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cl15876 wrote:Here is the problem that I was challenged with providing a clear explanation with as to how one comes up with the solution! Can anyone help me out here?

Math Problem.jpg

OZZIE4DUKE wrote:

cl15876 wrote:Here is the problem that I was challenged with providing a clear explanation with as to how one comes up with the solution! Can anyone help me out here?

Math Problem.jpg

I can't read any of the text - it shows up as gibberish.

OZZIE4DUKE wrote:The answer is 75% is shaded. Each triangle pair makes up 1/4 of the large square, so that is half. The inner square makes up another 25%.

DukeUsul wrote:Let's calculate the area of the outer blue square, then the white square that's overlayed on top of it, then the smaller blue square overlayed on that (that's how I think of this one). If we take the area of the large square, subtract the white square and then add the little blue square we'll have the area that's shaded.

Let's set the length of the sides of the outer cube to be 1. So the distance from A to B or A to D is 1.

This means the distance form A to H is 0.5

This means the distance from H to E is sqrt (0.5^2 + 0.5^2) (think pythagorean)

Now you know the length of the sides of the "white" square. Take the lengths of the sides of the white square and do the same thing again (pythagorean) to get the lengths of the sides of the inner square. The distance IJ is equal to sqrt (IE^2 + EJ^2)

Now take the blue area (1 squared) subtract the white area and add the small blue area. Since the area of big square is one, then the answer you get will be the fractional shaded area.

OZZIE4DUKE wrote: Look at the four blue shaded triangles. They are four equal area triangles since their base and height are all the same length (the midpoints of the outer square). They are all "1, 1, square root of 2" triangles. If you put "fold over" the adjacent triangle so the hypotenuses meet, you'll have a square, the area of each is 1/4 of the total area of the outer square. You have two of those, so they together equal 1/2 the area of the bigger square. The area of the inner blue square is also equal to that. The sides of the inner square ILJK are all also 1/2 the length of the outer square sides since they are at the midpoint of each "hypotenuse", ie, they have to be equal and there is as much length to the outside of the outer square as there is along the lines IL, LK or JK etc. I'm being somewhat visual/intuitive, but I'm sure you can prove it mathematically on paper.

CameronBornAndBred wrote:I don't see how EFGH and ABCD can both be squares, if only two opposite corners of EGFH overlap ABCD. For both to be squares, all 4 corners would have to overlap if they are at the midpoints.

CameronBornAndBred wrote:

OZZIE4DUKE wrote: Look at the four blue shaded triangles. They are four equal area triangles since their base and height are all the same length (the midpoints of the outer square). They are all "1, 1, square root of 2" triangles. If you put "fold over" the adjacent triangle so the hypotenuses meet, you'll have a square, the area of each is 1/4 of the total area of the outer square. You have two of those, so they together equal 1/2 the area of the bigger square. The area of the inner blue square is also equal to that. The sides of the inner square ILJK are all also 1/2 the length of the outer square sides since they are at the midpoint of each "hypotenuse", ie, they have to be equal and there is as much length to the outside of the outer square as there is along the lines IL, LK or JK etc. I'm being somewhat visual/intuitive, but I'm sure you can prove it mathematically on paper.

That sounds great BUT....

CameronBornAndBred wrote:I don't see how EFGH and ABCD can both be squares, if only two opposite corners of EGFH overlap ABCD. For both to be squares, all 4 corners would have to overlap if they are at the midpoints.

That's the explanation I'm waiting on. One of those two HAS to be a rectangle. Or I've had too much beer.

CameronBornAndBred wrote:

OZZIE4DUKE wrote: Look at the four blue shaded triangles. They are four equal area triangles since their base and height are all the same length (the midpoints of the outer square). They are all "1, 1, square root of 2" triangles. If you put "fold over" the adjacent triangle so the hypotenuses meet, you'll have a square, the area of each is 1/4 of the total area of the outer square. You have two of those, so they together equal 1/2 the area of the bigger square. The area of the inner blue square is also equal to that. The sides of the inner square ILJK are all also 1/2 the length of the outer square sides since they are at the midpoint of each "hypotenuse", ie, they have to be equal and there is as much length to the outside of the outer square as there is along the lines IL, LK or JK etc. I'm being somewhat visual/intuitive, but I'm sure you can prove it mathematically on paper.

That sounds great BUT....

CameronBornAndBred wrote:I don't see how EFGH and ABCD can both be squares, if only two opposite corners of EGFH overlap ABCD. For both to be squares, all 4 corners would have to overlap if they are at the midpoints.

That's the explanation I'm waiting on. One of those two HAS to be a rectangle. Or I've had too much beer.

CameronBornAndBred wrote:

OZZIE4DUKE wrote: Look at the four blue shaded triangles. They are four equal area triangles since their base and height are all the same length (the midpoints of the outer square). They are all "1, 1, square root of 2" triangles. If you put "fold over" the adjacent triangle so the hypotenuses meet, you'll have a square, the area of each is 1/4 of the total area of the outer square. You have two of those, so they together equal 1/2 the area of the bigger square. The area of the inner blue square is also equal to that. The sides of the inner square ILJK are all also 1/2 the length of the outer square sides since they are at the midpoint of each "hypotenuse", ie, they have to be equal and there is as much length to the outside of the outer square as there is along the lines IL, LK or JK etc. I'm being somewhat visual/intuitive, but I'm sure you can prove it mathematically on paper.

That sounds great BUT....

CameronBornAndBred wrote:I don't see how EFGH and ABCD can both be squares, if only two opposite corners of EGFH overlap ABCD. For both to be squares, all 4 corners would have to overlap if they are at the midpoints.

That's the explanation I'm waiting on. One of those two HAS to be a rectangle. Or I've had too much beer.

cl15876 wrote:

CameronBornAndBred wrote:

OZZIE4DUKE wrote: Look at the four blue shaded triangles. They are four equal area triangles since their base and height are all the same length (the midpoints of the outer square). They are all "1, 1, square root of 2" triangles. If you put "fold over" the adjacent triangle so the hypotenuses meet, you'll have a square, the area of each is 1/4 of the total area of the outer square. You have two of those, so they together equal 1/2 the area of the bigger square. The area of the inner blue square is also equal to that. The sides of the inner square ILJK are all also 1/2 the length of the outer square sides since they are at the midpoint of each "hypotenuse", ie, they have to be equal and there is as much length to the outside of the outer square as there is along the lines IL, LK or JK etc. I'm being somewhat visual/intuitive, but I'm sure you can prove it mathematically on paper.

That sounds great BUT....

CameronBornAndBred wrote:I don't see how EFGH and ABCD can both be squares, if only two opposite corners of EGFH overlap ABCD. For both to be squares, all 4 corners would have to overlap if they are at the midpoints.

That's the explanation I'm waiting on. One of those two HAS to be a rectangle. Or I've had too much beer.

CB&B - it may be my drawing capabilities, they are effectively 3 squares ABCD, EFGH, IJKL laid ontop of each other. I tried hard to make them look exactly like the problem. Sorry if I confused you.

OZZIE4DUKE wrote: You've had too much beer. ABCD is defined as a square in the problem. EFGH has to be a square because all four sides are equal length and the corners are all right angles.

wilson wrote:

OZZIE4DUKE wrote: You've had too much beer. ABCD is defined as a square in the problem. EFGH has to be a square because all four sides are equal length and the corners are all right angles.

Maybe you should switch to Bud Light.
Never gets old.

CameronBornAndBred wrote:

cl15876 wrote:

CameronBornAndBred wrote:"quote="OZZIE4DUKE""
Look at the four blue shaded triangles. They are four equal area triangles since their base and height are all the same length (the midpoints of the outer square). They are all "1, 1, square root of 2" triangles. If you put "fold over" the adjacent triangle so the hypotenuses meet, you'll have a square, the area of each is 1/4 of the total area of the outer square. You have two of those, so they together equal 1/2 the area of the bigger square. The area of the inner blue square is also equal to that. The sides of the inner square ILJK are all also 1/2 the length of the outer square sides since they are at the midpoint of each "hypotenuse", ie, they have to be equal and there is as much length to the outside of the outer square as there is along the lines IL, LK or JK etc. I'm being somewhat visual/intuitive, but I'm sure you can prove it mathematically on paper."/quote"
That sounds great BUT....

"quote="CameronBornAndBred""I don't see how EFGH and ABCD can both be squares, if only two opposite corners of EGFH overlap ABCD. For both to be squares, all 4 corners would have to overlap if they are at the midpoints."/quote"
That's the explanation I'm waiting on. One of those two HAS to be a rectangle. Or I've had too much beer.

CB&B - it may be my drawing capabilities, they are effectively 3 squares ABCD, EFGH, IJKL laid ontop of each other. I tried hard to make them look exactly like the problem. Sorry if I confused you.

YOU drew them? Oh...well then you had too much beer. And made my brain hurt. Bad bunchanumbers!

OZZIE4DUKE wrote:

wilson wrote:

OZZIE4DUKE wrote: You've had too much beer. ABCD is defined as a square in the problem. EFGH has to be a square because all four sides are equal length and the corners are all right angles.

Maybe you should switch to Bud Light.
Never gets old.

It's the drinkability!

devildeac wrote:this is the kinda stuff one has to put up with as a mod...? :oops: :roll: