SOLUTION: this is a word problem. The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures sqrt6m. find the width and height. I'm not sure how to

Question 142744This question is from textbook
: this is a word problem. The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures sqrt6m. find the width and height.
I'm not sure how to set this up to solve it. Can you help? Thanks, Anne This question is from textbook

Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
Well, we need to use pythagorean theorem coupled with quadratic formula in solving this. Let's see,
The width, the height, and diag. brace forms a right triangle. By Pyth. theo., the given equates to: (sq rt6)^2 = h^2 + w^2 ------- eqn 1
And as it stated that the width is 2meters larger than the height,"w= h+2"--eq 2
Therefore we substitute this value of "w" to eqn 1. To show,
(sq rt6)^2 = h^2 + (h+2)^2
6=h^2 +h^2 +4h +4
0=2h^2 +4h -2 ---- divide the whole eqn by 2 becomes,
0=h^2 +2h -1
In here we use quadratic formula, and let's assinged "h" = "x" so you won't be misleading,
x = -b +- sqrt[b^2 -4ac]/ 2a
where,
x= h
a= 1
b= 2
c= -1
x= {-2 +- sqrt[2^2 - 4*1*-1]}/2
x= {-2 +- sqrt[4+4]}/2
x= {-2 +- sqrt(8)}/2 = (-2 +- 2.83)/2
It has 2 values,
x = (-2-2.83)/2= -2.415m, ther's no neg value for a gate???
x= (-2+2.83)/2 = 0.415 m, this is the one to used! = h
For w, as per eqn 2, = 0.415+2= 2.415m
In doubt? go back eqn 1, pyth.theor.
(sqrt6)^2 = (0.415)^2 + (2.415)^2
6=0.17+ 5.83
6m = 6m
Thank you,
Jojo

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