Proposed Problem

Click the figure below to see the complete problem 403 about Circular sector, 90 degrees, Circle, Semicircle, Area.

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Complete Problem 403

Level: High School, SAT Prep, College geometry

## Wednesday, December 16, 2009

### Problem 403. Circular sector, 90 degrees, Circle, Semicircle, Area

Labels:
90,
area,
circle,
circular sector,
semicircle,
triangle

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Let R radius OC and r radius of circle D.

ReplyDeleteLet x distance of D to segment OB. Then x^2 = (R+r)^2-(R-r)^2=4Rr.

Radius OB is equal to 2R, and then:

2R=r+sqrt(x^2+r^2) -> R=2r -> S = (pi/4)*R^2 (1)

A(AOB) = (pi/4)*(2R)^2 - (pi/2)*R^2 = (pi/2)*R^2 = S + S1 + S2 + S3 (2)

By eq (1) and (2), we've: S = A(AOB)/2 = S1 + S2 + S3

MIGUE

Let the radius of circle D be r, that of the semicircle be R so that the radius of the quarter circle is 2R

ReplyDeleteWriting the altitude of Tr. CDO in 2 ways we have

(2R-r)^2 - r^2 = (R+r)^2 - (R-r)^2 which yields upon simplification that R = 2r

So S1 + S2 +S3 = 1/4pi (2R)^2 -1/2piR^2 - pir^2 which when simplified using R = 2r is = to pir^2 = S

Sumith Peiris

Moratuwa

Sri Lanka