Geometry Problem 1494. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

## Tuesday, November 2, 2021

### Geometry Problem 1494: Parallelogram, Midpoints, Octagon, Areas

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https://photos.app.goo.gl/Rp4TcgH5ZYPfWj878

ReplyDeletewill u explain your proof in detail? How u arrived at "a" "p" "x" areas etc?

Deletehttps://photos.app.goo.gl/tZ4btHGLpeq1AY1D7

ReplyDeleteLet AD=2a and AB=2b

Define points R,S,T,U,V,X,Y,W,A1,B1,C1 and D1 as shown in the sketch

Note that R,T,V and V are the midpoints of OM,ON,OP and OQ

Note that E, F, G, H, I, J, K,L are midpoints of AG,LB,BI,....

Ratios of similar triangles give GI=2.IP ; HF=2FM ; KI=2KQ; JH=2HN….

Ratios of similar triangles give XK/KD=XQ/CD= ¼ => LK= ⅖. a

Similarly IJ= ⅖ .b

Ratios of similar triangles give PP1/PO=⅖=> PP1=⅖.a and P1D1= ⅗ .b ; OP1=⅗. a

Let k=sin(AD,AB)=sin(A1L,A1E)= sin(D1J, D1K)

We have Area of [ABCD]= k. AB. AD= 4k.a.b

Area of [A1B1C1D1]= k. A1B1. A1D1= 36/25 . k.a.b

Area of [A1EL]+ [KD1J]+[IC1H]+[B1FG]= 8/25. k.a.b

So area of [EFGHIJKL]= 28/25 . k. a.b

And ratio of [EFGHIJKL]/[ABCD]= 7/25

http://www.xente.mundo-r.com/ilarrosa/GeoGebra/GoGeometry1494.gif

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