Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Saturday, March 30, 2019

### Geometry Problem 1430: Regular Octagon, Area, Equivalent Figures

Labels:
area,
equivalent,
geometry problem,
octagon,
regular polygon

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OE meet DF at P, extend DF, draw from G // OE, S intersection

ReplyDelete=> OPSG is a rectangle

good easy proof

Deletehttps://photos.app.goo.gl/xk498LjP5akrivYf8

ReplyDeleteDraw points L, K per sketch

Note that triangle OKD is a right isosceles triangle

And Triangle CLD congruent to DKE ( case ASA)

S(CKD)+ S(DKE)=1/2.DK(CL+KE)= ½.OK.(CO)= S( COK)

So red areas= yellow area

Let AE=x and AP=x1

ReplyDeleteArea of Yellow region = (x-2x1)*x/2 ----------(1)

Area of the Octagon = 4*S(ABH)+S(BDFH)= (x-2x1)x --------(2)

From (1) and (2) It can be observed that Area of yellow region is half of Area of Octagon

Q.E.D