a. 17.9 b. 22 c. 13.3 d. 9.1
DESCRIPTION
Find the perimeter of quadrilateral WXYZ with vertices W (2, 4), X (–3, 3), Y (–1, 0), and Z (3, –1). A. 17.9 B. 22 C. 13.3 D. 9.1. Unit 1-Lesson 4. Area of two dimensional figures. Objective. I can recall area formulas for parallelograms, trapezoids, and triangles. - PowerPoint PPT PresentationTRANSCRIPT
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A. 17.9
B. 22
C. 13.3
D. 9.1
Find the perimeter of quadrilateral WXYZ with verticesW(2, 4), X(–3, 3), Y(–1, 0), and Z(3, –1).
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Unit 1-Lesson 4
Area of two dimensional figures
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Objective
• I can recall area formulas for parallelograms, trapezoids, and triangles
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Before Area…
• We need to recall some important mathematical tools
• Pythagorean Theorem
• Right Angles / Triangles
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Right Angles
• Right angles measure 90 degrees• Two segments / lines that form a right angle are
perpendicular
This symbol indicates a right angle
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Right Triangle
• A right triangle has exactly one right angle
Can you find the right triangles?
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A Right Triangle’s Best Friend
• The Pythagorean Theorem
• A2 + B2 = C2
• The sum of the squares of the legs of a right triangle is equal to the square of its hypotenuse
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ExamplePythagorean Theorem
• Find the value of x
x
15
6
A = 6 B = 15 C = x
62 + 152 = x2
36 + 225 = x2
261 = x2
x = 261 293
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Why?
• In order to find surface area, you need to use know the height of the figure
• Height – perpendicular distance (you will see a right angle)
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Area of a Parallelogram
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Area of a Parallelogram
• Be careful!
height
Base
Yes! Notice the height intersects the base
heightBaseNO! Base and
height do not intersect at a right angle
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Area of a Parallelogram
Find the area of
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Area of a Parallelogram
Area Find the height of the parallelogram. The height forms a right triangle with points S and T with base 12 in. and hypotenuse 20 in.
c2 = a2 + b2 Pythagorean Theorem
202 = 122 + b2 c = 20 and a = 12
400 = 144 + b2 Simplify.
256 = b2 Subtract 144 from each side.
16 = b Take the square root of each side.
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Area of a Parallelogram
Continued
A = bh Area of parallelogram
= (32)(16) or 512 in2 b = 32 and h = 16
The height is 16 in. UT is the base, which measures 32 in.
Answer: The area is 512 in2.
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A. AB. BC. CD. D
A. 88 m; 255 m2
B. 88 m; 405 m2
C. 88 m; 459 m2
D. 96 m; 459 m2
A. Find the perimeter and area of
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Area of a Triangle
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Application
Perimeter and Area of a Triangle
SANDBOX You need to buy enough boards to make the frame of the triangular sandbox shown and enough sand to fill it. If one board is 3 feet long and one bag of sand fills 9 square feet of the sandbox, how many boards and bags do you need to buy?
What is it asking for?
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Application
Perimeter and Area of a TriangleStep 1 Find the perimeter of the sandbox.
Perimeter = 16 + 12 + 7.5 or 35.5 ft
Step 2 Find the area of the sandbox.
Area of a triangle
b = 12 and h = 9
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Application
Perimeter and Area of a Triangle
Step 3 Use unit analysis to determine how many ofeach item are needed.
Boards
Bags of Sand
boards
Answer: You will need 12 boards and 6 bags of sand.
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Area of a Trapezoid
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Area of a Trapezoid
SHAVING Find the area of steel used to make
the side of the trapezoid shown below.
Area of a trapezoid
h = 1, b1 = 3, b2 = 2.5
Simplify.
Answer: A = 2.75 cm2
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Area of a Trapezoid
OPEN ENDED Miguel designed a deck shaped like the trapezoid shown below. Find the area of the deck.
Read the Problem
You are given a trapezoid with one base measuring 4 feet, a height of 9 feet, and a third side measuring 5 feet. To find the area of the trapezoid, first find the measure of the other base.
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Solve the Test Item
Draw a segment to form a right triangle and a rectangle. The triangle has a hypotenuse of 5 feet and legs of ℓ and 4 feet. The rectangle has a length of 4 feet and a width of x feet.
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Use the Pythagorean Theorem to find ℓ.
a2 + b2 = c2 Pythagorean Theorem42 + ℓ2 = 52 Substitution
16 + ℓ2 = 25 Simplify.ℓ2 = 9 Subtract 16 from each side.ℓ = 3 Take the positive square root
of each side.
By Segment Addition, ℓ + x = 9. So, 3 + x = 9 and x = 6. The width of the rectangle is also the measure of the second base of the trapezoid.
Area of a trapezoid
Substitution
Simplify.Answer: So, the area of the deck is 30 square feet.
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Area of a Rhombus or Kite
The area A of a rhombus or a kite is one half the product of the length of its diagonals, d1 and d2
A= d1d2
25
12
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Area of Regions
8
10
12
4 14
8
The area of a region is the sum of all of its non-overlapping parts.
A = ½(8)(10)
A= 40A = (12)(10)
A= 120
A = (4)(8)
A=32
A = (14)(8)
A=112
Area = 40 + 120 + 32 + 112 = 304 sq. units
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Other Types of Polygons
• Regular Polygon – a polygon with sides that are all the same length and angles that are all the same measure
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Areas of Regular Polygons
Perimeter = (6)(8) = 48 apothem =
Area = ½ (48)( ) = sq. units
8
If a regular polygon has an area of A square units, a perimeter of P units, and an apothem of a units, then A = ½ (a)(p).
4 34 3
4 3 96 3