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 parallelogram

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

Area = base x height

A = b x h


Use the bottom picture as an example.

1. A = b x h Copy formula.

2. A = 10 x 8 Input correct numbers for height (h) and base (b).

3. A = 80 cm^2 Multiply and label.

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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.

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    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

    The picture to the right shows where to find radius.

    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.

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    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
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    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

    The picture to the right is a net of a prism and it shows where to find the dimensions needed.

    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.

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    Surface area of a cylinder

    The picture to the right is a net of a cylinder telling where the right dimensions are.
    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.
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    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

    The picture to the right shows where to find the dimensions needed.
    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.
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    Surface area of a cone

    The picture to the right shows where to find the correct dimensions.
    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.
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    Real life examples

    Chapter 10 section 6

    vocabulary

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

    Volume of a prism

    The picture to the right shows where to find all the correct dimensions.
    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.
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    Volume of a cylinder

    The picture to the right shows where to find the correct dimensions.

    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.

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    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

    The picture to the right shows where to find the correct dimensions.

    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.

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    all chapter 10 formulas

    Area Parallelogram - A=bh
    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