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

2. Area: Parallelograms
Lesson 10-1
New Terms: area – the area of a figure is the number of square units it encloses
2. altitude – an altitude is a line segment perpendicular to the line containing a base of the figure and drawn from the side opposite that base
Tips: look for two segments that form a right angle when determining a base and height

3. Find the area of the rectangle.
Area: Parallelograms
Lesson 10-1
Find the area of the rectangle.
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 and 1.5.
= 6 Simplify.
The area of the rectangle is 6 m2.

4. Find the area of each parallelogram.
Area: Parallelograms
Lesson 10-1
Find the area of each parallelogram. a. b.
A = bh area formula
A = bh
= (8)(2) Substitute.
= (2.5)(6)
= 16 Simplify.
= 15
The area is 16 m2.
The area is 15 in.2.

5. Objectives: 1. to find areas of triangles
Area: Triangles and Trapezoids
Lesson 10-2
Objectives: 1. to find areas of triangles
2. to find areas of trapezoids

6. Area: Triangles and Trapezoids
Lesson 10-2
New Terms: altitude of a triangle – is the perpendicular segment from a vertex of a triangle to the line containing the opposite side. The height is the length of the altitude.
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)

7. Area: Triangles and Trapezoids
Lesson 10-2
Find the area of the triangle.
A = bh Use the formula for area of a triangle. 1 2
= • 13 • 6 Replace b with 13 and h with 6. 1 2
= 39 Simplify.
The area is 39 in.2.

8. Area: Triangles and Trapezoids
Lesson 10-2
Find the area of the figure.
Area of triangle
A = bh 1 2
= • 45 • 20
= 450
Area of rectangle
A = bh
= 45 • 30
= 1,350
Add to find the total: ,350 = 1,800.
The area of the figure is 1,800 cm2.

9. Area: Triangles and Trapezoids
Lesson 10-2
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.
A = h(b1 + b2)
Use the formula for the area of a trapezoid. 1 2
A = • 3( )
Replace h with 3, b1 with 31, and b2 with 40. 1 2
= • 3(71)
Simplify. 1 2
= • 213 1 2 =
The area of the cross section is ft2.

10. Objectives: 1. to find areas of circles
Area: Circles
Lesson 10-3
Objectives: 1. to find areas of circles
2. to find areas of irregular figures that include parts of a circle

11. 2. radius – is half the diameter.
Area: Circles
Lesson 10-3
New Terms: circumference – is the measurement around the outside of the circle. 2 times the radius times pi ~ 2πr or πd.
2. radius – is half the diameter.
3. diameter – segment that goes from one point of the circle to another point, and goes through, contains, the center.
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.

12. Find the exact area of a circle with diameter 20 in.
Area: Circles
Lesson 10-3
Find the exact area of a circle with diameter 20 in.
A = r 2
= (10)2 r = d; r = 10 1 2
= Simplify.
The area is in.2.

13. The area of the region is about 7,850 mi2.
Area: Circles
Lesson 10-3
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.
A = r 2
= (50)2 r = d; r = 50 1 2
= 2,500 exact area
(2,500)(3.14) Use 3.14 for .
= 7,850 approximate area
The area of the region is about 7,850 mi2.

14. Area of region that is one fourth of a circle: area of circle = r 2
Area: Circles
Lesson 10-3
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.
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.
= ft2 1 4

15. Area of region that is a rectangle: area of rectangle = bh
Area: Circles
Lesson 10-3
(continued)
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.
1,302 ÷ Divide to find the number of pounds of seed.
You need to buy one 3-lb bag of grass seed.

16. Objectives: 1. to identify common space figures
Lesson 10-4
Objectives: 1. to identify common space figures
2. to identify space figures from nets
New Terms: space figure – are three-dimensional figures, solids
net – is a pattern (2-D) you can form into a space figure
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.

17. Space Figures
Lesson 10-4
A prism has two parallel bases that are congruent polygons, and lateral faces that are parallelograms
A pyramid has a base that is a polygon. The lateral faces are triangles.

18. A cylinder has two parallel bases that are congruent circles.
Space Figures
Lesson 10-4
A cylinder has two parallel bases that are congruent circles.
A cone has one circular base and one vertex.

19. Space Figures
Lesson 10-4
A sphere is the set of all points in space that are a given distance from a given point called the center.

20. Describe the bases and name the figure.
Space Figures
Lesson 10-4
Describe the bases and name the figure.
The bases are circles.
The figure is a cylinder.

21. Name the space figure you can form from the net.
Space Figures
Lesson 10-4
Name the space figure you can form from the net.
With two hexagonal bases and rectangular sides, you can form a hexagonal prism.

22. Surface Area: Prisms and Cylinders
Lesson 10-5
Objectives: 1. to find surface areas of prisms
2. to find surface areas of cylinders

23. Surface Area: Prisms and Cylinders
Lesson 10-5
New Terms: Surface Area (SA) – is the sum of the areas of the base(s) and the lateral faces of a space figure.
Lateral Area (LA) – is the sum of the areas of the lateral faces.
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

24. Surface Area: Prisms and Cylinders
Lesson 10-5
Find the surface area of the rectangular prism using a net.
Find the area of each rectangle in the net.
Draw and label a net.
= Add the areas.
The surface area is 600 cm2.

25. Surface Area: Prisms and Cylinders
Lesson 10-5
Find the surface area of the rectangular prism.
Step 1: Find the lateral area.
L.A. = ph Use the formula for lateral area.
= ( ) p = and h = 20
= 440
Step 2: Find the surface area.
S.A. = L.A. + 2B Use the formula for surface area.
= (5 • 6) L.A. = 440 and B = 5 • 6 =
= 500
The surface area of the rectangular prism is 500 in.2.

26. Surface Area: Prisms and Cylinders
Lesson 10-5
Find the surface area of the cylindrical water tank.
Step 1: Find the lateral area.
L.A. = rh Use the formula for lateral area.
2(3.14)(8)(15)
754
Step 2: Find the surface area.
S.A. = L.A B Use the formula for surface area.
= L.A ( r 2)
(3.14)(8)2
1,156
The surface area of the water tank is about 1,156 ft2.

27. Surface 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

28. Surface Area: Pyramids, Cones, and Spheres
Lesson 10-6
New Terms: slant height (l )– is the height of a face, used to find the area of the lateral faces.
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).

29. Surface Area: Pyramids, Cones, and Spheres
Lesson 10-6
Find the surface area of the square pyramid.
Step 1: L.A. = p Use the formula for lateral area. 1 2
= • 20 • 8 = p = 4(5) and = 8.
Step 2: S.A. = L.A. + B Use the formula for surface area.
= Lateral area = 80 and B = 52.
= = 105
The surface area of the pyramid is 105 m2.

30. Surface Area: Pyramids, Cones, and Spheres
Lesson 10-6
Find the surface area of the cone.
Step 1: L.A. = r Use the formula for lateral area.
3.14(3)(7) r = 3 and = 7. =
Step 2: S.A. = L.A. + B Use the formula for surface area.
(3) L.A and B = (3)2. = =
The surface area of the cone is about 94 m2.

31. Surface Area: Pyramids, Cones, and Spheres
Lesson 10-6
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.
S.A. = r 2 Use the formula for surface area.
4(3.14)(3,963)2 r ,963
= 197,259, Multiply.
197,259,000 Round to nearest 1,000.
The surface area of Earth is about 197,259,000 mi2.

32. Volume: Prisms and Cylinders
Lesson 10-7
Objectives: 1. to find volumes of prisms
2. to find volumes of cylinders

33. Volume: Prisms and Cylinders
Lesson 10-7
New Terms: Volume (V) – is the number of cubic units needed to fill a 3- dimensional figure.
cubic unit – is the space occupied by a cube with edges 1 unit long. (ie. in3)
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.

34. Volume: Prisms and Cylinders
Lesson 10-7
Find the volume of the triangular prism.
V = Bh Use the formula for volume.
= 63 • B = • 9 • 14 = 63 cm2 1 2
= 1, Simplify.
The volume is 1,260 cm3.

35. Volume: Prisms and Cylinders
Lesson 10-7
Find the volume of the juice can, to the nearest cubic centimeter.
V = Bh Use the formula for volume.
V = r 2h B = r 2
3.14 • 3.42 • Replace r with 3.4, and h with 16.
= Simplify.
The volume is about 581 cm3.

36. Volume: Pyramids, Cones, and Spheres
Lesson 10-9
Objectives: 1. to find volumes of pyramids and cones
2. to find volumes of spheres

37. Volume: Pyramids, Cones, and Spheres
Lesson 10-9
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.

38. Volume: Pyramids, Cones, and Spheres
Lesson 10-9
Find the volume of the cone.
V = Bh Use the formula for volume. 1 3
V = r 2h B = r 2 1 3
(3.14)(2)2(12) Replace r with 2 and h with 12. 1 3
= Simplify.
The volume of the cone is about 50 in.3.

39. Volume: Pyramids, Cones, and Spheres
Lesson 10-9
Find the volume of the square pyramid.
V = Bh Use the formula for volume. 1 3
V = s 2h B = s2 1 3
= (8)2(12) Replace s with 8 and h with 12. 1 3
= Simplify.
The volume of the pyramid is 256 in.3.

40. Volume: Pyramids, Cones, and Spheres
Lesson 10-9
Additional Examples
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.
V = r Use the volume formula. 4 3
(3.14)(3,963)3 Replace r with 3,963. 4 3
260,579,713,159 Simplify.
The volume of the Earth is about 260,580,000,000 mi3.