Geometry Problem.

Post your solution in the comments box below.

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

Click the figure below to see the complete problem 1000.

## Friday, April 4, 2014

### Geometry Problem 1000. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Distance, Equilateral, Congruence, Perpendicular, Metric Relations

Labels:
120,
angle,
congruence,
equilateral,
isosceles,
midpoint,
perpendicular,
scalene,
triangle

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CD=AE=2GH=8, GH=4, JH=2.

ReplyDeleteBy problem 998, FH=4.

Hence, x=√12=2√3.

http://s22.postimg.org/iff0mclox/problem_1000.png

ReplyDeleteLet P is the midpoint of DE

As the result of problem 998 we have

Triangle DBC congruent to ABE=> AF=DC=8

And triangle PFH and FHG are equilateral

Since G and H are midpoints of AC and EC => GH= 1/2AE=4

In equilateral triangle FGH altitude FJ= GH.sqrt(3)/2= 2.sqrt(3)

Following on from my proof of Problem 998, FGH is equilateral too and is of side 4 so

ReplyDeletex = sqrt(16 - 4) = 2sqrt3

Sumith Peiris

Moratuwa

Sri Lanka

Tr. ABE and CBD are congruent (SAS) => AE=8

ReplyDeleteSince G and H are mid-points of AC and EC => GH//AE and GH=4 -----(1)

Similarly GF//CD and GF=4 -----------------(2)

As ABD is isosceles=> m(BDF)=30 and Tr.DBF is 30-60-90 and BF=DB/2.

Similarly BH=EB/2

Tr. BFH similar to ABE with each side of BFH half of corresponding sides of ABE

Hence HF=AE/2=4 ---------(3)

From (1) ,(2) & (3) GFH is an equilateral triangle with each side = 4 units

=> FH=2Sqrt(3)