Review Surface Area SA of Right Rectangular Prisms and Cylinders A face of the prism Surface Area of Right Rectangular Prisms There are 6 faces Surface Area of Right Rectangular Prisms ID: 767351
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Review: Surface Area (SA) of Right Rectangular Prisms and Cylinders
A face of the prism Surface Area of Right Rectangular Prisms
There are 6 faces Surface Area of Right Rectangular Prisms
SA is the sum of the areas of its faces 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 Surface Area of Right Rectangular Prisms
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
So, SA = (2 x area of top face) + (2 x area of front face) + (2 x area of right face) Surface Area of Right Rectangular Prisms
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 2 circular bases Base 2 Curved surface is a rectangle Curved Surface Surface Area of a Cylinder
Area of Curved surface = 2 π r x h Area of a base = π r ² Area of 2 bases = 2 π r ² Surface Area of a Cylinder
Area of Curved surface = 2 π r x h Area of a base = π r ² Area of 2 bases = 2 π r ² SA of a cylinder = (2 π r x h) + 2 π r ² Surface Area of a Cylinder
What is the surface area, SA, of this right cylinder? SA = 2π r ² + 2 π rh = (2 × π × 2²) + (2 × π × 2 × 5) ≈ 87.9645 cm²
A face of the prism Review: Volume (V) of Right Rectangular Prisms and Cylinders
Volume is a space occupied by the prism Volume (V) of Right Rectangular Prisms
= (Area of the bottom face) x (height) = A x h Volume (V) of Right Rectangular Prisms
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 = (Area of a base) x (height) = ( π r²) x h Volume (V) of Right Cylinder
Determine the volume of this right cylinder V = area of a base x height = π r ² × h = π(5)² × 8 ≈ 628.3 cm³
HomeworkA Worksheet:Finish both sides by tomorrow
1.4 Surface Areas of Right Pyramids and Right Cones
Build Your Pyramids!
Right Pyramidis 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. WHAT IS A POLYGON ?
Polygons are 2-dimensional shapes. They are made of straight lines, and the shape is "closed" (all the lines connect ). WHAT IS A POLYGON ? Polygon (straight sides ) Not a Polygon (has a curve ) Not a Polygon (open, not closed)
The shape of the base determines the NAME of the pyramidRight Pyramid
MUST know the vocabulary!Apex = a point where triangular faces meetSlant height = a height of a triangular face Right Pyramid
Regular Polygon = Same sides and same angles If a base is Regular Polygon , then the triangular faces are congruent (same) 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
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 + Area of the base
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
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 SA = Area of faces + Area of the base
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 = 4A + BSA = 4· (80) + 64SA = 384 The surface area of the pyramid is 384cm². Surface Area of a Right Pyramid
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 ?
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 cm²The surface area of the tetrahedron is approximately 140 cm². Surface Area of a Right Pyramid
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 ! How can you calculate the slant height of each of the triangular faces?
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 How can you calculate the slant height of ∆EFH ?
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) SA = Area of faces + Area of the base Also, AREA of ∆EAB = 5√272 How can you calculate the slant height of ∆EFH ?
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 How can you calculate the slant height of ∆EFG ?
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) SA = Area of faces + Area of the base Also, AREA of ∆EAD = 4√281
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 ft2
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
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 AL
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 A rea + Area of base = A L + Area of base = (½ s)(perimeter of base) + Area of base Surface Area of any Right Pyramid with a Regular Polygon Base
Right Circular Coneis 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.
HOMEWORKPAGE: 34 - 35PROBLEMS: 4, 6, 8, 9, 10, 11, 18