Question 142744
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