Find the length of the segment x #mathpuzzles #thinkoutsidethebox #geometryskills

Find the length x. #geometryskills #mathpuzzles #thinkoutsidethebox

We are required to find the length X.

This will be so much appreciated.

grayyeonmath
Important Geometry skills explained
Rectangles
circles
Right Triangle
similarity of triangles
ratio and proportion
ratio
geometry
Find the length X
quadratic equations
completing the squares,quadratic equations
algebra
maths
line segment
midsegment theorem
brian mclogan
area
find the length
satmath
midpoint
length
width
triangles
similar triangles
math
igko
pythagorean theorem problem solving,
pythagorean theorem word problems,
right triangles
mathematics
gre
premath
ixl
prmo

#grayyeonmath
#ImportantGeometryskillsexplained
#rectangles
#circles
#righttriangles
#similarityoftriangles
#triangles
#ratioandproportion
#proportional
#ratio
#geometryskills
#Findthelength
#quadraticequations
completing the squares,
#quadraticequation
#equations
#algebra,
#mathpuzzles
#maths
#linesegment
#midsegmenttheorem
#brianmclogan
#area
#findthelength
#trigonometricratios
#mathpuzzles
#midpoint
#length
#width
#triangles
#similartriangles
#triangles
#math
#igko
#pythagoreantheoremproblem solving,
#pythagorean theorem word problems,
#righttriangles
#triangles
#mathematics
#green
#premath
#satmath
#ixl
#prmo

[solimana-soli]

Once drawn point F, as GreyYon made, I found AF drawing a circle with center in B and radius =3 that intersects line DF in A and F, Being DE tangent, so we can apply the tangent secant theorem doing:
(9+AF) : x = x : 9
AF = (x^2 - 81)/9
Then applying Pythagorean theorem on DEF we can write:
X ^2 + 6^2 = (9 + (x^2-81)/9)^2
X^4 - 81X^2 - 2916 = 0
Setting X^2 = t
t^2 - 81t - 2916 = 0
t = 108=X^2
X = 6sqrt3

[rabotaakk-nwnm]

7:55 α=opp/hyp ???
11:00 C <--> D !!!

[pk]

The original diagram is misleading . The line segment AB should actually be drawn parallel to line segment DC . Interesting problem though . We initially don't know for sure what the two equal angles are . An interesting variation would be to change the length of line segment AB . I think I will try it with AB = 2 .

[thelightning]

Very nice solution!
There i have mine, line AE is perpendicular to DF, so 9*AF=AE^2 and AF^2+AE^2=6^2 there we have that 9AF+AF^2=36 => AF*(9+AF) =36 => AF=3 and beta is 60° because FE=2AF, at last X=6sqrt3, greetings from Paraguay!

[geometryexpressions]

A second (slightly degenerate) solution would be where ABC are collinear, then ACD is an isosceles triangle and x = 3*sqrt(5)