Help. Please.

Problem \(1\):

 

The formula for the surface area of a cone is \(πrl+π{r}^{2}\), where \(r\) is the radius and \(l\) is the slant height. For problem one, \(r = 6\) and \(l = 14\). By plugging in the variable values, we can get \(π(6)(14)+π{6}^{2} = 84π + 36π = 120π\). 

 

Problem 2:

 

The formula for the surface area of a cylinder is \(2πrh+2π{r}^{2}\), where \(r\) is the radius and \(h\) is the height. \(r = 13\) and \(h = 17\). When you plug in the variable values, you can get \(2π(13)(17)+2π{13}^{2} = 442π + 338π = 780π\), and it is asking for the nearest whole number, so multiply \(3.14\) by \(780\). The product is \(2449.2\), which rounds down to \(2449\).

 

Problem 3:

 

The formula for the surface area of a cone is \(πrl+π{r}^{2}\), where \(r\) is the radius and \(l\) is the slant height. \(r = 7\) and \(l = 24\). By plugging in the variable values, we can get \(π(7)(24)+π{7}^{2} = 168π + 49π = 217π\). Since they are asking for a decimal to the nearest tenth, we can get \(681.4\).

 

Problem 4:

 

The formula for the surface area of a cylinder is \(2πrh+2π{r}^{2}\), where \(r\) is the radius and \(h\) is the height.  \(r = 25\) and \(h = 36\). When you plug in the variable values, you can get \(2π(25)(36)+2π{25}^{2} = 1800π + 1250π = 3050π\).

 

- Daisy