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)
We know that
A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. ( Inscribed Quadrilateral Theorem)
so
m∠B+m∠D=180°
2x+(3x-5)=180
5x=180+5
5x=185
x=185/5
x=37°
m∠A=x+5-----> 37+5------> 42°
the answer is
m∠A is 42°
Answer: D (bottom option)
Not Congruent
Step-by-step explanation:
Assuming the trapezoids are to scale, they are not the same shape. This means that they could not be congruent even if they were centered on top of each other.
Answer:
Step-by-step explanation:
by definition : The inverse of a relation consisting of points of the form (x,y) is the set of points (y,x)
if (1,7) is a point belong to f(x), then the inverse has to be (7,1) not (8,1)
