Review:
Surface Area (SA) of Right Rectangular
Prisms and Cylinders

Surface Area of Right Rectangular Prisms
A face of the prism

Surface Area of Right Rectangular Prisms

Surface Area of Right Rectangular Prisms
SA

Surface Area of Right Rectangular Prisms
SA = Area of top + Area of bottom + Area of front + Area of back + Area of left + Area of right

Opposite faces are congruent (same):
• Area of top and bottom are the same
• Area of front and back are the same
• Area of left and right are the same

Surface Area of Right Rectangular Prisms
So, SA = (2 x area of top face) + (2 x area of front face) + (2 x area of right face)

What is the surface area of this right rectangular prism?
SA =
(2 × 35 × 80) + (2 × 80 × 45) + (2 × 35 × 45)
=
5600 + 7200 + 3150
=
15 950 cm²

Base 1
Surface Area of a Cylinder
Curved surface is a rectangle
2 circular bases
Base 2
Curved Surface

Surface Area of a Cylinder
Area of Curved surface = 2πr x h
Area of a base = πr²
Area of 2 bases = 2πr²

Surface Area of a Cylinder
Area of Curved surface = 2πr x h
Area of a base = πr²
Area of 2 bases = 2πr²
SA of a cylinder =
(2πr x h) +
2πr²

What is the surface area, SA, of this right cylinder?
SA =
2πr² + 2πrh
=
(2 × π × 2²) + (2 × π × 2 × 5)
≈
87.9645 cm²

Review:
Volume (V) of Right Rectangular Prisms and
Cylinders
A face of the prism

Volume (V) of Right Rectangular Prisms
Volume is a space occupied by the prism

Volume (V) of Right Rectangular Prisms
= (Area of the bottom face) x
(height)
= A x h

Determine the volume of this right rectangular prism
V =
A × h
=
(4.0 × 6.0) x 1.5
=
24.0
× 1.5
= 36.0 m³

Volume (V) of Right Cylinder
Volume = (Area of a base) x (height)
= ( π r²) x h

Determine the volume of this right cylinder
V = area of a base x height
=
πr² × h
=
π(5)² × 8
≈ 628.3 cm³

A Worksheet:
Finish both sides by tomorrow

1.4 Surface Areas of Right
Pyramids and Right Cones

• is a 3-dimensional (3-D) object that has triangular faces and a base that is a polygon .

• is a 3-dimensional (3-D) object that has triangular faces and a base that is a polygon .

• Polygons are 2-dimensional shapes .
• They are made of straight lines , and the shape is
"closed" ( all the lines connect ).
Polygon
(straight sides)
Not a Polygon
(has a curve)
Not a Polygon
(open, not closed)

Right Pyramid
• The shape of the base determines the
NAME of the pyramid

Right Pyramid
• MUST know the vocabulary!
• Apex = a point where triangular faces meet
• Slant height = a height of a triangular face

Surface Area of a Right Pyramid
If a base is Regular Polygon , then the triangular faces are congruent (same)
Regular Polygon
=
Same sides and same angles

Surface Area of a Right Pyramid
The surface area of a right pyramid is the sum of the areas of the triangular faces and the base
SA = Area of faces + Area of the base

Surface Area of a Right Pyramid
The surface area of a right pyramid is the sum of the areas of the triangular faces and the base
SA = Area of faces + Area of the base
A review question!!!
WHAT IS THE AREA OF A
TRIANGLE?

Surface Area of a Right Pyramid
• The surface area of a right pyramid is the sum of the areas of the triangular faces and the base
• SA = Area of faces x Area of the base
• REVIEW!!!
• WHAT IS THE AREA OF A TRIANGLE?

Surface Area of a Right Pyramid
SA = Area of faces + Area of the base
This right square pyramid has a slant height of 10 cm and a base side length of 8 cm. What is its surface area?

Surface Area of a Right Pyramid
Answer:
The area, A, of each triangular face is:
A = (8 )· ( 10)
A = 80
The area, B, of the base is:
B = (8 )· ( 8)
B = 64
So, the surface area, SA, of the pyramid is:
SA = 4 A + B
SA = 4 · (80) + 64
SA = 384
The surface area of the pyramid is 384cm².

Jeanne-Marie measured then recorded the lengths of the edges and slant height of this regular tetrahedron. What is its surface area to the nearest square centimetre?

Surface Area of a Right Pyramid
Answer:
The regular tetrahedron has 4 congruent faces . Each face is a triangle with base 9.0 cm and height 7.8 cm.
The area, A, of each face is:
A = ½( 9.0 cm)x (7.8 cm)
The surface area, SA, is:
SA = 4 x ½ (9.0 cm)x(7.8 cm)
SA = 140.4 cm²
The surface area of the tetrahedron is approximately
140 cm² .

POWERPOINT PRACTICE PROBLEM
Calculate the surface area of this regular tetrahedron to the nearest square metre.
(Answer: 43 m 2 )

A right rectangular pyramid has base dimensions 8 ft. by 10 ft., and a height of 16 ft. Calculate the surface area of the pyramid to the nearest square foot .
• There are 4 triangular faces and a rectangular base.
(What do you need to know to calculate the AREAS of each face?)
• Sketch the pyramid and label its vertices.
• Draw the slant heights on two adjacent triangles.
• Opposite triangular faces are congruent .
• In ∆EFH, FH is ½ the length of BC , so FH is 4 ft .
• EF is the height of the pyramid, which is 16 ft .
SA = Area of faces + Area of the base

To calculate AREA OF TRIANGULAR FACES (½ base x height) you need to know their SLANT HEIGHTS !

To calculate AREA OF TRIANGULAR FACES (½ base x height) you need to know their SLANT HEIGHTS !
Use the Pythagorean Theorem in right ∆EFH .
SA = Area of faces + Area of the base

To calculate AREA OF TRIANGULAR FACES (½ base x height) you need to know their SLANT HEIGHTS !
AREA of ∆EDC = ½ base x height (slant)
Also, AREA of ∆EAB = 5√272
SA = Area of faces + Area of the base

To calculate AREA OF TRIANGULAR FACES (½ base x height) you need to know their SLANT HEIGHTS !
Use the Pythagorean Theorem in right ∆EFG .
SA = Area of faces + Area of the base

To calculate AREA OF TRIANGULAR FACES (½ base x height) you need to know their SLANT HEIGHTS !
AREA of ∆EBC = ½ base x height (slant)
Also, AREA of ∆EAD = 4√281
SA = Area of faces + Area of the base

To calculate AREA OF TRIANGULAR FACES (½ base x height) you need to know their SLANT HEIGHTS !
AREA of the base □ DCBA = DC x CB
SA = Area of faces + Area of the base
SA = 5(√272) + 5(√272) + 54( √ 281) + 4( √ 281) + 80
= 379.0286
≈ 379 ft 2

POWERPOINT PRACTICE PROBLEM
A right rectangular pyramid has base dimensions 4 m by 6 m, and a height of 8 m. Calculate the surface area of the pyramid to the nearest square metre.

Surface Area of any
Right Pyramid with a
Regular Polygon Base
Each triangular face has base l and height s .
Area of each face:
A = ½ (base)(height)
A = ½(l)(s)
Area of 4 faces is:
= 4 [½ (l)(s)]
= 4(½ s)(l)
Area of 4 faces has a special name:
Lateral Area or
A
L

Surface Area of any Right Pyramid with a Regular
Polygon Base
A
L
= 4(½ s)(l) = (½ s)(4l)
• (4 l) is a perimeter of the base
So, SA of any pyramid with a polygon base:
SA = Lateral Area + Area of base
= A
L
+ Area of base
= (½ s)(perimeter of base) + Area of base

• is a 3-dimensional (3-D) object that has a circular base and a curved surface .
• MUST know the vocabulary!
• Height = the perpendicular distance from the apex to the base
• Slant height = the shortest distance on the curved surface between the apex and a point on the circumference of the base

A right cone has a base radius of 2 ft. and a height of 7 ft.
Calculate the surface area of this cone to the nearest square foot.

A right cone has a base radius of 2 ft. and a height of 7 ft.
Calculate the surface area of this cone to the nearest square foot.

POWERPOINT PRACTICE PROBLEM
A right cone has a base radius of 4 m and a height of 10 m. Calculate the surface area of this cone to the nearest square metre.

 PAGE: 34 - 35
 PROBLEMS: 4, 6, 8, 9, 10, 11, 18