CPhill

S1  = 5

S2 =  15

Then   

a1 =  5     and   a2  = 10         d =  5

So

a15 =  a1   + (15 - 1) (d)  =    5 + (15 - 1) (5)  =  5 + 14  * 5  =  75

S15  =[ a1 + a15 ] ( 15 / 2)  =    [  5 + 75 ] (15 / 2)  =     600

The  slant height of the cone =  21

The circumference of the  cone  =  (120 / 360)  * 2 pi * 21  =  14 pi

The  radius of the cone can  be  found as

14 pi  = 2 pi * r

r = 7  

The height of the cone =  sqrt [ l^2 - r^2 ]  = sqrt [21^2 - 7^2 ]

Volume of the  cone  is

(1/3) pi *  r^2 * h   =

(1/3) pi  * (7)^2 * sqrt [ 21^2 - 7^2 ]  ≈  1015.94

I don't think this is possible  if AB  =  10   ( it must be greater)

If we can....let us imagine this  figure ....  { AB  is  actually ≈ 13.45 }

Let DC = DE =  4

Let AB  = "10"

Let AE =  "x"

And BE  =  BC  =   "10 - x"

And AD  =   sqrt [ DE^2 + AE^2]  = sqrt [ 4^2 + x^2 ] 

Note that ...Triangle ADE ≈  Triangle ABC

So

DE / AD  =  BC / AB

4 / sqrt [ 4^2 + x^2  ] =  (10 - x )  /  10        

The only real solution to  this  is  x =   0  =   AE    which is  impossible

Angle  ABC  =105

Let AD = x    CD = 10 - x

BC / sin 30  = AC /sin 105

BC /sin30 = 10 /sin105

BC = 5/sin105

BA / sin 45 =  AC / sin 105

BA / sin45 = 10 /sin105

BA = (10/sqrt 2) / sin 105  =   20 / (1 + sqrt 3)

Since BD is a  bisector, then

BA / AD = BC / CD

(10/sqrt2)/ sin 105 / x  = 5/sin105 / (10-x)

(10/sqrt 2) (10-x) / sin 105  =  5x / sin105

(10/sqrt 2) (10 -x)  = 5x

100/sqrt 2  - (10/sqrt 2)x = 5x

100/sqrt 2  = (5 + 10/sqrt 2) x

x = (100/sqrt 2) / ( 5 + 10/sqrt 2)  =   (100) / (5sqrt 2 + 10)  =  20 / (2 + sqrt 2) = AD

[ABD ] = (1/2)BA * AD  * sin 30  =  (1/4) [20 / (1 + sqrt 3)] * [ 20 /(2 + sqrt 2) ]  ≈ 10.72

https://web2.0calc.com/questions/triangles_143

It's still a pretty good problem.....!!!

I think it must be  3x^3 - 20x^2 + kx + 12   ????

3 is one  root

3  [  3     -  20           k        12    ]

                   9        -33       3k - 99

    ____________________________

       3       -11      k - 33     3k - 87

3k - 87 =  0

3k =  87

k = 29

3  [  3      -20         29       12  ]   

                  9         -33     -12

   _________________________

      3        -11         -4         0

The remaining polynomial is

3x^2  -11x  - 4   = 0

(3x  + 1) ( x - 4)  = 0

The other two roots   are   -1/3   and   4

*****

a - b =  (-1 -3 , -4  -2)  =  (-4, -6)

https://web2.0calc.com/questions/geometry_7281

https://web2.0calc.com/questions/geometry_63597

https://web2.0calc.com/questions/trapezoid_139

                      A

                                      8

B        2          K            3            C

Because AK is an altitude

AK   =  sqrt [ AC^2 - CK^2 ]  = sqrt [ 8^2 - 3^2 ] = sqrt [ 55 ]

AB =  sqrt [ AK^2 + BK^2 ] =  sqrt [ 55 + 4 ] =  sqrt [ 59 ]

(r-5)^2  + (r- 5)^2  = r^2

2 [ r^2 - 10r + 25 ]  =  r^2

2r^2 - 20r + 50  = r^2

r^2 - 20r  =   - 50      complete the square on r

r^2 - 20r + 100  =  -50 +100

(r - 10^2  =  50          take the positive root

r  - 10 = sqrt 50

r -10  =  5sqrt 2

r =  5sqrt 2 + 10  ≈  17.07

G

2       8

F           E

FE  = sqrt [ 8^2 - 2^2]  = sqrt 60  =  2 sqrt 15

sin G =  FE / EG =    2sqrt (15) / 8  =  sqrt (15)  / 4

sec G  =  GE / FG  =  8 / 2  =   4

tan E  =  FG / FE  =  2  /  [ 2sqrt 15]  =  1 /sqrt 15 =  sqrt (15)  / 15

A

N

C      M       B

AC^2  +  (BC/2)^2  = 1

(AC/2)^2 + BC^2 = 1           simplify this system

AC^2 + BC^2 / 4  =  1

AC^2/4 + BC^2 = 1

4AC^2 + BC^2  = 4

-AC^2/4 - BC^2  =  -1       add these

(15/4)AC^2  =  3

AC^2  = 4/5

And

AC^2 + (BC/2)^2  = 1

4/5 + BC^2 / 4  =  1

BC^2 / 4  = 1/5

BC^2  = 4/5

Hypotenuse length =  sqrt [ AC^2 + BC^2 ] =   sqrt [ 4/5 + 4/5 ] =  sqrt (8/5) = 2sqrt (2) / sqrt (5)  =

2 sqrt (10) /  5

British Flag Theorem

AP^2  + CP^2 = BP^2 + DP^2

3^2  + 3^2  = 3^2  +  DP^2

9  = DP^2

DP  =  3