7.3

What If It Has One Base?

Pg. 8Surface Area of Pyramids and Cones

7.3 – What If It Has One Base?Surface Area of Pyramids and Cones

Today you will examine pyramids and cones. As you work today, you will discover ways to classify pyramids by their shape and will develop new tools of measurement.

7.10 – PYRAMIDSA pyramid is a solid with a polygonal base formed by connecting each point of the base to a single given point (apex) that is above or below the flat surface containing the base. Each triangle is a lateral face of a pyramid.

a. Examine the pyramid below. If surface area measures the total area of all of the faces, find the surface area of the pyramid.

Rectangle

(8)(8) + 4 (½(8)(5))64 + 4(20)

64 + 80144 un2

+ 4 triangles

b. Which number did you end up not using to find the surface area? Why not?

3, not part of the triangles or rectangle

c. This time the height of the triangle is missing. Find the height of the triangle, then find the surface area of the pyramid.

10

26

20 cm

102 + 242 = h2

100 + 576 = h2

676 = h2

26 = h

10

26

20 cm

Rectangle

(20)(20) + 4 (½(20)(26))

400 + 4(260)

400 + 1040

1440 cm2

+ 4 triangles

7.11 – DIFFERENT HEIGHTSA pyramid has two different heights. One is the actual height of a pyramid. The other is the height of the triangles. This triangle height is called the slant height or the lateral height of the pyramid. We use the cursive letter to represent this length.

HEIGHTS

lant height

a. Obtain the resource pages from your teacher. Cut out each net and fold along the lines to create the three-dimensional solids. Label the base and the lateral height of the pyramid.

BaseLateral height

Base

Lateral height

Base

Lateral height

Base

Lateral height

Base

Lateral height

b. With your team develop a formula that will give you the surface area of any pyramid.

SA = B +½ P

1

2SA B P

c. Explain how this formula is different than the formula for prisms.

SA = B +½ PSA = 2B + PH

Two bases One base

Rectangles for lateral faces

Triangles for lateral faces

7.12 – SURFACE AREAUse the new formula to find the surface area of each shape.

81

3610

SA = B + ½ PSA = 81 + ½(36)(10)SA = 81 + 180

261m2

100

40

SA = B + ½ PSA = 100 +SA = 81 +

5

2 2 25 15 2250

5 10

5 10

40 5 10½100 10

281 100 10in

7.13 – CIRCULAR BASEA cone is in the form of a pyramid, but has a circle as its base. a. Convert the formula you created in the previous problem to find the surface area of a cone.

1

2SA B P

2 12

2SA r r

2SA r r

b. Find the surface area of the following cones.

2SA r r

27SA 7 1549SA 105

2154SA in

2SA r r

29SA 9 1581SA 135

2216SA ft2 2 29 12

222515

7.14 – HAPPY BIRTHDAY!Your class has decided to throw your principal a surprise birthday party. The whole class is working together to create party decorations, and your team has been assigned the job of producing party hats. Each party hat will be created out of special decorative paper and will be in the shape of a cone.

Your Task: Use the sample party hat provided by your teacher to determine the size and shape of the paper that forms the hat. Then determine the amount of paper (in square centimeters) needed to produce one party hat and figure out the total amount of paper you will need for each person in your class to have a party hat.

Lateral area of each cone

2SA r r

2SA B PH 22 2SA r rH

1

2SA B P 2SA r r