then work out a formula for the area of each region, and you'll have three equations in three unknowns.
It turns out that there are a lot of shortcuts, so you won't have to fully carry out that process if you're observant. I haven't yet seen an "obvious" explanation for the absurdly easy calculation you'll end up doing, but perhaps if you take some time thinking about relationships between the areas, you'll discover something interesting.
I wouldn't want to take away the fun of discovering this! So give it a try, and let us know what you find.
[Edit: Now I have a nice way to explain what happens here; as a hint, you might draw in a line from each vertex of the square to the meeting point, and think about how the areas of various triangles are related.]
It turns out that there are a lot of shortcuts, so you won't have to fully carry out that process if you're observant. I haven't yet seen an "obvious" explanation for the absurdly easy calculation you'll end up doing, but perhaps if you take some time thinking about relationships between the areas, you'll discover something interesting.
I wouldn't want to take away the fun of discovering this! So give it a try, and let us know what you find.
[Edit: Now I have a nice way to explain what happens here; as a hint, you might draw in a line from each vertex of the square to the meeting point, and think about how the areas of various triangles are related.]
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My answer was 32. I am not sure if thats correct and how to do this properly
Dr.Peterson
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There are many ways to do it -- no one "proper" way. It's your job to solve the problem, right? So I can give hints, but I don't want to get in the way of the growth you will experience by discovering a solution yourself.
No, 32 is not the answer; it doesn't seem likely that two adjacent regions would have the same area, does it?
But maybe you had some good ideas, so please tell me how you got that, and I can make suggestions for things to change.
No, 32 is not the answer; it doesn't seem likely that two adjacent regions would have the same area, does it?
But maybe you had some good ideas, so please tell me how you got that, and I can make suggestions for things to change.
Otis
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Dr.Peterson
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Please try following my hint! It's much more fun to discover something and celebrate, than to be given the answer and groan that you didn't see it.
Here is a picture:
Look at any two opposite triangles, such as AEO and CGO (O being the point where they meet). Think about their bases and altitudes. Once you see that, look at a pair of opposite quadrilaterals, like the red and green.
I have to say I'm a little surprised at this being from a job exam. What job? Maybe if you'd given the context from the start I would have looked at it differently, assuming it must have a quick trick; but I'm pretty good at math, and it took me a while to find the answer the first time, and a little longer to see that the answer is "obvious". (Maybe someone else has a better explanation than mine.)
Here is a picture:
Look at any two opposite triangles, such as AEO and CGO (O being the point where they meet). Think about their bases and altitudes. Once you see that, look at a pair of opposite quadrilaterals, like the red and green.
I have to say I'm a little surprised at this being from a job exam. What job? Maybe if you'd given the context from the start I would have looked at it differently, assuming it must have a quick trick; but I'm pretty good at math, and it took me a while to find the answer the first time, and a little longer to see that the answer is "obvious". (Maybe someone else has a better explanation than mine.)
Its for an R&D position focused on computer graphics. This question was in the Artificial Intelligence part of the exam.
It still bothers me and trying to figure out using Dr. Peterson’s clue.
At the time I see this question I tried for a bit but ended up guessing.
It still bothers me and trying to figure out using Dr. Peterson’s clue.
wagchagwei
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I got the point. The sum area of the opposite two triangles is a constant value
P.K.Tandon
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I will like to solve this sum using coordinate geometry. (₁ ₂ , ₃ ₄)
If corner points in a quadrilateral are [ ( x1, y1 ) , ( x2, y2 ), ( x3, y3 ) and ( x4, y4 ) ] then its Area can be written as
(1/2) [ ( x1 y2 + x2 y3 + x3 y4 + x4 y1 ) - ( x2 y1 + x3 y2 + x4 y3 + x1 y4 ) .............. Formula
Refer to the following figure -
B is origin and BC is along x-axis . BA -s along y-axis. Let's suppose BC = 2a and F,G,H and I are mid poits of the given square ABCD. Coordinates of all the points have been shown in the figure. The Point E has coordinates = (h,k) ...... suppose
Area of AGEF (Ar1) = (1/2) [ ak + ah - 2a² - 2 ah ] = 20 .... given
=> ak + ah - 2a² - 2 ah = 40
=> a ( k - h ) - 2 a² = 40 ............... (1)
Similarly -
Ar2 = (1/2) [ ( 2 a² + 2 a² + 2 ak + 2 ah ) - ( 4 a² - 4 a² - ah - ak } ] = 32
=> 4 a² - a ( k + h ) = 64
=> 4 a² = a ( k + h ) + 64 ............... (2)
And Ar3 = a ( k + h ) = - 32 . But each of a, k and h is a positive number.
=> a ( h + k ) = 32 ................., (3)
From (2) and (3) we get --
Area of ABCD = 4 a² = 32 + 64 = 98 ........................ Incorrect
But Ar1 + Ar2 + Ar3 + Ar4 = Area ABCD
Ar4 = 98 - ( 20 + 32 + 16 ) = 30
Hence Un-known Area = 30 sqr units ...................... Answer
If corner points in a quadrilateral are [ ( x1, y1 ) , ( x2, y2 ), ( x3, y3 ) and ( x4, y4 ) ] then its Area can be written as
(1/2) [ ( x1 y2 + x2 y3 + x3 y4 + x4 y1 ) - ( x2 y1 + x3 y2 + x4 y3 + x1 y4 ) .............. Formula
Refer to the following figure -
B is origin and BC is along x-axis . BA -s along y-axis. Let's suppose BC = 2a and F,G,H and I are mid poits of the given square ABCD. Coordinates of all the points have been shown in the figure. The Point E has coordinates = (h,k) ...... suppose
Area of AGEF (Ar1) = (1/2) [ ak + ah - 2a² - 2 ah ] = 20 .... given
=> ak + ah - 2a² - 2 ah = 40
=> a ( k - h ) - 2 a² = 40 ............... (1)
Similarly -
Ar2 = (1/2) [ ( 2 a² + 2 a² + 2 ak + 2 ah ) - ( 4 a² - 4 a² - ah - ak } ] = 32
=> 4 a² - a ( k + h ) = 64
=> 4 a² = a ( k + h ) + 64 ............... (2)
And Ar3 = a ( k + h ) = - 32 . But each of a, k and h is a positive number.
=> a ( h + k ) = 32 ................., (3)
From (2) and (3) we get --
Area of ABCD = 4 a² = 32 + 64 = 98 ........................ Incorrect
But Ar1 + Ar2 + Ar3 + Ar4 = Area ABCD
Ar4 = 98 - ( 20 + 32 + 16 ) = 30
Hence Un-known Area = 30 sqr units ...................... Answer
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Dr.Peterson
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You omitted the absolute value (and a closing bracket) in the formula; areas are positive, while the formula as you give it is a signed area, dependent on orientation. You appear to have used this fact later without stating it expressly.
P.K.Tandon
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Yes. I assumed it to be obvious. Don't you think a normal student cannot solve this problem ? That was why I solved it fully.
Dr.Peterson
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Did I say you shouldn't have solved this one? This is an old problem, long abandoned by the OP, and then resurrected by someone else a week ago who made my hint explicit. There's nothing wrong with answering it now, and in a different (more complicated but more "routine") way. In any case, don't worry; I don't consider myself the judge of what you should do.
But if a student is given a problem that a "normal student" can't be expected to solve, then I would think of it as a contest-type problem, which is intended to teach or test unusual problem-solving skills. and they need mere hints all the more, since the goal of such a problem is to learn ways of thinking that go beyond "normal"! If this were a new problem, I would have just given the formula as a hint first. But that's just me, not a uniform opinion of the whole community here.
In this case, it is not a student, and the problem is reportedly "a question on a job exam I took ... for an R&D position focused on computer graphics." I imagine they might like your method using coordinates; or they might prefer my more creative shortcut approach. Who knows?
And this isn't even trigonometry!
P.K.Tandon
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Did I say you shouldn't have solved this one?
Dr. Peterson. Plz take it easy. I am over 80 and participating just to keep myself fit.
Dr.Peterson
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The problem doesn't involve angles, which is what trigonometry is about; and none of the methods of solution that have been mentioned use trigonometric functions. It's a question of geometry.
For a perfect square
Total are = side x side
Table
If Side Then Area
1 cm 1 Sq.Cm
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81
In above problem figure.
Let missing Area is 'X'
Total area = 20+32+16+X
= 68 + X
As 68 is greater than 64 (above table)
The Total area must be 81
Then 81 = 68 + X
There fore X = 81-68
Answe X = 23 Sq. Cm
Total are = side x side
Table
If Side Then Area
1 cm 1 Sq.Cm
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81
In above problem figure.
Let missing Area is 'X'
Total area = 20+32+16+X
= 68 + X
As 68 is greater than 64 (above table)
The Total area must be 81
Then 81 = 68 + X
There fore X = 81-68
Answe X = 23 Sq. Cm
Dr.Peterson
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That would be an interesting (though flawed) method if we were told that the sides of the squares are integers.
But that assumption is wrong; in fact, the sides are about 9.8 cm long. Your answer also does not fit the hint that was given.
(The answer in #11 is also wrong; I don't know why I didn't say so explicitly at the time. The error is in saying "32 + 64 = 98".)
Total 98 is not perfect square.
Next perfect square is 100 only.
Three portions total area is 68.
If assumption is decimal, the side should be more than 8.2 min.( n number of decimals can be assumed and you can't get a single answer)
8.3x 8.3= 68.89 then missing Area 0.89
8.4x8.4 = 70.56 then .....2.56
8.5x8.5. = 72.25 then .....4.25
8.6x8.6 = 73.96 then .....5.96
8.7x8.7 = 75.69 then..... 7.69
8.8x8.8 = 77.44 then ..... 9.44
8.9x8.9 = 79.21 then .....11.21
Next perfect square is 100 only.
Assumption is integer is ok.
Three portions total area is 68.
If assumption is decimal, the side should be more than 8.2 min.( n number of decimals can be assumed and you can't get a single answer)
8.3x 8.3= 68.89 then missing Area 0.89
8.4x8.4 = 70.56 then .....2.56
8.5x8.5. = 72.25 then .....4.25
8.6x8.6 = 73.96 then .....5.96
8.7x8.7 = 75.69 then..... 7.69
8.8x8.8 = 77.44 then ..... 9.44
8.9x8.9 = 79.21 then .....11.21