A fold is formed on a 20 cm × 30 cm rectangular sheet of paper running from the short side to the long side by placing a corner over the long side. Find the minimum possible length of the fold?
1 Answer
L = 15sqrt(3) ~~ 25.98 \ cm
Explanation:
Let us set up the following variables:
{ (L, "Length of the fold", cm), (x, "DE", cm), (theta, "Angle " hat(DGE), "radians") :}
We are given that
By trigonometry for
sin hat(DGE) = (DE)/(EG) => sin theta = x/L ..... [A]
When folded the point
hat(HFG) = hat (FGD) = 2theta
Now:
hat(HFG) + hat(GFE) + hat(EFC) = pi
:. 2theta + pi/2 + hat(EFC) = pi
:. hat(EFC) = pi/2 - 2theta
By trigonometry for
sin hat(EFC) = (EC)/(EF )
:. sin (pi/2 - 2theta) = (CD-DE)/(EF )
:. sin (pi/2 - 2theta) = (20-x)/(x)
Using the sum of angle formula:
sin(A-B) -= sinAcosB - cosAsinB
we have:
sin(pi/2)cos(2theta) - cos(pi/2)sin(2theta) = (20-x)/(x)
:. cos(2theta) = 20/x - 1
Using the identity:
cos 2A -= 1-2sin^2A
we have:
1-2sin^2theta = 20/x - 1
Using
\ \ \ \ \ 1-2(x/L)^2 = 20/x - 1
:. 2 -2 x^2/L^2 = 20/x
:. x^2/L^2 = 1-10/x
:. x^2/L^2 = (x-10)/x
:. L^2/x^2 = x/(x-10)
:. L^2 = x^3/(x-10)
As
Differentiating wrt
d/dx L^2 = { (x-10)(d/dx(x^3)) - (d/dx(x-10))(x^3) } / (x-10)^2
\ \ \ \ \ \ \ \ \ \ = { (x-10)(3x^2) - (1)(x^3) } / (x-10)^2
\ \ \ \ \ \ \ \ \ \ = x^2{ 3(x-10) - x } / (x-10)^2
\ \ \ \ \ \ \ \ \ \ = x^2{ 3x-30 - x } / (x-10)^2
\ \ \ \ \ \ \ \ \ \ = x^2{ 2x-30 } / (x-10)^2
At a critical point this derivative vanishes, and so:
x^2{ 2x-30 } / (x-10)^2 = 0
x^2(2x-30 ) = 0
x=0,15
L^2 = 15^3/(15-10)
\ \ \ \ = 3375/5
\ \ \ \ = 675
:. L = 15sqrt(3) ~~ 25.98 \ cm