+ journal chapter 9 and 10 majo díaz-duran. + areas of a square, rectangle, triangle,...
TRANSCRIPT
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+
Journal Chapter 9 and 10 Majo Díaz-Duran
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+areas of a square, rectangle, triangle, parallelogram, trapezoid, kite and rhombus:Square A2, a = length of side
Rectangle w × h w = width h = height
Triangle ½b × h b = base h = vertical height
Parallelogram b × h b = base h = vertical height
Trapezoid ½(b1+b2) × h h = vertical height
Kite ½d1d2
Rhombus ½d1d2
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+examples
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+find the area of a composite figure. Explain what a composite figure is To find the area you need to break it into individual
pieces.
A composite figure is- any figure made up from 2 or more polygons/circles.
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+Examples: By adding:
By Subtracting:
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+area of a circle:
You can use the circumference of a circle to find the are
Circumference: 2πr
Area: πr2
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+Examples:
The radius of a circle is 3 inches. What is the area?
A=πr2 A= π(3 in)2 A= π(9 in2) A= 28.26 in2
The diameter of a circle is 8
centimeters. What is the area?
D=2r8 cm = 2r 8 cm ÷ 2 =r
r = 4 cmA=πr2
A= π(4 cm)2 A= 50.24 cm2
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+what a solid is:
A three dimensional figure, can be made up of flat or curved surfaces, each flat surfaces is called a face, an
edge is the segment that is the intersection of two faces, a vertex is the point that is the intersection of
three or more faces.
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+Examples:
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+find the surface area of a prism. What is a prism? Explain what a “Net” is A prism is formed by two parallel congruent polygonal
faces called bases connected by faces that are parallelograms.
The surface area of a prism = right prism with lateral area(L) and base area(B) L + 2B.
A net is a diagram of the surfaces of a three-dimensional figure that can be folded to form the three-dimensional figure
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+Examples:
Net: Pyramid
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+find the surface area of a cylinder
A cylinder is formed by two parallel congruent circular vases and a curved surface that connects the bases.
Surface Area of a Cylinder = 2 πr 2 + 2πrh
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+Examples: Find the surface area of a cylinder with
a radius of 2 cm, and a height of 1 cm
S=2pir2+2pirh
S=2pi22+2pi(2)(1)
S=6.28(4)+6.28(2)
S=25.12+12.56
Surface area = 37.68 cm2
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+find the surface area of a pyramid:
A pyramid is formed by a polygonal base and triangular faces that meet at a common vertex
Lateral area(L) and base area(B) is L+B or P(perimeter)l +B
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+Example:
a square pyramid with a base that is 20 m on each side and a slant height of 40 m Find the surface area of the base and the lateral faces.Base: A=s2 or (20)2A=400 SA=400+4(400) SA=2000 m2
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+find the surface area of a cone.
A cone is formed by a circular base and a curved surface that connects the base to a vertex.
lateral are L and Base are B L +B or πrl+πr2
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+Examples:
a cone with a radius of 4 cm and a slant height of 12 cm:SA=pir2+pirLSA=pi(4)2+pi(4)(12) SA=50.3 +150.8SA =201.1 cm2
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+find the volume of a cube
The volume of a cube is (length of side)3.
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+Examples:
Example #2
Find the volume if the length of one side is 2 cm
V = 23
V = 2 × 2 × 2
V= 8 cm3
Example #3:
Find the volume if the length of one side is 3 cm
V= 33
V = 3 × 3 × 3
V = 27 cm3
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+Cavalieri’s principle
If two objects have the same cross sectional area and the same height they have the same volume.
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+examples
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+find the volume of a prism
Base area (B) and height (h) is V= Bh
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+Examples:
What is the volume of a prism whose ends have an area of 25 in2 and which is 12 in long:Answer: Volume = 25 in2 × 12 in = 300 in3
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+find the volume of a cylinder
With base are (B) radius ( r) and height h is V: Bh or V= πr2h
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+Examples: What is the volume of the
cylinder with a radius of 2 and a height of 6?Volume= Πr2h Volume = Π2(6) = 24Π
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+find the volume of a pyramid
Base area (B) and height(h) V= 1/3Bh
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+examples A square pyramid has a
height of 9 meters. If a side of the base measures 4 meters, what is the volume of the pyramid?
Since the base is a square, area of the base = 4 × 4 = 16 m2
Volume of the pyramid = (B × h)/3 = (16 × 9)/3 = 144/3 = 48 m3
A rectangular pyramid has a height of 10 meters. If the sides of the base measure 3 meters and 5 meters, what is the volume of the pyramid?
Since the base is a rectangle, area of the base = 3 × 5 = 15 m2
Volume of the pyramid = (B × h)/3 = (15 × 10)/3 = 150/3 = 50 m3
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+find the volume of a cone
Base area (B), radius ( r ) and height (h) v= 1/3Bh or v= 1/3πr2h
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+Examples:
Calculate the volume if r = 2 cm and h = 3 cm
V = 1/3 × pi × 22 × 3
V= 1/3 × pi × 4 × 3
V = 1/3 × pi × 12
V = 1/3 × 37.68
V = 1/3 × 37.68/1
V = (1 × 37.68)/(3 × 1)
V = 37.68/3
V= 12.56 cm3
Calculate the volume if r = 4 cm and h = 2 cm
V= 1/3 × pi × 42 × 2
V= 1/3 × pi× 16 × 2
V = 1/3 ×pi× 32
V= 1/3 × 100.48
V = 1/3 × 100.48/1
V= (1 × 100.48)/(3 × 1)
V= 100.48/3
V = 33.49 cm3
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+find the surface area of a sphere
surface area = 4πr2
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+examples Find the surface area of a sphere
with a radius of 6 cm
SA = 4 × pi × r2
SA = 4 × pi × 62
SA = 12.56 × 36
SA = 452.16
Surface area = 452.16 cm2 Find the surface area of a sphere with a radius of 2 cm
SA = 4 × pi × r2
SA = 4 × pi × 22
SA = 12.56 × 4
SA = 50.24
Surface area = 50.24 cm2
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+find the volume of a sphere
V= 4/3πr3
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+Examples:
If r = 300 mi (the moon), then the volume would be V = 4πr3/3 = 4(pi)(300 mi)3/3 = 4(pi)(27,000,000 mi3)/3 = 113,040,000 mi3.
If r = 4 cm (a marble), then the volume would be V = 4πr3/3 = 4(pi)(4 cm)3/3 = 4(pi)(64 cm3)/3 = 267.9 cm3