CPhill

Let TG be a median of triangle BMT

And triangle BMT is similar to triangle DET

So FT is a median of  triangle DET

Then the area of traingle FDT = (1/2) [ DET ]  = 7

And triangle AED  similar to triangle ABM

So ED  =   1/2  BM

And FD  = 1/2 GM

So the height of triangle FDT  = 1/2 height of triangle GTM

So FT / FG =  FT / [ FT + 2FT] = 1/3

So FT = (1/3)FG 

FG = 3FT

And FG = AF

So AF = 3 FT

So  the height of triangle AFD  = 3 times the height of triangle FDT

And  since they're on the same  base (ED)..... the area of triangle AFD =  3 [ FDT]  =  3 * 7 = 21

[ ADT ] = [ AFD ] + [ FDT ] =  21 + 7  =  28 

x + y  = 10   square both sides

x^2 + y^2 + 2xy = 100

x^2 + y^2  = 100 - 2xy

x^3 + y^3 =  300

(x + y) ( x^2 + y^2 - xy)  = 300

(10) ( 100 - 3xy) = 300

100 - 3xy = 30

100 - 30 =3xy

70 / 3  = xy

y = 70 / (3x)

x + y  =10

x + 70 / (3x) = 10     multiply through by x

x^2 + (70/3) = 10x

x^2 -10x + 70/3 = 0

3x^2 - 30x + 70  = 0

x =  [30 +/- sqrt [ 900 - 840 ] ] / 6  =  [ 30 +/- sqrt 60 ] / 6  =  5 + sqrt (15) / 3  or 5 - sqrt (15) / 3

(x,y)  = ( 5 + sqrt (15) / 3 , 5 -sqrt (15) / 3 )   or (5 - sqrt (15) / 3  , 5 + sqrt (15) / 3 )

https://web2.0calc.com/questions/help-geometry_23995

A line through (-5.6)  perpendicular to the given  equation has the equation

y = -(1/3) (x + 5) + 6  →  -(1/3)x + 13/3

Set this equal to the given equation  to find the x intersection of these lines

3x - 2 = -(1/3)x + 13/3

(10/3)x = 19/3

x = 1.9

y = 3(1.9) - 2 = 3.7

The reflected point B  is

( 1.9 + (1.9 -  - 5) , 3.7 + (3.7 - 6) =  (1.9 + 6.9 , 3.7 - 2.3)  = (8.8 , 1.4)

Area =  pi r^2 * (83/360)

60 pi = pi* r^2 * (83/360)

r^2  = (60 * 360) / 83

r = sqrt [ 60 * 360 / 83 ]  ≈ 16.13

https://web2.0calc.com/questions/square_38

We already know the three sides of PQR.....the other info is extraneous

semi-perimeter   =  [ 36 + 22 + 26 ] / 2 =  42

Heron's Formula

[ PQR ]  = sqrt [ 42 * (42 - 36) * (42 - 26) (42 - 22) ]    ≈ 284

(120 + x) / (60 + x)  = (140 + x) / (120 + x)

(120 + x) (120 + x) = (60 + x) (140 + x)

14400 + 240x + x^2  =  8400 + 200x + x^2

14400 - 8400  = 200x - 240x

6000 = - 40x

-6000/40 = x =  -150

The ratio is  [ 120 - 150 ] / [ 60 -150 ] =  [ -30] / -90 ] =   1 / 3

https://web2.0calc.com/questions/help-geometry_23995

https://web2.0calc.com/questions/geometry_50437