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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