It is given that the diameter of the base of the cone is AB=20, and the slant height is BC=10.
Recall that the radius is one-half of the diameter, it follows that:
![r=(1)/(2)\cdot20=10](https://img.qammunity.org/2023/formulas/mathematics/college/njyph1qdrzg1zp9d2rqd43grnjihgy7lcc.png)
Note that the radius (r), vertical height (h), and slant height (l) of a cone form a right triangle which satisfies the Pythagorean Theorem:
![h^2+r^2=l^2](https://img.qammunity.org/2023/formulas/mathematics/college/1ckc78zxvbsdm1jocd32n137wzngilg17n.png)
Substitute r=10 and l=10 into the equation:
![\begin{gathered} h^2+10^2=10^2 \\ \Rightarrow h^2=10^2-10^2 \\ \Rightarrow h^2=0 \\ \Rightarrow h=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/erp6zn9x38r5sav3ah1lxwq4gvylf67qfj.png)
Notice that the height of the cone is calculated to be zero which is not possible for a cone.
The Surface Area of a cone is:
![S=\pi r^2+\pi rl](https://img.qammunity.org/2023/formulas/mathematics/college/pr8myfxyzpkwwgud5wpp16e9r7xcs3yrgn.png)
Substitute r=10 and l=10 into the formula:
![\begin{gathered} S=\pi(10^2)+\pi(10)(10) \\ \Rightarrow S=100\pi+100\pi=200\pi\text{ square units} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rx4dukxcwu1ztffco6adgwihs7yka01lj4.png)
The answer is 200π square units.