Geometry Problem

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

Click the figure below to see the complete problem 971.

## Wednesday, January 29, 2014

### Geometry Problem 971: Equilateral Triangle, Rectangle, Common Vertex, Sum of Right Triangles Areas

Labels:
area,
common vertex,
equilateral,
rectangle,
right triangle

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http://imagizer.imageshack.us/v2/800x600q90/835/8nxu.png

ReplyDeleteDraw rectangular FDNM as per sketch

Note that area ( EBC) = area(ENC)

And Area(EAF)= area(FME)

This problem becomes problem 969.

Let z(P) be the complex number representing P.

ReplyDeleteLet z(C)=0, z(B)=−a, z(E)=−a+bi.

Then

z(F) = (−a+bi)(1/2 + √3/2 i) = (−1/2 a − √3/2 b) + (−√3/2 a + 1/2 b)i

z(A) = −a + (−√3/2 a + 1/2 b)i

z(D) = (−√3/2 a + 1/2 b)i

Thus

BE = b, BC = a

AF = −1/2 a + √3/2 b, AE = √3/2 a + 1/2 b

DC = √3/2 a − 1/2 b, DF = 1/2 a + √3/2 b

Area of ΔEBC = 1/2 ab

Area of ΔFAE = 1/8 (−√3 a² + 2 ab + √3 b²)

Area of ΔFDC = 1/8 (√3 a² + 2 ab − √3 b²)

Hence,

Area of ΔEBC = Area of ΔFAE + Area of ΔFDC

Sir, can you give me the idea how to get Z(F) Z(B) and Z(D)

Deletethx for a lot

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