angle 5 and angle 4 are exterior angles
from the diagram, we can see that the height or line perpendicular to the parallel sides is 8.5.
likewise we can see that the parallel sides or "bases" are 24.3 and 9.7, so
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=8.5\\ a=24.3\\ b=9.7 \end{cases}\implies \begin{array}{llll} A=\cfrac{8.5(24.3+9.7)}{2}\\\\ A=\cfrac{8.5(34)}{2}\implies A=144.5~in^2 \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%20h%3D8.5%5C%5C%20a%3D24.3%5C%5C%20b%3D9.7%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B8.5%2824.3%2B9.7%29%7D%7B2%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B8.5%2834%29%7D%7B2%7D%5Cimplies%20A%3D144.5~in%5E2%20%5Cend%7Barray%7D)
It is 4170 I believe if my calculations are correct
2⃣, -4⃣ = x; no radical necessary.
Answer:
See below.
Step-by-step explanation:
7 ) 29.0
= 4 1/7
= 4.142857142857.........
The 142857 repeats itself without bounds.