Find the area of the triangle#mathpuzzles #geometryskills #importantgeometryskillsexplained

Find the area of the triangle#mathpuzzles #geometryskills #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]

At about 6;13, the equation r² + 26r = 174 has been derived but is not solved for r. It takes some foresight to know that this equation does not need to be solved for r. I lacked the foresight and solved for r, as follows. I subtracted 174 from both sides so r² + 26r - 174 = 0 and used the quadratic equation. After discarding the negative value for r, I found r = √(343) - 13. AC = r + 6 = √(343) - 13 + 6 = √(343) - 7 and AB = r + 20 = √(343) - 13 + 20 = √(343) + 7. Using AB as the base and AC as the height, A = (1/2)(√(343) + 7)(√(343) - 7). I used the identity x² - y² = (x + y)(x - y). A = (1/2)((√343)² - (7)²) = (1/2)(343 - 49) = (1/2)(294) = 147 sq. units, as GrayYeon Math also found.

[RAG]

Neat ending with a fortuitous result from the quadratic for r. r = 7rt7 - 13 by the way.

[santiagoarosam]

6^2=CD(CD+9) ; CD=3
BE(BE+9)=20^2. ; BE=16
(6+r)^2+(20+r)^2=(3+9+16)^2. ; r=-13+7√7~5.52
Área ABCD =(6+r)(20+r)/2=147.
Saludos.

[StephenRayWesley]

(20)^2=400 (6)^2=36 (9)^2=81 {400+81}/6=481/6=√81 √9^√2 3^2^√2^1 3^2√1^√1 3^2 ((ABCDEFG+2ABCDEFG-3)