Posted by hollistergurl on Thursday, January 27, 2011 at 5:43pm.
If the sum of the sides of a right triangle is 49 inches and the hypotenus is 41 inches, then what would be the other two sides?

7th grade math: Pythagorean Thereom  Anonymous, Thursday, January 27, 2011 at 5:51pm
4941=8
7+1
6+2
5+3
4+4

7th grade math: Pythagorean Thereom  helper, Thursday, January 27, 2011 at 5:57pm
Do 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 6:01pm
That 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 6:03pm
41+4+4=49

7th grade math: Pythagorean Thereom  helper, Thursday, January 27, 2011 at 6:13pm
It 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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