9-2

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9-2 Tuesday 3/19 or Wednesday 3/20

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9-2. Tuesday 3/19 or Wednesday 3/20. Bell Work. Find the unknown side lengths in each special right triangle . 1. a 30°-60°-90° triangle with hypotenuse 2 ft. 2. a 45°-45°-90° triangle with leg length 4 in. 3. a 30°-60°-90° triangle with longer leg length 3m. The irrational number  - PowerPoint PPT Presentation

Transcript of 9-2

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9-2

Tuesday 3/19 or Wednesday 3/20

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Bell WorkFind the unknown side lengths in each special right triangle.

1. a 30°-60°-90° triangle with hypotenuse 2 ft

2. a 45°-45°-90° triangle with leg length 4 in.

3. a 30°-60°-90° triangle with longer leg length 3m

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A circle is the locus of points in a plane that are a fixed distance from a point called the center of the circle. A circle is named by the symbol and its center. A has radius r = AB and diameter d = CD.

The irrational number is defined as the ratio ofthe circumference C tothe diameter d, or

Solving for C gives the formulaC = d. Also d = 2r, so C = 2r.

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You can use the circumference of a circle to find its area. Divide the circle and rearrange the pieces to make a shape that resembles a parallelogram.

The base of the parallelogram is about half the circumference, or r, and the height is close to the radius r. So A r · r = r2.

The more pieces you divide the circle into, the more accurate the estimate will be.

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Example 1A: Finding Measurements of Circles

Find the area of K in terms of .

A = r2 Area of a circle.

Divide the diameter by 2 to find the radius, 3.A = (3)2

Simplify.A = 9 in2

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Example 1B: Finding Measurements of Circles

Find the radius of J if the circumference is (65x + 14) m.

Circumference of a circle

Substitute (65x + 14) for C.

Divide both sides by 2.

C = 2r

(65x + 14) = 2r

r = (32.5x + 7) m

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Example 1C: Finding Measurements of Circles

Find the circumference of M if the area is 25 x2 ft2

Step 1 Use the given area to solve for r. Area of a circle Substitute 25x2 for A.

Divide both sides by .

Take the square root of both sides.

A = r2

25x2 = r2

25x2 = r2

5x = r

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Example 1C Continued

Step 2 Use the value of r to find the circumference.

Substitute 5x for r.

Simplify.

C = 2(5x)

C = 10x ft

C = 2r

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On your own! # 1

Find the area of A in terms of in which C = (4x – 6) m.

A = r2 Area of a circle.

A = (2x – 3)2 m Divide the diameter by 2 to find the radius, 2x – 3.

A = (4x2 – 12x + 9) m2 Simplify.

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The key gives the best possible approximation for on your calculator.Always wait until the last step to round.

Helpful Hint

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Example 2: Cooking ApplicationA pizza-making kit contains three circular baking stones with diameters 24 cm, 36 cm, and 48 cm. Find the area of each stone. Round to the nearest tenth.

24 cm diameter 36 cm diameter 48 cm diameter

A = (12)2

≈ 452.4 cm2

A = (18)2

≈ 1017.9 cm2

A = (24)2

≈ 1809.6 cm2

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On your own! # 2

A drum kit contains three drums with diameters of 10 in., 12 in., and 14 in. Find the circumference of each drum.

10 in. diameter 12 in. diameter 14 in. diameter

C = d

C = (10)C = 31.4 in.

C = d

C = (12)C = 37.7 in.

C = d

C = (14)C = 44.0 in.

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The center of a regular polygon is equidistant from the vertices. The apothem is the distance from the center to a side. A central angle of a regular polygon has its vertex at the center, and its sides pass through consecutive vertices. Each central angle measure of a regular n-gon is

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Regular pentagon DEFGH has a center C, apothem BC, and central angle DCE.

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To find the area of a regular n-gon with side length s and apothem a, divide it into n congruent isosceles triangles.

The perimeter is P = ns.

area of each triangle:

total area of the polygon:

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Example 3A: Finding the Area of a Regular Polygon

Find the area of regular heptagon with side length 2 ft to the nearest tenth.

Step 1 Draw the heptagon. Draw an isosceles triangle with its vertex at the center of the heptagon. The central angle is .

Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.

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Example 3A Continued

Step 2 Use the tangent ratio to find the apothem.

The tangent of an angle is . opp. legadj. leg

Solve for a.

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Example 3A Continued

Step 3 Use the apothem and the given side length to find the area.

Area of a regular polygon

The perimeter is 2(7) = 14ft.

Simplify. Round to the nearest tenth. A 14.5 ft2

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The tangent of an angle in a right triangle is the ratio of the opposite leg length to the adjacent leg length. See page 525.

Remember!

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Example 3B: Finding the Area of a Regular Polygon

Find the area of a regular dodecagon with side length 5 cm to the nearest tenth.

Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.

Step 1 Draw the dodecagon. Draw an isosceles triangle with its vertex at the center of the dodecagon. The central angle is .

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Example 3B Continued

Solve for a.

The tangent of an angle is . opp. legadj. leg

Step 2 Use the tangent ratio to find the apothem.

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Example 3B Continued

Step 3 Use the apothem and the given side length to find the area.

Area of a regular polygon

The perimeter is 5(12) = 60 ft.

Simplify. Round to the nearest tenth. A 279.9 cm2

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On your own! # 3 Find the area of a regular octagon with a side length of 4 cm.

Step 1 Draw the octagon. Draw an isosceles triangle with its vertex at the center of the octagon. The central angle is .

Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.

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On your own! #3 Continued

Step 2 Use the tangent ratio to find the apothem

The tangent of an angle is . opp. legadj. leg

Solve for a.

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On your own! #3 Continued

Step 3 Use the apothem and the given side length to find the area.

Area of a regular polygon

The perimeter is 4(8) = 32cm.

Simplify. Round to the nearest tenth.

A ≈ 77.3 cm2

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Lesson Quiz: Part I

Find each measurement.

1. the area of D in terms of

2. the circumference of T in which A = 16 mm2

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Lesson Quiz: Part IIFind each measurement.

3. Speakers come in diameters of 4 in., 9 in., and 16 in. Find the area of each speaker to the nearest tenth.

Find the area of each regular polygon to the nearest tenth.

4. a regular nonagon with side length 8 cm

5. a regular octagon with side length 9 ft

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9-3

Tuesday 3/19 or Wednesday 3/20

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Bell WorkFind the area of each figure.

1. a rectangle in which b = 14 cm and h = 5 cm

2. a triangle in which b = 6 in. and h = 18 in.

3. a trapezoid in which b1 = 7 ft, b2 = 11 ft, andh = 3 ft

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A composite figure is made up of simpleshapes, such as triangles, rectangles,trapezoids, and circles. To find the area of a composite figure, find the areas of the simple shapes and then use the Area Addition Postulate.

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Example 1A: Finding the Areas of Composite Figures by Adding

Find the shaded area. Round to the nearest tenth, if necessary.

Divide the figure into parts.

area of half circle:

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Example 1A Continued

area of triangle:

area of the rectangle:

A = bh = 20(14) = 280 mm2

shaded area:

50 + 280 + 84 ≈ 521.1 mm2

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Example 1B: Finding the Areas of Composite Figures by Adding

Find the shaded area. Round to the nearest tenth, if necessary.

Divide the figure into parts. area of parallelogram:A = bh = 8(5)= 40ft2

area of triangle:

shaded area: 40 + 25 = 65 ft2

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On your own! #1 Find the shaded area. Round to the nearest tenth, if necessary.

Area of rectangle: A = bh = 37.5(22.5)

= 843.75 m2

Area of triangle:

= 937.5 m2Total shaded area is about 1781.3 m2.

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Example 2: Finding the Areas of Composite Figures by Subtracting

Find the shaded area. Round to the nearest tenth, if necessary.

area of a triangle:

area of the half circle:

Subtract the area of the half circle from the area of the triangle.

area of figure:234 – 10.125 ≈ 202.2 ft2

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Example 2: Finding the Areas of Composite Figures by Subtracting

Find the shaded area. Round to the nearest tenth, if necessary.

area of circle:A = r2 = (10)2 = 100 cm2

area of trapezoid:

area of figure: 100 –128 186.2 cm2

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On your own! # 2 Find the shaded area. Round to the nearest tenth, if necessary.

area of circle:A = r2 = (3)2 28.3 in2

area of square:

A = bh (4.24)(4.24) 18 in2

area of figure: 28.3 – 18 = 10.3 in2

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Example 3: Fabric Application

A company receives an order for 65 pieces of fabric in the given shape. Each piece is to be dyed red. To dye 6 in2 of fabric, 2 oz of dye is needed. How much dye is needed for the entire order?

To find the area of the shape in square inches, divide the shape into parts.

The two half circles have the same area as one circle.

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Example 3 ContinuedThe area of the circle is (1.5)2 = 2.25 in2.

The area of the square is (3)2 = 9 in2.

The total area of the shape is 2.25 + 9 ≈ 16.1 in2.

The total area of the 65 pieces is 65(16.1) ≈ 1044.5 in2.

The company will need 1044.5 ≈ 348 oz of dye for the entire order.

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On your own! # 3 The lawn that Katie is replacing requires 79 gallons of water per square foot per year. How much water will Katie save by planting the xeriscape garden?

Area times gallons of water

375.75(79) = 29,684.25

Subtract water used

29,684.25 – 6,387.75 =23,296.5 gallons saved.

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To estimate the area of an irregular shape, you can sometimes use a composite figure.

First, draw a composite figure that resembles the irregular shape.

Then divide the composite figure into simple shapes.

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Example 4: Estimating Areas of Irregular Shapes

Use a composite figure to estimate the shaded area. The grid has squares with a side length of 1 ft.

Draw a composite figure that approximates the irregular shape. Find the area of each part of the composite figure.

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Example 4 Continued

area of triangle a:

area of triangle b:

area of rectangle c:area of trapezoid d:

A = bh = (2)(1) = 2 ft2

Area of composite figure: 1 + 0.5 + 2 + 1.5 = 5 ft2

The shaded area is about 5 ft2.

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On your own! #4

Use a composite figure to estimate the shaded area. The grid has squares with side lengths of 1 ft.

Draw a composite figure that approximates the irregular shape. Find the area of each part of the composite figure.

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On your own! #4 Continued

area of triangle:

area of half circle:

area of rectangle:A = lw = (3)(2) = 6 ft2

The shaded area is about 12 ft2.

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Lesson Quiz: Part IFind the shaded area. Round to the nearest tenth, if necessary.

1.

2.

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Lesson Quiz: Part II

3. Mike is remodeling his kitchen. The countertop he wants costs $2.70 per square foot. How much will Mike have to spend on his remodeling project?

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9-4

Thursday 3/21

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Bell WorkUse the slope formula to determine the slope of each line.

1.

2.

3. Simplify

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Example 1A: Estimating Areas of Irregular Shapes in the Coordinate Plane

Estimate the area of the irregular shape.

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Example 1A Continued

Method 1: Draw a composite figure that approximates the irregular shape and find the area of the composite figure.

The area is approximately 4 + 5.5 + 2 + 3 + 3 + 4 + 1.5 + 1 + 6 = 30 units2.

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Example 1A Continued

Method 2: Count the number of squares inside the figure, estimating half squares. Use a for a whole square and a for a half square.

There are approximately 24 whole squares and 14 half squares, so the area is about

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On your own! #1

Estimate the area of the irregular shape.

There are approximately 33 whole squares and 9 half squares, so the area is about 38 units2.

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Example 2: Finding Perimeter and Area in the Coordinate Plane

Draw and classify the polygon with vertices E(–1, –1), F(2, –2), G(–1, –4), and H(–4, –3). Find the perimeter and area of the polygon.

Step 1 Draw the polygon.

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Example 2 Continued

Step 2 EFGH appears to be a parallelogram. To verify this, use slopes to show that opposite sides are parallel.

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Example 2 Continued

slope of EF =

slope of FG =

slope of GH =

slope of HE = The opposite sides are parallel, so EFGH is a parallelogram.

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Example 2 Continued

Step 3 Since EFGH is a parallelogram, EF = GH, and FG = HE.

Use the Distance Formula to find each side length.

perimeter of EFGH:

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Example 2 Continued

To find the area of EFGH, draw a line to divide EFGH into two triangles. The base and height of each triangle is 3. The area of each triangle is

The area of EFGH is 2(4.5) = 9 units2.

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On your own! #2 Draw and classify the polygon with vertices H(–3, 4), J(2, 6), K(2, 1), and L(–3, –1). Find the perimeter and area of the polygon.

Step 1 Draw the polygon.

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On your own! #2 Continued

Step 2 HJKL appears to be a parallelogram. To verify this, use slopes to show that opposite sides are parallel.

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On your own! #2 Continued

are vertical lines.

The opposite sides are parallel, so HJKL is a parallelogram.

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On your own! #2 Continued

Step 3 Since HJKL is a parallelogram, HJ = KL, and JK = LH.

Use the Distance Formula to find each side length.

perimeter of EFGH:

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On your own! #2 Continued

To find the area of HJKL, draw a line to divide HJKL into two triangles. The base and height of each triangle is 3. The area of each triangle is

The area of HJKL is 2(12.5) = 25 units2.

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Example 3: Finding Areas in the Coordinate Plane by Subtracting

Find the area of the polygon with vertices A(–4, 1), B(2, 4), C(4, 1), and D(–2, –2).

Draw the polygon and close it in a rectangle.

Area of rectangle:

A = bh = 8(6)= 48 units2.

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Example 3 Continued

Area of triangles:

The area of the polygon is 48 – 9 – 3 – 9 – 3 = 24 units2.

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On your own! #3

Find the area of the polygon with vertices K(–2, 4), L(6, –2), M(4, –4), and N(–6, –2).

Draw the polygon and close it in a rectangle.Area of rectangle:A = bh = 12(8)= 96 units2.

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On your own! #3 Continued

Area of triangles:

a b

d c

The area of the polygon is 96 – 12 – 24 – 2 – 10 = 48 units2.

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Example 4: Problem Solving Application

Show that the area does not change when the pieces arerearranged.

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Find the area of each shape.

top triangle: top triangle:

top rectangle: top rectangle:A = bh = 2(5) = 10 units2

Left figure Right figure

A = bh = 2(5) = 10 units2

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Lesson Quiz: Part I

1. Estimate the area of the irregular shape.

2. Draw and classify the polygon with vertices L(–2, 1), M(–2, 3), N(0, 3), and P(1, 0). Find the perimeter and area of the polygon.

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Lesson Quiz: Part II 3. Find the area of the polygon with vertices

S(–1, –1), T(–2, 1), V(3, 2), and W(2, –2).

4. Show that the two composite figures cover the same area.