Need Math Explaination HELP? Please!

[cl15876]

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?

[OZZIE4DUKE]

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.

[TillyGalore]

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.

Did you click on the image? It opens the image making clearer.

I couldn't figure out how to solve the problem.

[OZZIE4DUKE]

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%.

[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.

[DukeUsul]

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.

[cl15876]

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%.

Ding, Ding, Ding! That is correct! 3/4's! But now I have to figure out how to explain this to my daughter. I understand the 1/4 triangles making up half which is kind of visual, but how did you assume the inner square made up the other 1/4 or 25%?

DukeUsul - that is what I was looking for to explain to her, but was at a loss for how to explain to her. She now knows pythagorean theorem like the back of her hand!

Oz and DukeUsul - thanks!

[OZZIE4DUKE]

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.

You're complicating the solution using the pythagorean theorem.

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.

[DukeUsul]

I really hope the math in my answer works out to 75%...... :lol: :lol: :lol: :lol:

And of course I'm complicating it. It's what I tend to do.

[CameronBornAndBred]

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....

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.

[DukeUsul]

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....

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.

I think it's just crudely drawn.

[cl15876]

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....

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]

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....

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.

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.

[CameronBornAndBred]

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....

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.

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

[wilson]

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.

[OZZIE4DUKE]

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!

[cl15876]

"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!

[cl15876]

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!

I agree! I have lemons for that water!!!

[devildeac]

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

[CameronBornAndBred]

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

Tell them if you are going to draw a square it damn well better be square.