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

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

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