Answer:
![r =(D)/(2)=(18mm)/(2)= 9mm](https://img.qammunity.org/2021/formulas/mathematics/college/781vjnj8n0rhdo7seerjbi9fo4o5li6yny.png)
The area is given by:
![A= \pi r^2](https://img.qammunity.org/2021/formulas/physics/college/p0oc8rfnvjxlb0r8rpjer9bun2e9lw810y.png)
And replacing we got:
![A=\pi (9mm)^2 =81\pi mm^2](https://img.qammunity.org/2021/formulas/mathematics/college/e0txcqjjln4vao0ha2q76ujtjpbveq5fl5.png)
So then we can conclude that the area of the coin is
mm^2
Explanation:
For this case we know that we have a coin with a diamter of
, and by definition the radius is given by:
![r =(D)/(2)=(18mm)/(2)= 9mm](https://img.qammunity.org/2021/formulas/mathematics/college/781vjnj8n0rhdo7seerjbi9fo4o5li6yny.png)
The area is given by:
![A= \pi r^2](https://img.qammunity.org/2021/formulas/physics/college/p0oc8rfnvjxlb0r8rpjer9bun2e9lw810y.png)
And replacing we got:
![A=\pi (9mm)^2 =81\pi mm^2](https://img.qammunity.org/2021/formulas/mathematics/college/e0txcqjjln4vao0ha2q76ujtjpbveq5fl5.png)
So then we can conclude that the area of the coin is
mm^2