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Thanks. It is spot on

                               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

https://web2.0calc.com/questions/polygon_72

https://web2.0calc.com/questions/circles_146

                   Base_upper + Base_lower   

Median  =  ––––––––––––––––––––   

                                 2    

                   12 + (12 + 16)          12 + 28   

Median  =  ––––––––––––––  =  ––––––––   

                               2                         2   

Median  =  20   

The length of the median of a trapezoid is    

the average of the lengths of the two bases.    

The area has nothing to do with this problem.    

_.

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

3 - 3 - 1 - 1

3 - 1 - 1 - 1 - 1 - 1

1 -1- 1 -1- 1 -1 - 1 -1

PQ  = sqrt [ 2^2 + 3^2 ] = sqrt [ 13]

AQ  = sqrt [1^2 + 4^2 ] sqrt [17]

PA  = sqrt  [2^2 + 4^2] = sqrt [ 20]  

Law of Cosines

PQ^2  = AQ^2 + PA^2  - 2 ( AQ * PA) (cos PAQ)

13 = 17 + 20 - 2( sqrt 340) * cos (PAQ)

[ 13 - 17 -20 ] / [ -2sqrt 340 ]  = cos (PAQ) = 6/sqrt 85

sin PAQ =   sqrt [ 85 - 6^2 ] / sqrt 85  = sqrt [ 49] / sqrt 85  =  7 / sqrt 85

ABC  ≈ XYZ

BC / AB = YZ / XY

8 / 14  = YZ / 18

4/ 7   = YZ / 18

YZ =  [ 18 / 4 ] / 7  =  32 / 7

That problem has already been solved here: https://web2.0calc.com/questions/please-help-asap_39806

THANK YOU

Hey...you got it  !!!!

Here's the same problem with a side of 12

https://web2.0calc.com/questions/square_34   (proyaop's answer )

See if you can follow the reasoning with your problem

Hint :  I believe that your "PR"   will be  (10/12)  of what was found in this link 

So...is it 20/(sqrt(2))?

When we have this situation..

Area =   (1/2) ( product of sides between the included angle) * sine (included angle)

See what you come up with....if you get stuck....let me know

British Flag Theorem

AX^2  + CX^2 = BX^2  + DX^2

Can you take it  from  here ???

Using the square of the distance formula = r^2....let the center = (x,y)

So we have

(x -22)^2  + (y-15)^2 = (x -22)^2 + (y + 29)^2   simplify as

(y -15)^2  = (  y - 29)^2

y^2 -30y + 225  =  y^2 -58y + 841

28y =  616

y = 616 / 28 =  22

So...using  (22,15)  and (-25,0)   we can  find x  as

(x -22)^2  + ( 22-15)^2  = (x + 25)^2  + ( 22 - 0)^2

(x -22)^2  + 49  = ( x + 25)^2  + 484

x^2 - 44x + 484  + 49  =  x^2 + 50x + 625 + 484

-44x  + 49  =  50x + 625

- 94x  =  576

x = 576 / -94 = -288/47

Coordinates are  ( -288/47 , 22)

I solved it.

Let x represent BX.

15^2 + x^2 (AB) + 6^2 + x^2 (BC) = 21^2 (AC)

Solving this equation gives us x = 3sqrt(10)

Therefore, BX = 3sqrt(10)

Find the area using Heron's Formula

A =  sqrt [ 18.5 (3.5) (6.5) (8.5) ]    ≈  59.8

The shortest altitude is drawn to the longest side

59.8 =  (1/2) (15) (altitude)

59.8 / [ 15/2 ] =  altitude   ≈  7.97  

         B

     6           8

A                  C 30

We have a SSA  situation......there may be 1 triangle formed, 2 triangles formed or no triangles formed

Law of Sines

sin A / 8  =  sin 30 / 6

sin A / 8 = (1/2) / 6

sinA = (8/6) (1/2) = 2/3

arcsin (2/3) ≈  41.81°    or    180 - 41.81  = 138.19

So   angle B  =  180  - 30  - 41.81   =  108.19°

or    angle B =  180 - 30 - 138.19  =    11.81°

If B =  108.19°   area of ABC =  (1/2) (8)(6)sin (108.19°)     ≈  22.8

If B  =  11.81°  area of ABC  = (1/2) (8)(6) sin ( 11.81°)  ≈  4.9

Hey, I just wanna know why are you using the same pfp and name as me...

    B

30    30

90C      D         A30

            12

BC = 12/sqrt 3  =  4sqrt 3

BA = 24/sqrt 3  = 8 sqrt 3

Since BD is  a bisector

BC / BA =   DC / DA

[4sqrt 3]/ [8/sqrt 3]  = DC /DA

4/8  =  DC/ DA

1/2 = DC / DA

Therefore   DA  =   2 / [1 + 2 ] AC   =  (2/3)AC = (2/3) 12 = 8

[ABD] = (1/2) (BA) ( DA)  sin (30)  =    (1/2) (8sqrt 3)(8) (1/2)  =  16 sqrt 3