Pre-Algebra Area: Parallelograms Lesson 10-1 Objectives: 1. to find areas of rectangles 2. to find areas of parallelograms

Pre-AlgebraPre-Algebra

Area: ParallelogramsArea: Parallelograms

Lesson 10-1

Objectives: 1. to find areas of rectangles

2. to find areas of parallelograms



Lesson 10-1

Tips: look for two segments that form a right angle when determining a base and height



Find the area of the rectangle.

Lesson 10-1

Step 1: Change the units so that they are the same.

150 cm = 1.5 m Change 150 centimeters to meters.

Step 2: Find the area.

A = bh Use the formula for area of a rectangle.

= (4)(1.5) Replace b and h with the dimensions 4 and 1.5.

= 6 Simplify.

The area of the rectangle is 6 m2.



Find the area of each parallelogram.

Lesson 10-1

a. b.

A = bh area formula

= (8)(2) Substitute.

= 16 Simplify.

The area is 16 m2. The area is 15 in.2.

A = bh

= (2.5)(6)

= 15


Area: Triangles and TrapezoidsArea: Triangles and Trapezoids

Lesson 10-2

Objectives: 1. to find areas of triangles

2. to find areas of trapezoids



Lesson 10-2

Tips: bases of a trapezoid are the two sides parallel to each other, even if the figure is turned

to find the area of an irregular figure, you may need to use more than one area formula. (make sure to know all the area formulas)



Find the area of the triangle.

Lesson 10-2

The area is 39 in.2.

A = bh Use the formula for area of a triangle.12

= • 13 • 6 Replace b with 13 and h with 6.12

= 39 Simplify.



Find the area of the figure.

Lesson 10-2

Add to find the total: 450 + 1,350 = 1,800.

Area of triangle

A = bh12

= • 45 • 2012

= 450

Area of rectangle

A = bh

= 45 • 30

= 1,350

The area of the figure is 1,800 cm2.



Suppose that, through the years, a layer of silt and mud

settled in the bottom of the Erie Canal. Below is the resulting cross

section of the canal. Find the area of the trapezoidal cross section.

Lesson 10-2

The area of the cross section is 106.5 ft2.

A = h(b1 + b2) Use the formula for the area of a trapezoid.12

A = • 3(31 + 40) Replace h with 3, b1 with 31, and b2 with 40.12

= • 3(71) Simplify.12

= • 21312

= 106.5


Area: CirclesArea: Circles

Lesson 10-3

Objectives: 1. to find areas of circles

2. to find areas of irregular figures that include parts of a circle



Lesson 10-3

Tips: remember to follow the order of operations when finding the area of a circle or semi-circle. Square the radius first, and then multiply by π.

when asked to find the “exact area” of a circle (or circular shape) do not substitute 3.14 in for π, leave π as π.

make sure not to confuse the radius as the diameter…know the difference.



Find the exact area of a circle with diameter 20 in.

Lesson 10-3

A = r 2

= (10)2 r = d; r = 1012

= 100 Simplify.

The area is 100 in.2.



A TV station’s weather radar can detect precipitation in a circular region having a diameter of 100 mi. Find the area of the region.

Lesson 10-3

A = r 2

The area of the region is about 7,850 mi2.

= (50)2 r = d; r = 5012

= 2,500 exact area

(2,500)(3.14) Use 3.14 for .

= 7,850 approximate area



A pound of grass seed covers approximately 675 ft2. Find the

area of the lawn below. Then find the number of bags of grass seed

you need to buy to cover the lawn. Grass seed comes in 3-lb bags.

Lesson 10-3

Area of region that is one fourth of a circle:

area of circle = r 2

area of quarter circle = r 2

A (3.14)(15)2 Replace with 3.14 and r with 15.

= 176.625 ft2

14

14



(continued)

Lesson 10-3

Area of region that is a rectangle:

area of rectangle = bh

A = 45 • 25 Replace b with 45 and h with 25.

= 1,125

The area of the lawn is about 177 ft2 + 1,125 ft2 = 1,302 ft2.

You need to buy one 3-lb bag of grass seed.

1,302 ÷ 675 1.93 Divide to find the number of pounds of seed.


Space FiguresSpace Figures

Lesson 10-4

Objectives: 1. to identify common space figures

2. to identify space figures from nets

Tips: a “cube” is a rectangular prism with six congruent square faces

Lateral means “on the side”. The lateral faces of a prism or pyramid are the surfaces that connect with a base.



Lesson 10-4



Lesson 10-4



Lesson 10-4



Describe the bases and name the figure.

Lesson 10-4

The bases are circles.

The figure is a cylinder.



Name the space figure you can form from the net.

Lesson 10-4

With two hexagonal bases and rectangular sides, you can form a hexagonal prism.


Surface Area: Prisms and CylindersSurface Area: Prisms and Cylinders

Lesson 10-5

Objectives: 1. to find surface areas of prisms

2. to find surface areas of cylinders


Tips: SA = surface Area, LA = lateral area, B = area of the base, p= perimeter of the base

One way to find the surface area of a space figure is to find the area of its net.

Surface Area – is measured in square units.

SA of a cylinder can also be written as SA= 2πrh + 2πr2


Lesson 10-5



Find the surface area of the rectangular prism

using a net.

Lesson 10-5

60 + 60 + 150 + 90 + 150 + 90 = 600 Add the areas.

The surface area is 600 cm2.

Draw and label a net.Find the area of each rectangle in the net.



Find the surface area of the rectangular prism.

Lesson 10-5

The surface area of the rectangular prism is 500 in.2.

Step 1: Find the lateral area.

L.A. = ph Use the formula for lateral area.

= (5 + 6 + 5 + 6)20 p = 5 + 6 + 5 + 6 and h = 20

= 440

Step 2: Find the surface area.S.A. = L.A. + 2B Use the formula for surface area. = 440 + 2(5 • 6) L.A. = 440 and B = 5 • 6 = 440 + 60 = 500



Find the surface area of the cylindrical water tank.

Lesson 10-5

The surface area of the water tank is about 1,156 ft2.

Step 1: Find the lateral area.

L.A. = 2 rh Use the formula for lateral area.

2(3.14)(8)(15)

754

Step 2: Find the surface area. S.A. = L.A. + 2B Use the formula for surface area. = L.A. + 2( r 2) 754 + 2(3.14)(8)2

1,156


Surface Area: Pyramids, Cones, and SpheresSurface Area: Pyramids, Cones, and Spheres

Lesson 10-6

Objectives:

1. to find surface areas of pyramids

2. To find surface areas of cones and spheres


Tips: the slant height is not perpendicular to the base of a pyramid or cone. In a pyramid, it is perpendicular to the base of a triangular face. In a

cone, it is the shortest segment that joins the vertex to a point on the circle base

there are a number of analogies that one can use to remember not only the lateral-area formulas, but also the surface-area formulas, each of which involves adding lateral area and base area(s).


Lesson 10-6



Find the surface area of the square pyramid.

Lesson 10-6

The surface area of the pyramid is 105 m2.

Step 1: L.A. = p Use the formula for lateral area.12

= • 20 • 8 = 80 p = 4(5) and = 8.12

Step 2: S.A. = L.A. + B Use the formula for surface area. = 80 + 52 Lateral area = 80 and B = 52. = 80 + 25 = 105


Step 1: L.A. = r Use the formula for lateral area.

3.14(3)(7) r = 3 and = 7.

= 65.94


Find the surface area of the cone.

Lesson 10-6

The surface area of the cone is about 94 m2.

Step 2: S.A. = L.A. + B Use the formula for surface area.

65.94 + 3.14(3)2 L.A. 65.94 and B = (3)2.

= 65.94 + 28.26

= 94.2



Earth has an average radius of 3,963 mi. What is Earth’s approximate surface area to the nearest 1,000 mi2? Assume that Earth is a sphere.

Lesson 10-6

The surface area of Earth is about 197,259,000 mi2.

S.A. = 4 r 2 Use the formula for surface area.

= 197,259,434.64 Multiply.

197,259,000 Round to nearest 1,000.

4(3.14)(3,963)2 r 3,963


Volume: Prisms and CylindersVolume: Prisms and Cylinders

Lesson 10-7

Objectives: 1. to find volumes of prisms

2. to find volumes of cylinders


Tips: read the question carefully, you can calculate both the volume and surface area of a 3-D figure, make sure to understand the concepts of both and use the correct formula.


Lesson 10-7



Find the volume of the triangular prism.

Lesson 10-7

The volume is 1,260 cm3.

V = Bh Use the formula for volume.

= 63 • 20 B = • 9 • 14 = 63 cm212

= 1,260 Simplify.



Find the volume of the juice can, to the nearest cubic centimeter.

Lesson 10-7

The volume is about 581 cm3.

V = Bh Use the formula for volume.

V = r 2h B = r 2

= 580.7744 Simplify.

3.14 • 3.42 • 16 Replace r with 3.4, and h with 16.


Volume: Pyramids, Cones, and SpheresVolume: Pyramids, Cones, and Spheres

Lesson 10-9

Objectives: 1. to find volumes of pyramids and cones

2. to find volumes of spheres


Tips: the formula for the volumes of cones and pyramids uses each figure’s height, not slant height. Remember, the height of a cone or pyramid is the length of the segment from the vertex perpendicular to the base.

the volume of a sphere is the only one (studied so far) that involves a third power.


Lesson 10-9



Find the volume of the cone.

Lesson 10-9

V = Bh Use the formula for volume.13

V = r 2h B = r 213

(3.14)(2)2(12) Replace r with 2 and h with 12.13

= 50.24 Simplify.

The volume of the cone is about 50 in.3.



Find the volume of the square pyramid.

Lesson 10-9

The volume of the pyramid is 256 in.3.

V = Bh Use the formula for volume.13

V = s 2h B = s213

= (8)2(12) Replace s with 8 and h with 12.13

= 256 Simplify.



Earth has an average radius of 3,963 mi. What is Earth’s approximate volume to the nearest 1,000,000 mi3? Assume that Earth is a sphere.

Lesson 10-9

The volume of the Earth is about 260,580,000,000 mi3.

V = r 3 Use the volume formula.43

(3.14)(3,963)3 Replace r with 3,963.43

260,579,713,159 Simplify.

Additional Examples

Documents

Pre-Algebra Area: Parallelograms Lesson 10-1 Objectives: 1. to find areas of rectangles 2. to find areas of parallelograms