Geometry Problem
Click the figure below to see the complete problem 462 about Square, Arcs, 90 Degrees, Circle, Tangent, Radius, Measurement.
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Complete Problem 462
Level: High School, SAT Prep, College geometry
Wednesday, May 26, 2010
Problem 462: Square, Arcs, 90 Degrees, Circle, Tangent, Radius, Measurement
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FE=x;AE=a-x;DE=a+x
ReplyDeletesin(FAE)=cos(EAD)=x/(a-x)
in triangle ADE
DE²=AE²+AD²-2AE.AD.cos(EAD)
(a+x)²=(a-x)²+a²-2ax
a²-6ax=0
x=a/6
.-.
AE= a-x DE=a+x. drop a perpendicular from E toAD meeting AD @ J
ReplyDelete(a-x)^2= EJ^2+x^2
(a+x)^2= EJ^2+(a-x)^2
subtract
(a-x)^2 -(a+x)^2 = X^2- (a-x)^2
Simplify
a^2-6ax=0
a^2 - 6ax =0
x= a/6