Find the perimeter of quadrilateral
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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.

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A. 17.9 B. 22 C. 13.3 D. 9.1

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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).

A.17.9

B.22

C.13.3

D.9.1


Unit 1 lesson 4

Unit 1-Lesson 4

Area of two dimensional figures


Objective

Objective

  • I can recall area formulas for parallelograms, trapezoids, and triangles


Before area

Before Area…

  • We need to recall some important mathematical tools

  • Pythagorean Theorem

  • Right Angles / Triangles


Right angles

Right Angles

  • Right angles measure 90 degrees

  • Two segments / lines that form a right angle are perpendicular

This symbol indicates a right angle


Right triangle

Right Triangle

  • A right triangle has exactly one right angle

Can you find the right triangles?


A right triangle s best friend

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 squareof its hypotenuse


Example pythagorean theorem

ExamplePythagorean Theorem

  • Find the value of x

A = 6 B = 15 C = x

62 + 152 = x2

36 + 225 = x2

261 = x2

x =


A 17 9 b 22 c 13 3 d 9 1

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)


Area of a parallelogram

Area of a Parallelogram


Area of a parallelogram1

Area of a Parallelogram

  • Be careful!

NO! Base and height do not intersect at a right angle

Yes! Notice the height intersects the base


A 17 9 b 22 c 13 3 d 9 1

Find the area of

Area of a Parallelogram


A 17 9 b 22 c 13 3 d 9 1

Area of a Parallelogram

AreaFind the height of the parallelogram. The height forms a right triangle with points Sand T with base 12 in. and hypotenuse 20 in.

c2= a2 + b2Pythagorean Theorem

202= 122 + b2c = 20 and a = 12

400= 144 + b2Simplify.

256= b2Subtract 144 from each side.

16= bTake the square root of each side.


A 17 9 b 22 c 13 3 d 9 1

The height is 16 in. UT is the base, which measures 32 in.

Area of a Parallelogram

Continued

A= bhArea of parallelogram

= (32)(16) or 512 in2b = 32 and h = 16

Answer: The area is 512 in2.


A 17 9 b 22 c 13 3 d 9 1

A. Find the perimeter and area of

  • A

  • B

  • C

  • D

A.88 m; 255 m2

B.88 m; 405 m2

C.88 m; 459 m2

D.96 m; 459 m2


Area of a triangle

Area of a Triangle


A 17 9 b 22 c 13 3 d 9 1

Application

Perimeter and Area of a Triangle

SANDBOXYou 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?


A 17 9 b 22 c 13 3 d 9 1

Application

Perimeter and Area of a Triangle

Step 1Find the perimeter of the sandbox.

Perimeter = 16 + 12 + 7.5 or 35.5 ft

Step 2Find the area of the sandbox.

Area of a triangle

b = 12 and h = 9


A 17 9 b 22 c 13 3 d 9 1

boards

Application

Perimeter and Area of a Triangle

Step 3Use unit analysis to determine how many ofeach item are needed.

Boards

Bags of Sand

Answer: You will need 12 boards and 6 bags of sand.


Area of a trapezoid

Area of a Trapezoid


A 17 9 b 22 c 13 3 d 9 1

Area of a Trapezoid

SHAVINGFind 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


Area of a trapezoid1

Area of a Trapezoid

OPEN ENDEDMiguel 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.


A 17 9 b 22 c 13 3 d 9 1

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.


A 17 9 b 22 c 13 3 d 9 1

Use the Pythagorean Theorem to find ℓ.

a2 + b2=c2Pythagorean Theorem

42 + ℓ2=52Substitution

16 + ℓ2=25Simplify.

ℓ2=9Subtract 16 from each side.

ℓ=3Take the positive square rootof 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

Answer: So, the area of the

deck is 30 square feet.

Simplify.


Area of a rhombus or kite

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


Area of regions

8

10

4

14

8

12

Area of Regions

The area of a region is the sum of all of its non-overlapping parts.

A = ½(8)(10)

A= 40

A = (12)(10)

A= 120

A = (4)(8)

A=32

A = (14)(8)

A=112

Area = 40 + 120 + 32 + 112 = 304 sq. units


Other types of polygons

Other Types of Polygons

  • Regular Polygon – a polygon with sides

    that are all the same length and angles that are all the same measure


Areas of regular polygons

Areas of Regular Polygons

Perimeter = (6)(8) = 48

apothem =

Area = ½ (48)( ) = sq. units

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).

8


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