A rectangle ABCD which measures 18 by 24 units is folded once, perpendicular to
diagonal AC, so that the opposite vertices A and C coincide. Find the length of the fold.
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
AC=AB2+BC2AC=182+242=30, then OC=30/2=15AC = \sqrt{AB^2 + BC^2}\\ AC = \sqrt{18^2 + 24^2} = 30,\:then\:OC = 30/2 = 15AC=AB2+BC2AC=182+242=30,thenOC=30/2=15
From the triagle ABC
∠BCA=θtanθ=ABBC=1824=34\angle BCA = \theta\\ tan\theta = \frac{AB}{BC} = \frac{18}{24} = \frac{3}{4}∠BCA=θtanθ=BCAB=2418=43
From the triagle MCO
∠MCO=θtanθ=OMOC=x15,thenx15=344x=15×3x=45/4=11.25, thenMN=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=θtanθ=OCOM=15x,then15x=434x=15×3x=45/4=11.25,thenMN=2×x=2×11.25=22.5
Answer:
Lenght of fold = 22.5
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