CPhill

https://web2.0calc.com/questions/parallelogram_45

A(-6,2)                             B   (12,2)

D(-6, -2)                          C  (12, -2)

AB  =  12 -  -6 =   18

BC =  2 - - 2 =  4

AB * BC =  18 * 4   =   72   = the area

Triangle BDH  similar to triangle BEC

BD/ DH  = BE  /EC

BD / 5  = BE / EC

BD * EC / 5    = BE

Triangle BEC similar to triangle ADC

BE / EC  = AD / DC

BE / EC  = 16 / DC

BE  / EC = 16/ BD

BE  = 16EC / BD

BD * EC  / 5 = 16 EC / BD

BD ^2  =  80

BD =  4 sqrt 5

[ABC ]  =  BD * AD  =   (4sqrt 5) * 16   =  64  sqrt 5   ≈ 143.1

CORRECTED  !!!

https://web2.0calc.com/questions/triangle_41241

https://web2.0calc.com/questions/triangles_142

Just a slight mistake by eramsby

The quadratic should be

26x^2 - 354x + 1204 = 0

The solutions  are  x  = 7    and x = 86/13

So y =  5(7) - 28  =  7

And  y = 5(86/13)  - 28  = 66/13

The points are  (7,7)  and (86/13 , 66/13)

Let the center of the circle  = O

Let the midpoint of the large semi-circle = F

Let the midpoint of the  the right semi-circle = D

Let the center of the semi-circle on the left = E

Form triangles EOF   and  DOF

DO =  (5 + r)

FO = (8 - r)

EO = (3 + r)\

DF = 3

EF = 5

By the Law of Cosines  we have two equations

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 - 2(DF * FO) cos ( DFO)

Because  angles EFO  and DFO  are supplementary   cos (DFO) = - cos (EFO)

So re-writing....we have

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 - 2(DF * FO) -cos ( EFO)

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 + 2(DF * FO) cos ( EFO)

(3+r)^2  = 5^2 +(8-r)^2  - 2 (5 * (8-r)) cos (EFO)

(5+ r)^2  = 3^2 + (8-r)^2  + 2( 3 (8-r)) cos EFO

cos EFO  =   [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ]

cos EFO  = [ (5 + r)^2 - 9 - (8-r)^2] / [ 2 *(24 - 3r) ]

Then

  [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ] =  [ (5 + r)^2 - 9 - (8-r)^2] / [2 *(24 - 3r) ] 

This is a little sticky to simplify, but WolframAlpha  helps 

We get that

272 / (r-8)  + 49 = 0

272 / (r -8)  = -49

272  = -49 ( r -8)

272 = -49r + 392

-120  =-49r

r =  120 / 49  ≈ 2.449

Let D= (0,0)   Let  B= (40,10)

Slope of line  through DB = ( 10 - 0 ) / (40 -0)  =  1/4

Equation of  this line

y =(1/4)x

Let E = (5,10)  Let F = (8,0)

Slope of line through EF  = (0 -10)/ (8-5)  = -10/3

Equation of this line

y = (-103)(x -8)

y = (-10/3)x + 80/3

To find the x coordinate of P

(1/4)x = (-10/3)x + 80/3

(1/4 + 10/3)X = 80/3

(3/12 + 40/12)x = 80/3

(43/12)x = 80/3

x = (80/3) (12/43)  =  320/43

y coordinate of P  =  (1/4) (320/43)  = 80/43

Height of triangle EBP  =  10 - 80/43 =  350/43

Base of EBP = 40 - 5 = 35

Area of EBP  = (1/2) (35) (350/43)  = 6125/43

[EBP ] / [ ABCD]  =  (6125/43) / [400]  = 245 / 688

                               P

             28                     Y     17              

Q                     X                      R

                  35

Because  PX is a bisector,  then

QX / PQ = RX / PR

PR / PQ = RX / QX

17/28  = RX / QX

So

RX = 17 / ( 17+ 28) * QR  =  (17 / 45) * 35  =  119/9

Law of Cosines

PQ^2  = QR^2 + PR^2  - 2(QR * PR) cos (PRQ)

28^2 = 35^2 + 17^2 - 2 (35 * 17) cos (PRQ)

cos (PRQ)  =  [ 28^2 - 35^2 - 17^2 ] /  [-2 * 35 * 17 ] = 73/119

sin (PRQ)  = sqrt [ 119^2 - 73^2 ] / 119  =    8sqrt (138) / 119

Law of Sines   (angle XYR = 90 )

XY /sin (PRQ)  = XR / sin(XYR)

XY / [8sqrt (138) / 119 ] =  (119/9) / sin 90        (sin 90  =  1)

XY  =  [ 8sqrt (138) / 119 ] ( [ 119/ 9]   =   (8/9)sqrt (138) ≈  10.44

Triangle HQN  is similar to triangle FQG        Triangle  HPG is similar to triangle FPM

HQ / NH  = FQ / GF                                              PH / HG = PF / FM                                        

HQ / 2  = FQ / 4                                                    PH / 4 = PF / 2

HQ / FQ = 2/4  = 1/2                                             PF / PH = 2/4 =1/2

HQ = (1/3)FH

PF  = (1/3) FH

PQ =  (FH - PF - HQ)  =  (FH - (1/3)FH - (1/3)FH)  =  (1/3)FH

So

PQ / FH  = 1/3