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

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).

Unit 1-Lesson 4

Area of two dimensional figures

Objective

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

Before Area…

• We need to recall some important mathematical tools

• Pythagorean Theorem

• Right Angles / Triangles

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

• A right triangle has exactly one right angle

Can you find the right triangles?

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

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

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

• Be careful!

height

Base

Yes! Notice the height intersects the base

heightBaseNO! Base and

height do not intersect at a right angle

Area of a Parallelogram

Find the area of

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.

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.

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

Area of a Triangle

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?

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

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.

Area of a Trapezoid

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

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.

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.

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.

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

26

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

Other Types of Polygons

• Regular Polygon – a polygon with sides that are all the same length and angles that are all the same measure

28

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