if the sum of the right triangle is 49 inches and the hypotenuse is 41 what are the two sides?

Question

i’m not sure what the answer is and would like help

Answer 1

"sum" can't mean perimeter, so maybe it means area, then:

A =bh/2 = 49

bh = 98

b=98/h

where b=base and h=height

41^2= b^2 + h^2 = (98/h)^2 + h^2

41h)^2 = 98^2 + h^4

h^4 - 1641h^2 - 98^2 = 0

let x=h^2

x^2 -1641h - 9604 = 0

use the quadratic formula to solve for x

x=164/2 + or - (1/2)sqr(1641^2+4(9604))

then take the square root of x to solve for h=height of the right triangle

plug that value into b=98/h to solve for the base of the right triangle

or maybe you meant the sum of the other two sides =41

then the two other sides are 40 and 9

41^2 = 40^2 + 9^2

1681 = 1600+ 81

that came out so evenly, that's probably what you want

b+h=41

b=41-h

41^2 =h^2 +b^2 = h^2 + (49-h)^2

solve much the same way, expand, combine like terms, use the quadratic formula

41^2 = h^2 +49^2-98h+ h^2

2h^2 -98h + 49^2- 41^2 = 0