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#