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)
Answer:
Step-by-step explanation:
Hence, the surface area of a triangular pyramid is 140 square units.
Answer:
1.) FI, AD, FA, DI. (Neither of them are parallel or right angles.)
2.) 95 (Same angle just upside down so it'd be the same size.)