CPhill

OK.....I'll have to think a bit about that....!!!!!!

Angle ABC  = Angle ACB...so....AB = AC.....so 

4x + 4  = 6x - 14         add 14 to both sides, subtract 4x from both sides

18  = 2x    ⇒    x  = 9

So

BC  =   2x + 7   =    2*9 + 7  =  25

This triangle is isosceles   with  AB  = AC  ....so....the angles opposite these sides are also =

So  angle   (5x)°  = 40...so these two angles sum to 80°

So...this means that    2y + 20  must = 100

So

2y =  80

So....y  =  40

x + 2x + 3x  =  180

6x  =  180  divide both sides by 6

x = 30°

And   2x  =  60°  = angle C

If these are corresponding angles, the lines will be parallel.....and corresponding angles are equal.....so.....

3x + 5  =  65     subtract 5 from each side

3x  = 60   divide both sides by 3

x =  20

(12 * 7)  ( 4 + 1)  =

84   *   5   =

420    ????

Mmmmm....I'll try rosala ....!!!

Let's suppose that we have a partial series like this  where     a  = the first term   and d is the common difference between terms

a  + (a + d)   + (a + 2d)  + (a + 3d) + (a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)   + (a + 8d)

Let  n  be the number of even place terms

And    n + 1  the number of odd place terms

Then the sum of the even place terms is :

n *a   +   n^2* d     =  n (a + n*d)    

And the sum of  the odd place terms is : 

(n + 1) * a  + ( n+1) (n)* d   =  (n + 1) (a + n*d)     for  n ≥ 0  

So....the ratio of the sum of the odd place terms to the even place terms is :

(n + 1) (a + n*d)               ( n + 1)

_____________   =         _______

   n    (a + n*d)                       n

V  =  (1/3) (17.2)^2 * 24.4   ≈  2406.17 cm^3

Thanks, sm!!!!.... we could use Heron's here, but here's a slightly easier approach...IMHO...!!!

Draw an altitude of the triangle......and the altitude of an isosceles triangle will divide the base in half

So....we will have two right triangles 

Looking at either of these.....they will have a hypotenuse of 13cm and one leg = 1/2 of the base   = 5cm

And the other leg will  be   sqrt (13^2 - 5^2)  = sqrt (144)  = 12cm  = the height (altitude) of the triangle

So....the area  is  

1 / 2  base  * height  =

5 cm  *  12 cm  =

60 cm^2