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

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

Click the figure below to view more details of problem 1194.

## Wednesday, March 2, 2016

### Geometry Problem 1194: Equilateral Triangle, Perpendicular, Distance, Side, Metric Relations

Labels:
distance,
equilateral,
metric relations,
perpendicular,
triangle

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Draw MN middle line of ADEC (= 12)Apply theorem of pythagore MNB, ADB, BEC

ReplyDeletewe get x=14

Draw CF ⊥ DA ..( F on DA)

ReplyDeleteLet z= x^2

Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get

BE^2=x^2-121=z-121

BD^2=z-169

AC^2=CF^2+AF^2 => (BE+BD)^2+4=z

Or 2.BE.BD=z-BE^2-BD^2-4= 286-z

4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2

Simplify this equation we get 3.z^2-588.z= 0

We get 2 solutions z= 0 and z= 196

Only z=196 is acceptable so x= 14

See below for geometry solution of problem 1194

Deletehttp://s30.postimg.org/urgsbgadt/pro_1194.png

Draw circles diameters BA and BC

Draw altitudes BG of tri. ABC and GH of triangle DGE

Observe that H is the midpoint of DE => DGE is isosceles tri.

∡ (DGE)= ∡ (DGB)+ ∡ (BGE)= ∡ (DAB)+ ∡ (BCE)= 180-120= 60 degrees

So DGE is an equilateral tri.

HG= ½( DA+CE)= 12= DE. √ (3)/2

So DE= 8. √3=CF

In right triangle ACF we have x^2= AF^2+CF^2= 4+(8√3)^2=196 => x= 14

DB + BE = √(x²-（13-11）²)

ReplyDelete√(x²-13²) + √(x²-11²) = √(x²-（13-11）²)

x = 14

Sumith Peiris has left a new comment on your post "Geometry Problem 1194:

ReplyDeleteEquilateral Triangle, Perpe...":

Applying Pythagoras 3 times we see

DE = sqrt(x^2-13^2) + sqrt(x^2-11^2)=sqrt(x^2-2^2)

One can solve this equation by algebra but the equation becomes somewhat

unwieldy

By trial and error x = 14 (check: sqrt27 + sqrt75 = sqrt(27 + 75 +

2Xsqrt(27X75)= sqrt(102 + 2 X 45)=sqrt192