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]

Very creative and clever! You avoided having to solve for the side length of the square and other intermediate values! Doing it "the hard way", drop a perpendicular from the triangle's right angle vertex to the bottom side of the square. It will also be perpendicular to the top side of the square and will divide the yellow triangle into 2 smaller right triangles, which can be proved to be similar to each other and to the yellow triangle. Let the bottom, shorter, side of the smaller triangle on the left have length y. Then, from similarity to the yellow triangle and ratios of corresponding sides, the longer side will have length 7y/5 and the other side of the other small triangle will have length (7/5)(7y/5) = 49y/25. The side of the square, call it s, will have length y + 49y/25 = 74y/25. The length of our dropped perpendicular is (7y/5) + 74y/25 = 109y/25. We now have a right triangle with hypotenuse X, one side with length y and other side the length of our dropped perpendicular, 109y/25. From Pythagoras, X² = y² + (109y/25)² = 12506y²/625. Also, applying Pythagoras to the yellow triangle, whose hypotenuse is a side of the square and, therefore, has length s, s² = 5² + 7² = 74 and s = √(74). However, we found s = 74y/25, so √(74) = 74y/25, y = 25/(√(74)) and y² = 625/74. Substitute for y² in X² = 12506y²/625: X² = (12506)(625/74)/625 = 12506/74 = 169. So, X = √(169) = 13, as GrayYeon Math also found.

[vierinkivi]

Hieno tehtävä. En keksinyt tätä helppoa tapaa.

[rabotaakk-nwnm]

a=5; b=7; c=v¯(a^2+b^2)=v¯74
cos(90°+β)=-sinβ=-b/c=-7/v¯74

[misterenter-izrz]

h=35/sqrt(74), h^2=(35^2/74), w^2=(25-(35^2/74)),