CPhill

Yep......I used fiora's  approach  [more or less ].......but.....I like the way heureka did it, too.....

LOL!!!!!......nah.......it's just something I thought up that might make a nice problem.......

Let me add that

10^(1834097.297)  =

10^(1834097)  * 10^(.297)  =

10^(183407) * 1.9815270258050982 =

About   19815270258050982  followed by 183391 zeroes

Yep....correct....nice work, fiora and heureka.....!!!!!

2x-4=5x+8   subtract 2x and 8 from each side

-12  = 3x      divide both sides by 3

-4 = x

a= {1,2,3,4,5,6,7,8} and b= {2,4,6,8,10,12}

The intersection is just the set of elements that are common to both sets....so we have.....

{2, 4, 6, 8 }

                                    a^n     - 9

a^n + 8        a^(2n)    - a^n     - 6

                    a^(2n)   +8a^n

                    --------------------------

                                -9a^n       - 6

                                -9a^n      -72

                               -----------------

                                                 66

So....the result is     ...    ( a^n - 9 )  Remainder  (66)/ (a^n + 8)

{I'm not sure what "grid" we're talking about  ....!!!  }

Assuming that the lateral sides are the equal ones, we have

x + 2*(5/4)x  = 14

x + (10/4)x  = 14

(14/4)x  = 14     multiply both sides by 4

14x  = 56         divide both sides by 14

x = 4 cm      and this is the base......and (5/4)x = (5/4)(4) = 5 cm......and this is a lateral side

And by the Pythagorean Theorem, the height will be √[5^2 - 2^2] = √21 cm

So....the area of the triangle will be    (1/2)bh  =  (1/2)(4)√21   =  2√21 cm  ≈ 9.165 cm^2