Unit 11 Review By: Jason/Julia

Overview of Key Concepts ● Finding volume of… o Prisms and Cylinders o Pyramids and cones o Spheres ● Finding surface area of… o Prisms o Pyramids o Circular solids

Connections to Other Units Unit 10: ● Finding area and perimeter of 2-D shapes Unit 8: ● Properties of circles Unit 7 (part 1): ● Right triangles and the pythagorean theorem

Common Mistakes Cone ● Not dividing by 3 when finding volume ● Using the actual height instead of slant height when finding surface area Sphere ● using r^2 instead of r^3 when finding volume Prism ● Forgetting to find both bases when finding surface area One remedy would be to check to make sure that you have all of the parts that you need in your equation before solving so you won’t forget anything

Surface Area and Volume of a Cylinder SA formula: 2πrh+2(πr^2) Volume formula: (rπ^2)h (Base*height)

Example 1: Radius is 2, height is 4. Find total Surface Area and Volume of the figure. Apply the SA formula for the cylinder; 2πrh+2(πr^2) Radius is 2, the height is 4. Plug in the numbers in the formula: 2π(2)(4)+2(π2^2) =16π+8π 24π

Surface Area and Volume of Sphere SA formula: (4πr^2) Volume: 4/3(πr^3)

Surface Area and Volume of Cone SA formula: (πrl)+(πr^2) Volume Formula: ⅓ πr^2(h)

Example 2: ½ Sphere, Cone, and Cylinder With the information provided below, find the Volume Volume of ½ Sphere + Volume of Cone + Volume of Cylinder Volume of ½ Sphere formula: ( ⅔ ) πr^3 Radius= 10, so ( ⅔ ) π(10^3) Volume of ½ Sphere= 666 ⅔ π

Example 2 Continued (1) Volume of Cone: ( ⅓ ) πr^2(h) Diameter=20, So Radius= 10 Slant height= 26 *Special right triangle, so the height is 24. ( ⅓ ) π10^2(24) =800π or Approximately

Example 2 Continued (2) Total length of the shape= 70 Height of the Cone= 24, Radius, or “height” of the ½ sphere= = is the height of the cylinder Volume for cylinder πr^2(h) (π)10^2(36) =3600π or Approx Add all of the shapes together; 3600π+800π+666 ⅔ π =5066 ⅔ π or Approx:

Example 3: Frustum of a Cone Given, the small radius is 6, and the large Radius is 18. The h in the given shape is 20 Solve for the Surface Area of the frustum Big cone - Small cone + Bottom base + Top base

Example 3 Continued (1) To solve for this, there is need to draw a non existing top part of the cone. The new “Slant height” of the small cone created by drawing can be called X. Since the small Radius is 6, and the large one is 18, a proportion could be set up: = and solve for x x=10 X 20 r R 6 18 X x+20

Example 3 Continued (2) Slant height of the small non existant cone is 10. Slant height of the frustum is 20 through addition property, the new value is 30. Given; the radius of bottom base which is 18 LA of cone: πrl π*18*30 =540π 30 18

Example 3 Continued (3) Given, small radius=6 Solved the Slant Height of the cone; 10 πrh=π(6)(10) =60π Big Cone-Small cone; 540π-60π=480π 480π= Lateral Area of the frustum SA requires to add the 2 bases; 480π+πR^2+πr^2 π18^2+π6^2; 324π+36π+480π 840π, or approximately

Real Life Applications You need to paint a shed that looks like a cube with a triangular prism on top that is made with isosceles triangles with the congruent side lengths of 5m, it has walls that are 8m high, and the total height from ground to peak is 11m. It also has a door that is 2m wide and 3m tall. One can of paint covers 6m^2, how many cans do you need? 8 2 (Not to scale)

Solution To solve this problem, you have to find the surface area, excluding the bottom of the roof and the side touching the ground. One side of the house is 8x8=64 and you need 4 of those sides, 64x4=256, but then you have to subtract the area of the door, 2x3=6, so 256-6=250m^2 Then, because the triangle is isosceles, you know that the two sides of the roof are 8x5=40 and 40x2=80m^2 Finally, the height of the triangle is 11-8=3 so plug that into the formula 1/2 bxh-> ½(8)x(3)=12 then you multiply that by two to get 24. Then add everything together, =354. Then to find the paint you have to divide that by two, to get 59. So you need 59 cans of paint