since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)
A circle can be circumscribed about a quadrilateral if and only if the opposite angles of the quadrilater sum up to 180.
This is not the case, so you can't circumscribe a circle about the quadrilateral.
Set up an equation:
0.75x = 18
Divide both sides by 0.75:
x = 18/0.75
x = 24
Answer: 24
Answer:
(3x + 5)/2 = 7
Step-by-step explanation:
If you substitute 3 for x, you get the equation:
(3(3) + 5)/2 = 7
(9 + 5)/2 = 7
14/2 = 7
Answer: 42 degrees
Explanation: KFD=84 KFL is half of KFD(84)
84 divided by 2 = 42
