ABCD - rectangle
AB = CD = 18
BC = AD = 24
MN perpendicular of diagonal AC
AO = OC = AC/2
MO = ON = x
From the Pythagorean theorem AC2 = AB2 + BC2
A C = A B 2 + B C 2 A C = 1 8 2 + 2 4 2 = 30 , t h e n O C = 30 / 2 = 15 AC = \sqrt{AB^2 + BC^2}\\
AC = \sqrt{18^2 + 24^2} = 30,\:then\:OC = 30/2 = 15 A C = A B 2 + B C 2 A C = 1 8 2 + 2 4 2 = 30 , t h e n OC = 30/2 = 15
From the triagle ABC
∠ B C A = θ t a n θ = A B B C = 18 24 = 3 4 \angle BCA = \theta\\
tan\theta = \frac{AB}{BC} = \frac{18}{24} = \frac{3}{4} ∠ BC A = θ t an θ = BC A B = 24 18 = 4 3
From the triagle MCO
∠ M C O = θ t a n θ = O M O C = x 15 , t h e n x 15 = 3 4 4 x = 15 × 3 x = 45 / 4 = 11.25 , t h e n M N = 2 × x = 2 × 11.25 = 22.5 \angle MCO = \theta\\
tan\theta = \frac{OM}{OC} = \frac{x}{15}, then\\
\frac{x}{15} = \frac{3}{4}\\
4x = 15\times3\\
x = 45/4 = 11.25,\: then\\
MN = 2\times x = 2\times11.25 = 22.5 ∠ MCO = θ t an θ = OC OM = 15 x , t h e n 15 x = 4 3 4 x = 15 × 3 x = 45/4 = 11.25 , t h e n MN = 2 × x = 2 × 11.25 = 22.5
Answer:
Lenght of fold = 22.5
Comments