# 7th grade math: Pythagorean Thereom

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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 -

49-41=8

7+1
6+2
5+3
4+4

• 7th grade math: Pythagorean Thereom -

Do you mean that the two unknown sides total 49, or all three sides total 49?

• 7th grade math: Pythagorean Thereom -

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 -

41+4+4=49

• 7th grade math: Pythagorean Thereom -

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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