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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

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

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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.
Please help
May you help me? Please?????
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
Ok
So Sean's figure would be c^2, and Gina's figure would be (a+b)^2.
Ok those are the areas of the big shape that contains all the small shapes for each
Now find the areas of the smaller shapes inside the big shapes
Ok.
The square in Gina's figue would be c^2, and the triangles in Gina's figure will be 1/2 ab.
and there are four of those triangles right?
And there are 4 triangles in Gina's figure so it would be 2 ab.
\[bigger area=the \small \square+the triangles \]
Yep yep :) So we have \[(a+b)^2=c^2+2ab \text{ right?}\]
For gina's
Now expand (a+b)^2 by writing (a+b)(a+b) and then distributing (or multiplying)
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.
Ok yep you are right gina messed up there :)
Ok now looking at sean's
you said the area of big square is c^2 right?
now lets find the areas of the smaller shapes that are contained in this big square
Of the bigger square, yes.
Ok
the area of the small square is?
a-b^2
(a-b)^2 is right
(a-b)(a-b) ?
Now tell me the area of on those triangle and then we will multiply it by 4 since there are 4 triangles
right (a-b)^2=(a-b)(a-b) :)
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.
1/2 *ab right?
Yes... I forgot the 1/2 part... whoops. Thanks for helping me remember that! :)
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
\[c^2=4 \cdot \frac{1}{2} \cdot ab+(a-b)^2\]
c^2 = 2ab + (a-b)(a-b) c^2 =2ab+ a^2 -ab-ab+b
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?
mean because -ab-ab=-2ab and so you have c^2=2ab+a^2-2ab+b^2
no... he's not correct, because he didn't have the a^2, or the negative sign in front of 2ab.
Hmmm then I must be looking at something different
Are you sure? look again.
It says, "Step 3: Area of EFGH = c2 = Area of 4 triangles + area of ABCD = 2ab + a2 + b2 – 2ab Hence c2 = a2 + b2"
The ordering he has is a little different
Ohhh..... whoops.
Ok, then he's right. I just didn't see it correctly.... Thanks!!!
:) So how do you feel about this question? Do you understand it better?
Yes. Thank you so very much!!!!!
:)
Np. Have a great day.
You too!

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