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

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

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

[jimlocke]

We note that tan(α) = OC/AC = 2/4 = 1/2. Applying the tangent double angle formula, tan(2α) = (2tan(α))/(1 - tan²(α)) = (2)(1/2)/(1 - (1/2)²) = 4/3. Using the designations for points given in the video at 1:35 and constructing line segments as needed, PE/AP = tan(2α) so PE/4 = 4/3 and PE = 16/3. So, by Pythagoras, AE = √(4)² + (16/3)²) = √((144/9) + (256/9)) = √(400/9) = 20/3. ΔAPE and ΔAQD are similar, so AD/AE = AQ/AP, 8/(20/3) = X/4 and X = 24/5 units, as GrayYeon Math also found.

[solimana-soli]

Tracing the perpendicular radius CH to line AQ we know that radius splits the chord in two equal parts, so we can calculate AH as:
AH = AC*cos 2 alpha
knowing that:
cos alpha = AC/AO = 4/2√ 5 then cos 2 alpha = 3/5
and AH = X/2, AC =4
x/2 = 4*3/5
X = 24/5

[rabotaakk-nwnm]

5:35 AE=4+2b-4=2b=20/3
ΔAPE~ΔAQD => AQ/AD=AP/AE
x/8=4/(20/3); x=8×4×3/20=24/5