The length of a rectangle is 1 cm longer than the width. If the diagnol of the rectangle is 4cm, what are the dimensions ( the length and the width) of the rectangle?
Please help !
2007-02-01 06:18:05 · 3 answers · asked by CookFrNW 3 in Science & Mathematics ➔ Mathematics
The diaganol forms the hypotenuse of a right triangle with sides x and x+1.
So... by Pythagorean Theorem:
x^2 + (x+1)^2 = 4^2
x^2 + x^2 + 2x + 1 = 16
2x^2 + 2x - 15 = 0
x = -b +/- (b^2 - 4ac)^.5 / 2a
x = -2 +/- (4 - 4*2*-15)^.5 / 4
x = -2 +/- (124)^.5 / 4
x = -2 +/- 11.135 / 4
x can't be negative, so x = (-2 + 11.135) / 4
x = 2.28375 cm = width
x + 1 = 3.28375 cm = length
Test:
2.28375^2 + 3.28375^2 = 15.998952
4^2 = 16
Hope this helps!
2007-02-01 06:25:36 · answer #1 · answered by disposable_hero_too 6 · 0⤊ 0⤋
Set up a right triangle with one leg = x, the other leg = (x+1), and the hypotenuse = 4.
x^2 + (x+1)^2 = 4^2
x^2 + x^2 + 2x + 1 = 16
2x^2 + 2x - 15 = 0
x = (-2+(2^2-4*-15*2)^(1/2)) / 4
x = (-2 + (124)^1/2)/4
x = -0.5 + 0.5(31^(0.5))
x = 2.28388
2007-02-01 14:26:09 · answer #2 · answered by Ptrain 1 · 0⤊ 0⤋
x^2+(x+1)^2=4^2
2x^2+2x+1=16
2x^2+2x-15=0
x=[-2+/-sq.rt.(4+120)]/4
x=[-2+sq.rt.124]/4
so the width is 2.28 cm
length=3.28 cm
2007-02-01 14:28:01 · answer #3 · answered by raj 7 · 0⤊ 0⤋