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In-a-rectangle-ABCD-E-is-the-midpoint-of-AB-F-is-a-point-on-AC-such-that-BF-is-perpendicular-to-AC-and-FE-perpendicular-to-BD-Suppose-BC-8-3-Find-AB-




Question Number 20599 by Tinkutara last updated on 28/Aug/17
In a rectangle ABCD, E is the midpoint  of AB; F is a point on AC such that BF  is perpendicular to AC; and FE  perpendicular to BD. Suppose BC = 8(√3).  Find AB.
InarectangleABCD,EisthemidpointofAB;FisapointonACsuchthatBFisperpendiculartoAC;andFEperpendiculartoBD.SupposeBC=83.FindAB.
Answered by ajfour last updated on 29/Aug/17
Commented by Tinkutara last updated on 29/Aug/17
Thank you very much Sir!
ThankyouverymuchSir!
Commented by ajfour last updated on 29/Aug/17
EF=AE=BE (radii of circle  with E as centre)  so ∠AFE=θ  ∠DBF=90°−2θ , so ∠EBF=2θ  now ∠AFE+∠EFB=90°  so              θ+2θ=90°    ⇒      tan θ=((BC)/(AB)) =((8(√3))/(AB)) = (1/( (√3)))      or   AB=24 .
EF=AE=BE(radiiofcirclewithEascentre)soAFE=θDBF=90°2θ,soEBF=2θnowAFE+EFB=90°soθ+2θ=90°tanθ=BCAB=83AB=13orAB=24.

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