geometry section 11-1/11-2

EXTENDING AREA, SURFACE AREA, AND VOLUME

CHAPTER 11/12

AREAS OF PARALLELOGRAMS, TRIANGLES, RHOMBI, AND

TRAPEZOIDS

SECTION 11-1 AND 11-2

ESSENTIAL QUESTIONS• How do you find perimeters and areas of

parallelograms?

• How do you find perimeters and areas of triangles?

• How do you find areas of trapezoids?

• How do you find areas of rhombi and kites?

VOCABULARY

1. Base of a Parallelogram:

2. Height of a Parallelogram:

3. Base of a Triangle:

4. Height of a Triangle:

5. Height of a Trapezoid:

VOCABULARY






Can be any side of a parallelogram

VOCABULARY







The perpendicular distance between any two parallel bases of a parallelogram

VOCABULARY








Can be any side of a triangle

VOCABULARY









The length of a segment perpendicular to a base to the opposite vertex

VOCABULARY









The length of a segment perpendicular to a base to the opposite vertex

The perpendicular distance between bases

EXAMPLE 1

Find the perimeter and area of .!RSTU

EXAMPLE 1


P = 2l + 2w

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

a2 = 256

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

a2 = 256

a2 = 256

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

a2 = 256

a2 = 256

a = 16

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

a2 = 256

a2 = 256

a = 16h = 16

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

a2 = 256

a2 = 256

a = 16h = 16

A = 32(16)

EXAMPLE 1


P = 2l + 2w

P = 2(32)+ 2(20)

P = 64 + 40

P = 104 in.

A = bh a2 +b2 = c2

a2 +122 = 202

a2 +144 = 400−144 −144

a2 = 256

a2 = 256

a = 16h = 16

A = 32(16)

A = 512 in2

EXAMPLE 2

Matt Mitarnowski needs to buy enough boards to make the frame of the triangular sandbox shown and enough sand to cover the bottom. If one of the boards is 3 feet long and one bag of sand

covers 9 square feet of the sandbox, how many boards and bags does he need to buy?

EXAMPLE 2



P = a +b + c

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83 Matt needs 12 boards.

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

A = 12bh

Matt needs 12 boards.

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

A = 12bh


A = 12(12)(9)

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

A = 12bh


A = 12(12)(9)

A = 54 ft2

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

A = 12bh


A = 12(12)(9)

A = 54 ft2

549

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

A = 12bh


A = 12(12)(9)

A = 54 ft2

549

= 6

EXAMPLE 2



P = a +b + cP = 12+16+ 7.5

P = 35.5 ft

35.53

≈11.83

A = 12bh


A = 12(12)(9)

A = 54 ft2

549

= 6

Matt needs 6 bags of sand.

POSTULATE 11.2

If two figures are congruent, then they have the same area.

EXAMPLE 3

Find the area of the trapezoid.

EXAMPLE 3


A = 12h(b

1+b

2)

EXAMPLE 3


A = 12h(b

1+b

2)

A = 12(1)(3+ 2.5)

EXAMPLE 3


A = 12h(b

1+b

2)

A = 12(1)(3+ 2.5)

A = 12(5.5)

EXAMPLE 3


A = 12h(b

1+b

2)

A = 12(1)(3+ 2.5)

A = 12(5.5)

A = 2.75 cm2

EXAMPLE 4

Fuzzy Jeff designed a deck shaped like the trapezoid shown. Find the area of the deck.

EXAMPLE 4


a2 +b2 = c2

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9 b = 3

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9 b = 3

b1= 9; b

2= 9− 3 = 6

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9 b = 3

b1= 9; b

2= 9− 3 = 6

A = 12h(b

1+b

2)

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9 b = 3

b1= 9; b

2= 9− 3 = 6

A = 12h(b

1+b

2)

A = 12(4)(6+ 9)

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9 b = 3

b1= 9; b

2= 9− 3 = 6

A = 12h(b

1+b

2)

A = 12(4)(6+ 9)

A = (2)(15)

EXAMPLE 4


a2 +b2 = c2

42 +b2 = 52

16+b2 = 25−16 −16

b2 = 9

b2 = 9 b = 3

b1= 9; b

2= 9− 3 = 6

A = 12h(b

1+b

2)

A = 12(4)(6+ 9)

A = (2)(15)

A = 30 ft2

EXAMPLE 5

Find the area of each rhombus or kite.

EXAMPLE 5


A = 12d

1d

2

EXAMPLE 5


A = 12d

1d

2

A = 12(7)(12)

EXAMPLE 5


A = 12d

1d

2

A = 12(7)(12)

A = 42 ft2

EXAMPLE 5


EXAMPLE 5


A = 12d

1d

2

EXAMPLE 5


A = 12d

1d

2

A = 12(14)(18)

EXAMPLE 5


A = 12d

1d

2

A = 12(14)(18)

A = 126 in2

EXAMPLE 6

One diagonal of a rhombus is half as long as the other diagonal. If the area of the rhombus is 64 square inches, what are the lengths

of the diagonals?

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

4(64) = ( 14x2 )4

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

4(64) = ( 14x2 )4

256 = x2

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

4(64) = ( 14x2 )4

256 = x2

256 = x2

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

4(64) = ( 14x2 )4

256 = x2

256 = x2

x = 16

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

4(64) = ( 14x2 )4

256 = x2

256 = x2

x = 16

d1= 16 in.

EXAMPLE 6


of the diagonals?

A = 12d

1d

2

d1= x

d2= 1

2x

64 = 12(x )( 1

2x )

64 = 14x2

4(64) = ( 14x2 )4

256 = x2

256 = x2

x = 16

d1= 16 in.

d2= 8 in.

PROBLEM SET

PROBLEM SET

p. 767 #1, 2, 5-9 all; p. 777 #1-13 odd

“The best preparation for tomorrow is doing your best today.” - H. Jackson Brown, Jr.

geometry section 11-1/11-2

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