Challenging Geometry Puzzle: Problem 1593. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
Gain comprehensive insights! Click below to reveal the complete details.
Click for additional details.
Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.
Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.
< ABD = 90 - 2@ and < BAD = 2@
ReplyDeleteSo <BAD = 2@ and < AFB = @ and < AEB = 90 - @
Hence AF = AB = AE = 5 and <FAB = 90
So AB^2 = 10^2 - 4^2 = 84
and AB = V84 = 2V21
Sumith Peiris
Moratuwa
Sri Lanka
Note that triangle EBF is a right triangle
ReplyDeleteand ABE is isoceles( both BED and ABE complement to alpha)
so AB=AE=5 and A is midpoint of EF
so EF= 10
and BF^2= EF^2-BE^2= 84
BF= 3.sqrt(21)
on simple angle chasing, we can notice that triangle ABF is an isosceles triangle. Let's consider BD = x, and then apply sine rule on triangles abd and triangles bde.
ReplyDeletethen we get x/sin 2theta = 5/sin 90 and x/sin(90-theta) = 4/ sin 90
after that, we can easily find out the value of sin theta and hence the value of cos theta using standard formulas. then we can use cosine rule.
cos theta = (5^2 + BF^2 - 5^2)/ (2*5*BF)=root(21/25)
on solving this we get BF = root(84)