CPhill

Angle BDC  = 180 - 70 -80 = 30

Law of Sines

6 /sin 30  = DC /sin 80

DC= 6sin 80 / sin 30  =  6sin 80 / (1/2)  =12sin 80  cm

Area of triangle  BCD =  (1/2) ( 6) (12 sin 80) * sin 70  =  36 sin 80 * sin 70   ≈ 33.3  cm^2

Height of DBC   = Height of trapezoid

33. 3  = (1/2) (DC) * height  of trapezoid

66.6  = (12 sn 80) * h

66.6 / (12 sin 80)  ≈  5.64 cm

Area of trapezoid =  (1/2) h * ( sum of bases)  = (1/2) (5.64) ( 12sin 80  + 3)  ≈  41.79 cm^2

      A                 B

               16

                P

                25

D                                 C

The area =  81

To see why

Triangles PAB and PDC  are similar  because angle PDC = angle PBA   and angle PCD =angle PAB

Since the area of APB  =16  and  the area of  DPC = 25....the  scale factor is sqrt [ 16/25] =4/5 = 4 : 5

So  AB = (4/5)CD      and   the height of  APB = (4)/(4 + 5)*h =  (4/9) h   (where h is the height of  the trapezoid)

Let the base  of  DPC = b  = DC.....so the base of APB = (4/5)b = AB

Using triangle PAB

Area = (1/2) (AB) ( 4/9) h

16  = (1/2)(4/5)b * (4/9) h

32  = [ 16  / 45 ] bh

32 ( 45  /16)  = bh

90 = bh

b =  90/h = DC

Then the base  of PAB  = (4/5) (DC)    =   (4/5) (90/h)  =  72/h

So  the area  of the trapezoid is

(1/2) (h)  (AB + DC)   =

(1/2) (h) ( 72/h + 90/h)  =

(1/2) ( h) (72 + 90) / h  =

(1/2) (162)   =  

81

See EP's answer, here :

https://web2.0calc.com/questions/help-with-algebra_96

Simplify as

(a + 3) / 3 =    [ -a^2 + a + 1 ] /  a             cross-multiply

a ( a + 3)  =  3 (-a^2 + a + 1)

a^2 + 3a   = -3a^2 + 3a + 3

4a^2  = 3

a^2  = 3/4                 take both roots

a = sqrt (3)  / 2          or     a =   -sqrt (3)  / 2

I think you mean that EF =  3

Triangle  EFP has a base of 3 and a height of (1) / (1+2)  * 6   =  2

Its  area =  (1/2)(3)(2)  =  3  = [ EFP]

Triangle   PGH  has a base of 10  and a height of (2)/(1 + 2) * 6  = 4

Its area =  (1/2)(10)(4)  =  20 = [ PGH ] 

Area of trapezoid  = (1/2) height * (sum of the bases)  = (1/2) (6) (3 + 10)  = 39  = [ EFGH ] 

[ PFG ]  = [ EFGH ] - [ PGH ] - [EFP ]  =  39 - 20 - 3  =   16

                                A

                          7                 17

                   B                  M                 C

We can show that such a triangle  cannot exist

angles AMB and AMC are supplemental....so cos (AMB)  =  -cos (AMC) 

Since AM is a median,  then BM = CM  = x

Using the Law of Cosines

AC^2 = CM^2 + AM^2  - 2 (AM * MC)cos (AMC)

AB^2  = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC)

17^2  = x^2 + 12^2  - 2 (AM * x) cos (AMC)

7^2  = x^2 +  12^2  + 2(AM * x) cos (AMC)         add these

338 = 2x^2  + 288

50  = 2x^2

25  = x^2

5  = x

Then BM  + CM =  5 + 5   =  10  = BC

But, by the Triangle Inequality,   AB + BC    must be > AC =  17

However  AB + BC =   7 + 10  = 17

Which means  that no  such triangle  can exist

arc LN  =  3/ (3 + 4 +  5) * 360  =  (3/12) (360) =   90

arc KL = ( 5 / 12) *(360)  =   150

Angle LMK =  (1/2) ( arc KL - arc LN)  =  (1/2) (150 - 90 )  =   (1/2)(60)  =  30°

Assuming that arc AD = arc BC

Then   

2 arc BC  + arc AB  =  180

2arc BC +  80    =180

2 arc BC  =100

arc BC =100  / 2    = 50

f(g(x))  =  2(-3x + c) + 1 =  -6x + 2c + 1

g(f(x))  =  -3(2x + 1) + c  = -6x -3 + c

f(g(x))  = g(f(x))

-6x + 2c + 1  = -6x - 3 + c         simplify as

c = -4

OA  =  60 * 1/4  =  15