# Chapter 10 Formulas

### Kennedy E

## Chapter 10 Section 1

## Vocabulary

- base of a parallelogram - the length of any side of the parallelogram can be used as the base
- height of a parallelogram - the perpendicular distance between the base and the opposite side
- bases of a trapezoid - the two parallel sides
- height of a trapezoid - the perpendicular distance between the bases

## Area of a trapezoid

The picture to the right shows where to find base 1, base 2, and height.

Area = 1/2 x (base one + base two) x height

A = 1/2 x (b1 + b2) x height

Use the bottom picture as an example.

1. **A = 1/2 x ( b1 + b2 ) x h **Copy formula.

2. **A = 1/2 x ( 31 + 77 ) x 25** Input correct numbers for base1, base2, and height.

3. **A** **= 558 units^2 **Top down solving and label.

## Real life Examples

## Chapter 10 Sectioin 2

## Vocabulary

- area - the amount of surface the figure covers
- circle - the set of all points in a plane that are the same distance from a fixed point called the center
- radius - the distance from the center to any point on the circle
- diameter - the distance across the circle through the center, or twice the radius
- circumference -n the distance around the circle
- pi (π) - the constant used for every circle which is around 3.14 or 22/7

## Area of a circle

Area = pi x radius squared

A = π r2

Use the picture below as an example. Round to the nearest tenths if needed:

1.** A = π r^2** Copy the formula.

2. **A = π 14^2** Input correct number for radius.

3. **A = 615.7521601035995...** Solve.

4.** A = 615.8 m^2** Round and label.

## Real life Examples

## Chapter 10 Section 3

## 3D Figures

**Vocabulary:**

- solid- a 3D figure that encloses a part of space
- polyhedron - a solid that is enclosed by polygons
- face - the polygons that form a polyhedron
- prism - a polyhedron that have two congruent bases that lie in parallel planes and the other faces are all rectangles
- pyramid - a polyhedron that have only one base and all faces are triangles
- cylinder - a solid with two congruent circular bases that lie in parallel planes
- cone - a solid with one circular base
- sphere - a solid formed by all points in space that are the same distance from a fixed point called the center
- edge - segments where faces of a polyhedron meet
- vertex - a point where three or more edges meet

## Chapter 10 Section 4

## Vocabulary

- net - a 2D pattern that forms a solid when it is folded
- surface area - the sum of the areas of a polyhedrons faces

## Surface area of a prism

Surface area = 2 x Area of base + Perimeter of base x height

S = 2B + Ph

Use the picture below as an example. *Be careful the formula will change depending on the shape of the base!*

1. **S = 2B + Ph** Copy the formula.

2. **S = 2(bh) + Ph** Enter formula for the area of base.

3. **S = 2(10x5) + 30 x 6** Input the correct numbers.

4. **S =100 + 180** Top down solve.

5. **S = 280 m^2** Solve and label.

## Surface area of a cylinder

Surface Area = 2 x area of base + base circumference x height

S= 2B + Ch

Use the picture below as an example.

*Use 3.14 for π.*

1.

**S = 2B + Ch**Copy formula.

2.

**S = 2(πr^2) + (2πr)h**Insert formulas.

3.

**S = 2 ( 3.14 x 3 ^2 ) + ( 2 x 3.14 x 3 ) 7**Input the correct dimensions.

4.

**S = 56.52 + 131.88**Top down solve.

5.

**S = 188.4 cm^2**Solve and label.

## Real life examples

## Chapter 10 Section 5

## Vocabulary

- slant height - the height of the lateral face, that is, any face any face that is not the base

## Surface area of a pyramid

Surface area = area of base + .5 x base perimeter x slant height

S = B + .5Pl

Use the picture below as an example.

1.

**S = B + .5Pl**Copy formula

2.

**S= (bh) +.5Pl**Enter formula for area of base.

3.

**S = (8x8) + .5 x 32 x 10**Insert correct dimensions.

4.

**S = 64 + 160**Top down solve.

5.

**S = 224 in^2**Solve and label.

## Surface area of a cone

Surface area = pi x radius squared + pi x radius x slant height

S = πr^2 + πrl

Use the picture below as an example.

*Use 3.14 for π. You will also need to find the slant height. Round o the nearest hundred.*

1.

**a^2 + b^2 = c^2**Find slant height using Pythagorean Theorem.

2.

**6^2 + 10^2 = c^2**Input correct numbers.

3.

**36 + 100 = c^2**Top down solve.

4.

**11. 6619037896906...**Solve and round .

5.

**S = πr^2 +**

**πrl**Copy formula for surface area.

6.

**S= 3.14 x 6^2 + 3.14 x 6 x 11.7**Input correct dimensions.

7.

**S = 113.04 + 220.428**Top down solve.

8.

**S = 333.468 in^2**Solve, round, and label.

## Real life examples

## Chapter 10 section 6

## vocabulary

- volume - a measure of the amount of space an object occupies

## Volume of a prism

Volume = area of base x height

V = Bh

Use the picture below as an example.

*Formula for area of base changes based on the shape of the base*.

1.

**V = Bh**Copy down the formula.

2.

**V = (bh)h**Insert formula for base.

3.

**V = (10x5)6**Input correct dimensions.

4.

**V= 300 in^3**Solve and label.

## Volume of a cylinder

Volume = area of base x height

V = Bh

Use the picture below as an example.* Use 3.14 for π. *

1. **V=Bh** Copy formula.

2. **V=(πr^2)h **Put in the formula for area of base.

3. **V=(π5^2)10 **Input correct dimensions.

4. **V=785 cm^2** Solve and label.

## Real life Example

## Chapter 10 Section 7

## Vocabulary

- pyramid - a solid, formed by polygons, that has one base. The base can be any polygon, and the other faces are triangles
- cone - a solid with one circular base
- volume- the amount of space the solid occupies

## Volume of a pyramid

Volume = 1/3 x area of base x height

V = 1/3 Bh

Use the picture below as an example.

1. **V=1/3 Bh** Copy formula.

2. **V=1/3(bh)h **Put in the formula for area of base.

3. **V= 1/3(6x6)15** Input correct dimensions.

4.** V= 180m^2** Solve and label.

## all chapter 10 formulas

Area Trapezoid - A=.5(b1+b2)h

Area Circle - A=πr^2

Surface area prism - S=2B+Ph

Surface area cylinder - S=2B+Ch

Surface area cone - S=πr^2+πrl

Surface area pyramid - S=B+.5Pl

Surface area sphere - S=4πr^2

Volume prism - V=Bh

Volume cylinder - V=Bh

Volume pyramid - V=1/3Bh

Volume cone - V=1/3Bh

Volume Sphere = 4/3πr^3