Question

The figures below show two different ways of arranging four identical triangles of grey poster board on top of a white square. The square has sides equal to x + y, while the hypotenuse of each triangle is represented by the variable c.

Answer 1

Hazel wrote the following statements to prove that c2 = x2 + y2.

1. Area of the four grey triangles inside figure A =
2. Area of the white square inside figure A = c2
3. Area of figure A = c2 + 2xy
4. Area of the four grey triangles inside figure B = 4xy
5. Area of the two white squares inside figure B = x2+ y2
6. Area of figure B = x2+ y2 + 4xy
7. Area of figure A = area of Figure B, hence c2 + 2xy = x2+ y2 + 4xy
8. Therefore, c2 = x2+ y2

Which is the first incorrect statement in Hazel’s proof?

Answer 2

Which is the first incorrect statement in Hazel’s proof?

Statement 6

Statement 7

Statement 4

Statement 5

Answer 3

The first time I picked statement 5 and I was wrong

Answer 4

isnt the area of the 4 grey triangles in B, xy + xy = 2xy? not 4xy ....

Answer 5

I think your right, so statement 6 would be the right answer?

Answer 6

1. Area of the four grey triangles inside figure A =
2. Area of the white square inside figure A = c2
3. Area of figure A = c2 + 2xy
4. Area of the four grey triangles inside figure B = 4xy <--- ????

Answer 7

Yeah it would be 2xy and he wrote 4xy so that statement is wrong

Answer 8

there are others that are wrong, but that seems to be the first to me

Answer 9

Did you see the image of the equation btw? It was suppose to be before the statements in the actual question

Answer 10

i did, but this site is still acting up ....

Answer 11

whats wrong with OS?

Answer 12

@Mertsj

Answer 13

Does the first statement mean that the 4 gray triangles inside square A are congruent ?

Answer 14

first statement = the equation I posted as a picture

Answer 15

4 is wrong. It says that the area of the 4 gray triangles in Figure B is 4 xy. But look at it. Each triangle has base x and height y so the area of each triangle is:
\[A=\frac{1}{2}xy\]
4 triangles is certainly 4 times one triangle so:
\[A(4 \Delta's)=4\times \frac{1}{2}xy=2xy\]
So statement 4 is false.

Answer 16

ok so statement 4 would of been the right answer?

Answer 17

alright thanks merts! I'll tag u in the next one I need to review.

Answer 18

can't close my question >.<