Missing area of a square with no given dimensions

Gopi said:
Total 98 is not perfect square.
Next perfect square is 100 only.
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You're totally missing the point. Nothing in the problem says that the area must be a perfect square (and the sides must be integers). And what the problem does say can't be satisfied with such a square.

The answer is that the missing area is 28 cm2, and the total area is 96 cm2. Two approaches have been shown that lead to this one correct solution. In particular, I have shown that the sums of the two pairs of opposite regions must be equal, so that 16+32=20+x, making x=48-20=28 cm2.

If you honestly think your answer of 13 cm2 is correct, you are required to show how to place the middle point to get the three given areas; you have said absolutely nothing about how to get them.
 
Dr.Peterson said:
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".)
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Gopi said:
Total 98 is not perfect square.
Next perfect square is 100 only.
Click to expand...
Assumption integer is ok. And missing Area is 13 cm.

Given three portions having total area is 68. And hence Side must be more than 8.

If assumption is in decimal, the side should be more than 8.2 min because 8.2 x 8.2= 67.24
( Similarly n number of decimals can also be assumed and you can't get a single answer)
Example
8.3x 8.3= 68.89 less 68 then missing area is 0.89 Sq cm.
Similarly missing area for
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

Upto side 8.8 cm is not practically possible because two sides of quadrant are more than half of square side.

Example. If side of sq is 8.8 and hence given two right angle sides of quadrant are equal which is 4.4 each. Hence as per the drawing the area formed by missing quadrant must be more than area of right angle triangle formed by the two sides.
Right angle Triangle area is 4.4X4.4x(1/2)= 9.68

Hence Assuming side as integer will give correct answer side is 9 cm. Area of sq 81 sq. cm and missing area is 13 sq cm.

When you are getting answer at side of square is 9, then no need to add decimals or integers with the side 9cm.
 
Gopi said:
Hence Assuming side as integer will give correct answer side is 9 cm. Area of sq 81 sq. cm and missing area is 13 sq cm.

When you are getting answer at side of square is 9, then no need to add decimals or integers with the side 9cm.
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Here is the actual figure, showing exact size of the square and location of the point:

1654910395606.png​

Here is a square with sides of length 9; and with the indicated areas for two of the given regions:

1654910750043.png​

Observe that the third given region has the wrong area; this is because that area, CFPG, has to have area 81/2-16=24.5. It is impossible to get the given conditions with the square you want to use.

This problem is not what you think it is! There is only one solution, and it does not have integer sides. Your assumption is unjustified, and leads to a wrong answer.

But it is certainly fun to play with.
 
Subhotosh Khan said:
Incorrect.

The dimension of an area cannot be "cm". Area must have dimension of square of a length (cm2, m2, sq.ft., km2, in2, etc.)


Please justify your assumption.
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Sorry for the error.
Missing area is 13 sq cm is correct.
 
Dr.Peterson said:
Here is the actual figure, showing exact size of the square and location of the point:


Here is a square with sides of length 9; and with the indicated areas for two of the given regions:


Observe that the third given region has the wrong area; this is because that area, CFPG, has to have area 81/2-16=24.5. It is impossible to get the given conditions with the square you want to use.

This problem is not what you think it is! There is only one solution, and it does not have integer sides. Your assumption is unjustified, and leads to a wrong answer.

But it is certainly fun to play with.
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Are you telling as per actual location point formed by given areas, the problem by is wrong.?
 
Gopi said:
Are you telling as per actual location point formed by given areas, the problem by is wrong.?
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I'm telling you, as I have several times, that your answer is wrong. You are solving a different problem.

Nothing you have ever said dealt with the actual four areas and showed that you can obtain them for your 9-cm square, which is central to the problem. The mere fact that four numbers add up to the area of a square is not enough; and the idea that that square has integer sides is not even suggested by the problem.
 
Dr.Peterson said:
I'm telling you, as I have several times, that your answer is wrong. You are solving a different problem.

Nothing you have ever said dealt with the actual four areas and showed that you can obtain them for your 9-cm square, which is central to the problem. The mere fact that four numbers add up to the area of a square is not enough; and the idea that that square has integer sides is not even suggested by the problem.
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??
 
I finally figured this one out. It took for ever but I did it in the end (with Dr Peterson's hint)


At the end I finally saw that the sum of the area that contained point A and the area that contained point C is exactly half the area. The sum of the other two regions must then be the other half of the square

Anyone see a clever way to see this 'easily'
 
Steven G said:
At the end I finally saw that the sum of the area that contained point A and the area that contained point C is exactly half the area. The sum of the other two regions must then be the other half of the square

Anyone see a clever way to see this 'easily'
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It depends on what "easily" means. Look here:

1655168918652.png​

The sum of the two green triangles is the same regardless of the location of P (why?); the same is true of the red triangles. Therefore, so is the sum of all four. Now put P in any nice place, such as this:

1655169105193.png​
 
Steven G said:
I finally figured this one out. It took for ever but I did it in the end (with Dr Peterson's hint)
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It took me a long time to see this too. Whoever invented this question managed to hide the "shortcut" in plain sight. A very nice puzzle. Even the choice of areas 16 and 32 made me start thinking this had something to do with the ratio 1:2. A very clever deception indeed (Galaxy Quest quote)
 
Sorry but one side of the square is not 4.899.
One side of the square is sqrt(96) = 4sqrt(6) ~ 9.798
 
Kuldeep Matharu said:
Mr Steven G it’s half of the sq side
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I agree that nothing you show there is wrong.

My question for you is, how do you think this contributes to the discussion? You appear to have calculated the areas in a much more complicated way than others have, and I don't see that this provides a way to solve the problem (where the size of the square is not known).
 
Dr.Peterson said:
I agree that nothing you show there is wrong.

My question for you is, how do you think this contributes to the discussion? You appear to have calculated the areas in a much more complicated way than others have, and I don't see that this provides a way to solve the problem (where the size of the square is not known).
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Dr Peterson sorry pl the sum of four areas are 96 under roots will be one side of the square that is 9.798

Kuldeep Matharu said:
Dr Peterson sorry pl the sum of four areas are 96 under roots will be one side of the square that is 9.798
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Dr Peterson sorry it could be easy in this way also 16+32=48
48-20= 28 ans
 
Kuldeep Matharu said:
Dr Peterson sorry pl the sum of four areas are 96 under roots will be one side of the square that is 9.798


Dr Peterson sorry it could be easy in this way also 16+32=48
48-20= 28 ans
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All of that has already been said. But it isn't what you said!

Your work, if anything, looks like you are checking the answer, taking it as already known that the area is 96. But I can't tell for sure how you did all the area calculations, or why. I found the location of my point P (your O) in a much simpler way.
 
Kuldeep Matharu said:
View attachment 33096
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Ignoring all the decimal numbers, your drawing does suggest a nice way to see the key idea:

1655410301639.png​

Each of the four triangles is clearly half of the rectangle containing it (e.g. HIJE), so the sum of the greens, and the sum of the reds, are both exactly half of the area of EFGH. This implies that the sum of opposite regions in the original drawing is half of ABCD, so that area is 96, and we can see the missing area.

Also, from that, we could work out the areas of the individual triangles, and the location of P, in a different way than I did before (which used triangles PEG and PHF).
 
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