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Posted by **hollistergurl** on Thursday, January 27, 2011 at 5:43pm.

- 7th grade math: Pythagorean Thereom -
**Anonymous**, Thursday, January 27, 2011 at 5:51pm49-41=8

7+1

6+2

5+3

4+4

- 7th grade math: Pythagorean Thereom -
**helper**, Thursday, January 27, 2011 at 5:57pmDo you mean that the two unknown sides total 49, or all three sides total 49?

- 7th grade math: Pythagorean Thereom -
**Anonymous**, Thursday, January 27, 2011 at 6:01pmThat is possible combinations, but in right triangle c^2=a^2+b^2

41=a^2+b^2

a=4 b=5

41=5^2+4^2

41=25+16

Other two sides is 4 and 5

- 7th grade math: Pythagorean Thereom -
**Anonymous**, Thursday, January 27, 2011 at 6:03pm41+4+4=49

- 7th grade math: Pythagorean Thereom -
**helper**, Thursday, January 27, 2011 at 6:13pmIt could be,

9 + 40 = 49

a^2 + b^2 = 41^2

a + b = 49

Solving simultaneously

a = 9, b = 40 or

a = 40, b = 9

41^2 = 9^2 + 40^2

1681 = 81 + 1600

1681 = 1681

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