Gina and Sean drew the following figures to prove the Pythagorean Theorem c2 = a2 + b2. Both figures are made of two squares and four identical right triangles, as shown below.
Gina and Sean wrote the following proofs.
Gina’s Proof:
Step 1: Area of PQRS = (a + b) 2 = a2 + b2 + 4ab
Step 2: Area of triangle PKN =; hence the area of 4 triangles = 4ab
Step 3: Area of KLMN = c2 = area of PQRS – area of 4 triangles = a2 + b2 + 4ab – 4ab
Hence, c2 = a2 + b2

- anonymous

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- anonymous

Sean’s Proof:
Step 1: Area of triangle EAF =; hence the area of 4 triangles =
Step 2: Area of square ABCD = (a – b) 2 = a2 + b2 – 2ab
Step 3: Area of EFGH = c2 = Area of 4 triangles + area of ABCD = 2ab + a2 + b2 – 2ab
Hence c2 = a2 + b2

- anonymous

http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_1000_Subtest_02_05/image0054e8c5217.jpg

- anonymous

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- anonymous

Which statement gives the correct conclusion about the proofs given by Gina and Sean?
Sean’s proof is correct and Gina’s proof is incorrect.
Both the proofs are correct.
Gina’s proof is correct and Sean’s proof is incorrect.
Both the proofs are incorrect.

- anonymous

Please help

- anonymous

May you help me?
Please?????

- myininaya

Look at one square at a time.
Find the area of the bigger shape using the formula for area of square.
Then find the area of the smaller shapes that are contained in the bigger shape.
And then see if you get the Pythagorean thm c^2=a^2+b^2

- anonymous

Ok

- anonymous

So Sean's figure would be c^2, and Gina's figure would be (a+b)^2.

- myininaya

Ok those are the areas of the big shape that contains all the small shapes for each

- myininaya

Now find the areas of the smaller shapes inside the big shapes

- anonymous

Ok.

- anonymous

The square in Gina's figue would be c^2, and the triangles in Gina's figure will be 1/2 ab.

- myininaya

and there are four of those triangles right?

- anonymous

And there are 4 triangles in Gina's figure so it would be 2 ab.

- myininaya

\[bigger area=the \small \square+the triangles \]

- myininaya

Yep yep :)
So we have
\[(a+b)^2=c^2+2ab \text{ right?}\]

- myininaya

For gina's

- myininaya

Now expand (a+b)^2 by writing (a+b)(a+b) and then distributing (or multiplying)

- anonymous

Yes. But, looking at the diagram, we know she isn't right, because she put 4 ab, instead of 2ab. So it's either Sean's right, or neither of them are right.

- myininaya

Ok yep you are right gina messed up there :)

- myininaya

Ok now looking at sean's

- myininaya

you said the area of big square is c^2 right?

- myininaya

now lets find the areas of the smaller shapes that are contained in this big square

- anonymous

Of the bigger square, yes.

- anonymous

Ok

- myininaya

the area of the small square is?

- anonymous

a-b^2

- myininaya

(a-b)^2 is right

- anonymous

(a-b)(a-b) ?

- myininaya

Now tell me the area of on those triangle and then we will multiply it by 4 since there are 4 triangles

- myininaya

right (a-b)^2=(a-b)(a-b) :)

- anonymous

ab, since on the hypotenuse, there are two different line segments on the same line which is a-b+b, times b. or otherwise, a times b.

- myininaya

1/2 *ab right?

- anonymous

Yes... I forgot the 1/2 part... whoops. Thanks for helping me remember that! :)

- myininaya

And now there are four of those triangles so
We have :
for sean:
c^2=4*1/2*ab+(a-b)^2
bigsquare=thetriangles+smallsquare

- myininaya

\[c^2=4 \cdot \frac{1}{2} \cdot ab+(a-b)^2\]

- anonymous

c^2 = 2ab + (a-b)(a-b)
c^2 =2ab+ a^2 -ab-ab+b

- myininaya

well one little correction to what you have:
c^2 = 2ab + (a-b)(a-b)
c^2 =2ab+ a^2 -ab-ab+b^2
Step 3: Area of EFGH = c2 = Area of 4 triangles + area of ABCD = 2ab + a2 + b2 – 2ab
And that is what he has for step 3 correct?

- myininaya

mean because -ab-ab=-2ab and so you have
c^2=2ab+a^2-2ab+b^2

- anonymous

no... he's not correct, because he didn't have the a^2, or the negative sign in front of 2ab.

- myininaya

Hmmm then I must be looking at something different

- myininaya

Are you sure? look again.

- anonymous

It says, "Step 3: Area of EFGH = c2 = Area of 4 triangles + area of ABCD = 2ab + a2 + b2 – 2ab
Hence c2 = a2 + b2"

- myininaya

The ordering he has is a little different

- anonymous

Ohhh..... whoops.

- anonymous

Ok, then he's right. I just didn't see it correctly.... Thanks!!!

- myininaya

:)
So how do you feel about this question?
Do you understand it better?

- anonymous

Yes. Thank you so very much!!!!!

- anonymous

:)

- myininaya

Np. Have a great day.

- anonymous

You too!

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