## Thursday, May 20, 2010

### Problem 458: Square, Semicircle, Circular Sector, Internal Common Tangent, Measurement

Geometry Problem
Click the figure below to see the complete problem 458 about Square, Semicircle, Circular Sector, Internal Common Tangent, Measurement.

Complete Problem 458

Level: High School, SAT Prep, College geometry

1. Let M be the midpt. of AB. Draw GN perpendicular to AD. AMEF is concyclic. Ang. AME = Ang. EFN or GN/FN=AD/AM or FN = a/2 and FG^2 = FN^2 + GN^2 = a^2 + a^2/4 = 5a^2/4 or FG = √5(a/2)
Ajit

2. draw GH//CD
▲FED ~ ▲AOD ( 90° & common ADO)
=> AOD = GFH
=>▲AOD = ▲FGH ( right tr, GH = AD, AOD = GFH )
=> FH = a/2
=> FG² = a² + a²/4

FG = a√5 /2
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3. Shift FG all the way such that G coincides with C. Since this is only translation distance is preserved.

Suppose the semicircle has center O. Draw OD.

I now claim that Triangle F'CD is congruent to triangle ODA.

Proof: Angle AOD is congruent to angle ODC since AB and CD are parallel. Angle A and angle FDC are right angles, therefore they are also congruent. AD is congruent to CD by properties of squares. Therefore ASA will show that the two triangles are congruent.

This then implies FG = OD.

OD = a (sqrt 5)/2 as can be easily shown with Pythagorean Theorem. Therefore FG = a (sqrt 5)/2

4. Let say the centre of cir.AEB as O.
∴ OD= a/2 √5 (Pythagoras theorem)