12.3 SA of Cones and Pyramids

12.3 SA of Cones and Pyramids

Review 12.2

• Remember, to find the SA of a prism/cylinder, we would find the area of the base and lateral sides separately

• We are going to do that for pyramids and cones as well!

Surface Area of Cones

• Similarly to 12.2, we need to separate into base and lateral area.

• Notice with a cone, we only have one base (circle)

• The lateral area is similar to that of a cylinder, but we won't use circumference, like we did for the cylinders.

Surface Area

• Base Area: It's a circle, so:

   $\pi r^2$  

• Lateral Area:  $\pi rl$  

•  $l$  represents the slant height!!

• Remember to add the base and lateral areas together at the end!

Try this on your own

 $36\pi\ or\ 113.1cm^2$  = Answer

What if we have the true height instead of our slant height?

• Can we find the slant height somehow? (What kind of triangle does it make?)

• Use pythagorean theorem to find the hypotenuse (slant height)!

•  $4^2+9^2=$  

•  $\sqrt{97}$  

• Now you have this slant height, find the lateral area

• Add this LA to the base area

• SA = 174.02

Fill in the Blanks

SA of Pyramids

• Base Area: Find the area of the base; in this case, it's a square (might not always be a square though!)

• Lateral Area: Just like cones, we are still using the 'slant' height. Once we find the slant height, we multiple it to our perimeter of our base, and then divide by two.

• Base Area: It's a square, so:  $l\times w$  

• Lateral Area:  $\frac{pl}{2};\ p=perimeter,\ l=slant\ height$   $SA=70in^2$  

Fill in the Blanks

Find the Surface area of this pyramid

Notice the base is a hexagon, and to find the area of a polygon, we need the apothem and perimeter.
Notice the apothem and one of the sides are already given to you though!

 $\frac{ap}{2}a=apothem,\ p=perimeter$  
Answer: 806.1

When finished, show me your notes, then you can start on MathXL