CPhill

13 x^2  + nx  - 17

We can factor  this  as

(13x  + 17) ( x - 1)   so   n  =  17 - 13  =  4

or

(13x - 1) ( x + 17)  so   n =  17*13 - 1 =  220

or

(x - 17) (13x + 1)    so n =  -17(13) + 1 =  -220

or

(x + 1) (13x - 17)    so   n =  -17 + 13  =   -4

The height of the trapezoid  can be found as

400 = (1/2) (20 + 30)(height)

height = 400 / 25  =   16

Triangles  AXB  and DXC    are similar

And   AB  : DC  =  2 : 3

Therefore   the    height of   triangle DXC  =  (3/5) 16  =  9.6

And the height of triangle  AXB  =  (2/5)16  = 6.4

So  the combined  areas of these triangles  = 

(1/2) ( DC * 9.6  + AB * 6.4)  = 

(1/2) ( 30 * 9.6  + 20 * 6.4)   =  208

And because of symmetry  triangles AXD and BXC  have the same areas

So,  triangle  BXC has an  area of  [ 400 - 208 ]  / 2 =  96

See the image below   (not to  scale)

The area of (1/2)  of the octagon =   (1/2) [ 2 * 12^2 ( 1 + sqrt 2)]  = 144 (1 + sqrt 2)

Triangle CDE  has an  area of   (1/2) CD * DE * sin 135°  =  (1/2) 12^2 sin 135° =   72 / sqrt 2 =  36sqrt (2)

So  the area of the trapezoid BCEF   =   

area of (1/2) octagon  -  area of triangle CDE

 144 ( 1 + sqrt 2)  - 36 sqrt 2  = 

[144 + 108sqrt 2] cm^2   ≈ 296.74 cm^2 

See the image below

Because  AB and DC are parallels cut by transversal AC, then  angles DAQ and NCQ  are equal alternate interior angles

And angle  PQD = angle CQN (vertical angles)

So...by AA congruency , triangle   AQD is similar to triangle CQN

But  NC =  (1/2) of  BC   so it also  equals (1/2)  of AD

Then NC :  AD  =  1 : 2

But this also means that  QN : QD = 1 : 2

So        QD  =     [ 2 / ( 1 + 2) ]  DN =    (2/3) (35) =    70 / 3

    D

                                 Q

                M

E                       P                                 F

Let M  be the intersection point of the two medians

By the property  of medians    DM  = (2/3)18 =  12     MQ =  (1/3) 28  =  28 / 3

So....since  DP and EQ  are perpendicular.....then     DQ =  sqrt [ 12^2  + (28/3)^2 ]  = 

sqrt [ 144  + 784/9 ] =   sqrt    [ 1296 + 784 ]  / 3  =  sqrt [ 2080 ] / 3  =     4sqrt (130) / 3

And DF =   2 DQ  =      2 * 4sqrt (130) / 3    =   (8/3)sqrt (130)

sin 60 =   altitude / side length

side length =  alltitude / sin 60 =   ( 3sqrt 3) /  [ sqrt 3 / 2 ]   =   6

The longer base is  found as

5  +  2a

Where a  =   6/2  = 3

So the longer base =   11

So  the perimeter =   5 + 11 + 2*6   =  28

See the image below :

If we draw  median  BO, this will split triangle ABC into  six smaller triangles that all have the same area

And triangle APC will be composed of  two of these

So  area of APC =   (2/6)Area of ABC   = (1/3) Area of ABC  =  (12 * 6)  / 2   * (1/3)    =  12

Here's a similar problem......https://web2.0calc.com/questions/polynomials_15

This is possible to solve by hand  .....   but.....it's a lot of work  .....  I'd recommend  using some technology..as I did in the question cited

  .

m = the slope =  [ -3 - 1 ]  / [ 3  - - 2 ]  =   -4 / 5

Using this and  (3  -3)   we have

y = (-4/5) ( x - 3)   - 3

y =(-4/5)x + 12/5  - 3

y =(-4/5)x + 12/5 - 15/5

y = (-4/5)x  -  3/5

If we draw a radius bisecting this  chord it will do so at right angles

The distance from the origin to this line is  given by

  l 3(0) - 1(0)  - 10 l    /  [sqrt 3^2 + 1^2 ]   =    10   / sqrt  10  =  sqrt 10

So    we can form  a right triangle   with  ......   r = the hypotenuse   and legs of 5  and sqrt (10)

r^2  =  5^2  + (sqrt 10)^2  =  25 + 10  =  35

Area of the circle =  pi * r^2  = pi * 35  =   35 pi  units^2