# 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#