Not enough info; undefined.
Unless you assume something about the two opposite sharp angles.
Jan
I'm not sure about the area of the shaded rectangle, but it's greater on my laptop than on my phone.
When one office was getting larger computer monitors, someone asked " will we need larger mouse mats as well ?" - well I can see the logic.
When one office was getting larger computer monitors, someone asked " will we need larger mouse mats as well ?" - well I can see the logic.
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the area is defined. no further definition of geometry is needed.
(edit: there are two geometric solutions, but the area is always the same)
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12
The trick is:
Any rectangle with an area of 12 fits the bill.
The rectangle is undefined, however the area is defined.
One can draw with
3x4
4x3
3.4641x3.4641
2x6
1x12
From Y : 3 = 4 : X (similar triangles), both sides of the equation multiply by 3 and multiply by X, you get X x Y = 3 x 4 = 12 as previously said.
Assuming the two opposite sharp angles are the same( same slopes ) you end up with: The area of the shaded rectangle must be 12.
For a mathematician, this not enough to answer, because this is no proof the triangle exists.
Then more work shows that it does exist. Not only one but any as long as it's area is 12.
thick black line triangles are similar (area is the same).
red triangles are similar (area is the same).
yellow triangles are similar (area is the same).
green rectangles must have same area. one of them is 4 * 3.
thanks AllenB!
red triangles are similar (area is the same).
yellow triangles are similar (area is the same).
green rectangles must have same area. one of them is 4 * 3.
thanks AllenB!
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The rectangle area puzzle.
There is no specific need for the 4 and 3 values, this is to lead to messing with Pythagoras, that ends nowhere.
Proof
Instead of 4 and 3 let's use A and B.
Draw the figure with a rectangle which area is AB
X horizontal edge.
AxB/X vertical edge.
The upper small triangle has straight edges X and A.
The slope of it's hypothenus is A/X
The lower small triangle has straight edges B and AxB/X.
The slope of it's hypothenus is (AxB/X)/B.
We see the two slopes are equal.
Hence we do have a figure where the answer to the puzzle is AxB regardless of the chosen X.
QED
There is no specific need for the 4 and 3 values, this is to lead to messing with Pythagoras, that ends nowhere.
Proof
Instead of 4 and 3 let's use A and B.
Draw the figure with a rectangle which area is AB
X horizontal edge.
AxB/X vertical edge.
The upper small triangle has straight edges X and A.
The slope of it's hypothenus is A/X
The lower small triangle has straight edges B and AxB/X.
The slope of it's hypothenus is (AxB/X)/B.
We see the two slopes are equal.
Hence we do have a figure where the answer to the puzzle is AxB regardless of the chosen X.
QED
I'm not saying I'm a genius or anything but the puzzle box said 3-6 years and I got it done in less than a day.
My case is going before a jury. Great, now I will be judged by a dozen people not smart enough to even get out of jury duty.
