1. Chapter 11 Surface Area and Volume of Solids By: Andrew Li and Jonathan Wang

2. Key Concepts and Vocab ● Key Vocab -Polyhedra- a 3D solid with flat polygonal faces -Base- The 2 parallel and congruent polygonal faces -Lateral Edges- the edges that form the lateral faces -Lateral Faces- The faces that are not the bases -Lateral Area- Sum of all the areas of the lateral faces -Surface Area- Sum of all the areas of the bases and the lateral faces -Regular Polygon- A polygon that has all congruent sides and angles

3. Key Concepts and Vocab (cont.) ● Surface Area,Lateral Area, and Volume of a Prism -Surface Area of a Prism is the sum of areas of all faces; when finding the area of the bases use the formula A=½(A)P. -The Lateral Area of a Prism is sum of areas of all lateral faces and usually the lateral faces are rectangular so use the formula A=(B)H. -The Volume of a Prism is V=(Area of Base) Height

4. Key Concepts and Vocab (cont.) ●Surface Area, Lateral Area, and Volume of a Pyramid -The Surface Area of a pyramid is the sum of the area of the polygonal base and the area of the lateral area. -The Lateral Area of a pyramid is LA=n(½∙Height∙Base) where n is the number of how many triangles and H is the height of one triangle and B is the base of the one triangle. -The Volume of the Pyramid is V= ⅓ (Area of Base) Height and

5. Key Concepts and Vocab (cont.) ● Surface Area, Lateral Area, and Volume of a Cylinder and Cone -The Surface Area of a Cylinder is SA=2rh+2r 2 and the Surface Area of a Cone is SA=rℓ+r 2 -The Lateral Area of a Cylinder is LA=2rh and the Lateral Area of a Cone is LA=rℓ -The Volume of a Cylinder is V=Area of the Base (Height) and the Volume of a Cone is V= ⅓ ∙Area of the Base (Height)

6. Key Concepts and Vocab (cont.) ● Surface Area and Volume of a Sphere -The Surface Area of a Sphere is SA=4r 2 ; when finding the Surface Area of a Hemisphere you usually have to divide the surface area of the sphere by 2 and add the area of the base The Volume of a Sphere is V=4/3∙πr 3

7. Common Mistakes a.) Many have forgotten to cube the r in the formula 4/3πr. -This is probably because of the consistency of using a square instead of a cube in other formulas. Students should wrap their mind around this “3” sign instead of always using “2.” b.) When finding the surface area of a shape cut in half, students forget the cut open face. (i.e. when a cone is cut in half, the new triangle formed by the cut is forgotten when calculating surface area) -This issue can be solved if we visualize painting the surface of the solid. This way, no faces on the figure will be left out.

8. Common Mistakes (cont.) c.) When solving for the volume of cones and pyramids, many forget to add ⅓ to the formula. -By basing this formula off of cylinders, it can be known that a cone is equal to ⅓ of the smallest cylinder it can fit inside, so therefore you must add ⅓ to your calculations.

9. Connections to other key concepts/units. ● The 30-60-90 triangles - Unit 7 -The 30-60-90 triangle is usually used in order to find the apothem of a base on a hexagonal and triangular prism, but you have to have a right triangle with the angle measurements of 30-60-90 in order to use it. ●The 45-45-90 triangles - Unit 7 -The 45-45-90 triangle is usually used when you are trying to find the apothem of a square on a cube and only used if you have a right triangle and it is isosceles.

10. Connections to other key concepts/units (cont.) ●Pythagorean Theorem - Unit 7 -This is usually used when you have a right triangle and two side lengths and used when trying to find the slant height or height of a cone and a pyramid. ●Heron’s and Equilateral Area Formulas - Unit 10 -The Heron’s Formula is used when you have 3 different side lengths of a triangle and the Equilateral Area Formula is used when you have an equilateral triangle and one side length of that triangle. These formulas are used when finding the bases of the triangular prisms.

11. Real Life Problem Answer: He can’t fit all of the pencils in his pencil case because the volume of all of his pencils combined is greater than the volume of his pencil case. In order to find the volume of his pencils you had to make a 30-60-90 triangle and find the apothem.5 and half of the side length.25. Then you could find the area of the base ½.25√(3)x3=.375√(3) and the volume of the pencils by 20 x.375√(3)= 7.5√(3). Next multiply 12 by the volume of the pencil to get the volume of all the pencils 7.5√(3)x12=90√(3). The volume of the pencil case is (5x6)4=120. Finally compare both of the volumes and the pencils had a greater volume. Bill has 12 regular hexagonal wooden pencils with the radius of.5 cm and a height of 20 cm plus he has a rectangular pencil case with the width of 5 cm, length of 6 cm, and a height of 4 cm. Would he be able to put all 12 pencils in his pencil case?

12. Sample Questions Easy Question 1- Find the surface area of this figure. (Figure not drawn to scale) 12 6 Answer: 4(72)+2(144)= 576 squared units.

13. Sample Questions Answer- Find the area of each lateral face: 3(18)+4(18)+5(18)=216 cm squared. The 5 is obtained from Pythagorean Triples 3-4-5. This is the lateral area. Next, find the area of both triangular bases using 30-60-90 triangles. Drawing a line through the middle, use Pythagorean Theorem to find the height, then use A=bh to find the area. Then, simply double that value because there are 2 bases, and add that value to the lateral area. That’s the surface area. Medium Difficult Question- Find the lateral and surface area of a right triangular prism that has a length of 18 cm, and base lengths of 3 and 4 cm. The third base length is unknown.

14. Sample Questions Hard Question- The bases of a prism are rhombuses with diagonals 12 meters and 16 meters long. The height of the prism is 8 meters. Find the lateral area, surface area, and volume of the prism. Answer: First use Pythagorean Triples 6-8-10 or 3-4-5 to find the area of each lateral face. The diagonals cut in half form a quarter triangle with values of 6 and 8, so the hypotenuse is 10. Then you would multiply 10 by 8 and multiply that product by the number of sides the rhombus has which is four 4(10x8)= 320 and that would be the lateral area In order to find the surface area you need to find the area of the bases which are rhombuses using ½ d ₁ (d ₂ ). (½ 12(16))=96 square meters. Double 96 to find the area of both bases. Then you add the area of both bases to the lateral area and you would get the surface area which is 512 meters squared Finally to find the volume of the prism you would need to multiply the area of the base which is 96 by the height 8 and the volume would be 768 meters cubed

15. Conclusion ● Some key concepts that are important to remember are the formulas for the surface area, lateral area, and volume. ● Remember that special right triangles, Pythagorean Theorem, and the Equilateral and Heron’s Formulas will help you a lot. ● Don’t forget the common mistakes that often occur in this unit.

16. Thank you for watching!