Saturday, June 26, 2010

Problem 465: Square, Arc, Angle, 90 Degrees, Measurement

Geometry Problem
Click the figure below to see the complete problem 465 about Square, Arc, Angle, 90 Degrees, Measurement.

Complete Problem 465

Level: High School, SAT Prep, College geometry

1. Let /_EAF = α, /_FAD = θ & square side = s.
We've cos(α+θ)=cos(θ)cos(α) - sin(α)sin(θ) or a/2s=acos(θ)-bsin(θ)=acos(θ)-b(1-(cos(θ))^2)^(1/2). Solve for cos(θ) to obtain, cos(θ)=[a^2+b(3a^2+4b^2)^(1/2)]/2s^2 noting that s^2=a^2+b^2 and rejecting the negative root since θ < 90 deg.
Now from Tr. FAD, we've: x^2=2s^2-2s^2*cos(θ).
Insert the value of cos(θ) above to obtain:
x^2 = a^2 + 2b^2 - b(3a^2+4b^2)^(1/2)
Hence etc.
Ajit

2. Typo correction:
a/2s=[acos(θ)-bsin(θ)]/s or a/2=acos(θ)-b(1-(cos(θ))^2)^(1/2) which can be solved for cos(θ)
Ajit

3. Another way to do this would be to extend AD to G such that DG=AD. G, F & E will be collinear with AE^2+GE^2=AG^2 or a^2+(GF+b)^2 =4(a^2+b^2) giving us GF=-b+(3a^2+4b^2)^(1/2). Now apply Stewart's Theorem to triangle AFG to obtain:
(a^2+b^2)+(-b+(3a^2+4b^2)^(1/2))^2=2(a^2+b^2+x^2) which when simplified gives the desired result.
Ajit

4. clearly AF=AD=(a^2+b^2)^(1/2)...join DE and draw DG perpendicular to AE...<GDE=90-<GED=<DEF=<DAF...sin(DAF)=sin(GED)=(a/2DE)...but DE=AD=AF...so x=2AFsin(DAF/2)...get the result...