Check the picture below.
so the area of the hexagon is really just the area of two isosceles trapezoids.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ a=2\\ b=4\\ h=2 \end{cases}\implies \begin{array}{llll} A=\cfrac{2(2+4)}{2}\implies A=6 \\\\\\ \stackrel{\textit{twice that much}}{2A = 12} \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2%5C%5C%20b%3D4%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B2%282%2B4%29%7D%7B2%7D%5Cimplies%20A%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwice%20that%20much%7D%7D%7B2A%20%3D%2012%7D%20%5Cend%7Barray%7D)
Firstly look at the options carefully and we will get to know that two answer options are incorrect because 2 of them are representing AT MOST SIGN which we don't need. So in that way Options B and D are eliminated.
So Now we are left with 2 options A and C. Lets figure it out.
We know that Carlos has 5 complete set with 4 individual figures and josh has 3 complete sets with 14 individual figures.
So lets write in the numerical language:-
let us assume that complete sets are X.
ATQ,
5x + 4 ≥ 3x + 14
subtracting 3x from both sides.
5x -3x + 4 ≥ 14
2x + 4 ≥ 14
subtracting 4 from both sides
2x ≥ 14 - 4
2x ≥ 10
dividing both sides by 2
x ≥ 5 Answer
So correct answer option is C
First one (20 characters) hope this helped
Since the traversal intersects two parallel lines, the corresponding angles are congruent. Set both expressions equal to each other and you get x=6
