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 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

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 Curved Surface Base 2

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 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 is a space occupied by 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 = (Area of a base) x (height) Volume (V) of Right Cylinder Volume = (Area of a base) x (height) = (πr²) 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

Finish both sides by tomorrow Homework A Worksheet: Finish both sides by tomorrow

1.4 Surface Areas of Right Pyramids and Right Cones

Build Your Pyramids!

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

WHAT 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 ). 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 pyramid 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 = 4A + 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 m2)

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! How can you calculate the slant height of ∆EFH? 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! How can you calculate the slant height of ∆EFH? 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! How can you calculate the slant height of ∆EFG? Use the Pythagorean Theorem in right ∆EFG. SA = Area of faces + Area of the base

AREA of ∆EBC = ½ base x height (slant) 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

AREA of the base □ DCBA = DC x CB 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 Surface Area of any Right Pyramid with a Regular Polygon Base

Surface Area of any Right Pyramid with a Regular Polygon Base AL = 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 = AL + Area of base = (½ s)(perimeter of base) + Area of base

Right Circular Cone 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 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 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.

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