Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Problem submitted by Kadir Latintas, Math teacher in Emirdag, Turkey
Click the figure below to enlarge the figure.
Wednesday, October 22, 2014
Geometry Problem 1051: Triangle, Square, Semicircle, Arc, Area
Subscribe to:
Post Comments (Atom)
Assume square ABCD defines a square grid with A as origin and D = (1,0). Add two more grids: a blue grid such that AD = blue(2,-1) and a brown one where AD = brown(4,1). See http://bleaug.free.fr/gogeometry/1051.png
ReplyDeleteIn this notation E=brown(0,2) and F=blue(1,1). With grid calculation, we see that EF slope = brown(1,3) = blue(5,3).
Let AH1=brown(1,3) // EF therefore area(EFH1)=S1. Since EH1=brown(1,1) and EC=brown(3,3) thus S3=3*S1
Let FB'=blue(0,-1) therefore area(EFB')=S2. H2 on EC with B'H2//EF implies EH2=brown(2.5, 2.5) thus S2=2.5*S1
=> S1+S2+S3=6.5*S1
Take E'F' symmetric of EF relative square centre. See http://bleaug.free.fr/gogeometry/1051b.png
Since S1, S2, S3 share the same base, sum of their areas = base*sum of heights. From picture, we see that S1+S2+S3=S4
tan(ADE/2)=1/4 and tan(CDF/2)=1/2. By composing tangents we get tan(FDE)=13/84 => sin(FDE)=13/85
S4=S/2*sin(FDE)=13/170*S
S=85*S1=34*S2
bleaug
Assume A:(0,0), B:(0,4), C:(4,4) & D:(4,0) i.e a square of side 4 units.
ReplyDeleteSo E is the intersection of:(x-4)² + y²=16 & x²+(y-1)²=1 and thus E is (8/17.32/17).
Similarly, F is the intersection of: (x-4)² + y²= 16 and (x-2)²+(y-4)²=4 and thus F is (4/5.12/5)
Triangle Area: A=(1/2)[x1(y2- y3)+x2(y3- y1)+x3(y1- y2)]
(x1,y1)n can taken as the origin (0,0) while all other necessary points are known. Thus we can calculate S1 = 16/85, S2=40/85 & S3 = 48/85 while S4 = 104/85
It is easy to see that: S4 = S1 + S2 + S3 and that S = 85 * S1 = 34* S2.