what is the lengths of a right triangle if the hypotenuse is 20 and the area of the triangle 90?

Question

I have no clue how to figure this answer out.

Thanks

Answer 1

20^2 = h^2 + b^2

hb/2 = 90

hb=180

h= 180/b

400= (180/b)^2 + b^2

400b^2 = 180^2 + b^4

b^4 -400b^2 + 32400 = 0

let x = b^2

x^2 - 400x +32400=0

x= 400/2 + or - (1/2)sqr(400^2 -4(32400))

x = 200-(1/2)sqr160000-129600) =200-87.178= 112.822

b=sqrx = 10.62

hb/2 =90

h=180/b = 180/10.62 = 16.95

Its sides are 10.62 and 16.95 approximately, or very slightly less, given rounding

10.62^2 + 16.95^2 = 112.8 + 287.3 = 400.1

Answer 2

20^2 = h^2 + b^2

hb/2 = 90

hb=180

h= 180/b

400= (180/b)^2 + b^2

400b^2 = 180^2 + b^4

b^4 -400b^2 + 32400 = 0

let x = b^2

x^2 - 400x +32400=0

x= 400/2 + or - (1/2)sqr(400^2 -4(32400))

x = 200-(1/2)sqr160000-129600) =200-87.178= 112.822

b=sqrx = 10.62

hb/2 =90

h=180/b = 180/10.62 = 16.95

Its sides are 10.62 and 16.95 approximately, or very slightly less, given rounding

10.62^2 + 16.95^2 = 112.8 + 287.3 = 400.1= 20^2 approximately