What does the word “polygon” mean? Let's Discuss What does the word “polygon” mean? What is the smallest number of sides a polygon can have? What is the largest number of sides a polygon can have?

Polygons Poly means many and gon means angles. The word ‘polygon’ is a Greek word. Poly means many and gon means angles.

Polygons The word polygon means “many angles” A two dimensional object A closed figure

More about Polygons Made up of three or more straight line segments There are exactly two sides that meet at a vertex The sides do not cross each other

EXAMPLES

Triangle Octagon Quadrilateral Nonagon Pentagon Decagon Dodecagon Hexagon n-gon Heptagon Hendecagon

More Examples of Polygons

These are not Polygons

Terminology Side: One of the line segments that make up a polygon. Vertex: Point where two sides meet.

Vertex Side

Interior angle: An angle formed by two adjacent sides inside the polygon. Exterior angle: An angle formed by two adjacent sides outside the polygon.

Exterior angle Interior angle

Let us recapitulate Interior angle Diagonal Vertex Side Exterior angle

Types of Polygons Equiangular Polygon: a polygon in which all of the angles are equal Equilateral Polygon: a polygon in which all of the sides are the same length

Regular Polygon: a polygon where all the angles are equal and all of the sides are the same length. They are both equilateral and equiangular

Examples of Regular Polygons

A convex polygon: A polygon whose each of the interior angle measures less than 180°. If one or more than one angle in a polygon measures more than 180° then it is known as concave polygon. (Think: concave has a "cave" in it)

IN TERIOR ANGLES OF A POLYGON

Let us find the connection between the number of sides, number of diagonals and the number of triangles of a polygon.

3 x 180o = 540o 4 sides Quadrilateral 5 sides Pentagon 2 1 diagonal 2 diagonals 180o 180o 180o 180o 180o 180o 6 sides Hexagon 7 sides Heptagon/Septagon 4 4 x 180o = 720o 5 5 x 180o = 900o 3 diagonals 4 diagonals

Sum of the interior angles Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600

Sum of the interior angles Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600 Quadrilateral 4 2 2 x1800 = 3600 3600/4 = 900

Sum of the interior angles Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600 Quadrilateral 4 2 2 x1800 = 3600 3600/4 = 900 Pentagon 5 3 x1800 = 5400 5400/5 = 1080

Sum of the interior angles Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600 Quadrilateral 4 2 2 x1800 = 3600 3600/4 = 900 Pentagon 5 3 x1800 = 5400 5400/5 = 1080 Hexagon 6 4 x1800 = 7200 7200/6 = 1200

Sum of the interior angles Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600 Quadrilateral 4 2 2 x1800 = 3600 3600/4 = 900 Pentagon 5 3 x1800 = 5400 5400/5 = 1080 Hexagon 6 4 x1800 = 7200 7200/6 = 1200 Heptagon 7 5 x1800 = 9000 9000/7 = 128.30

Sum of the interior angles Each interior angle Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600 Quadrilateral 4 2 2 x1800 = 3600 3600/4 = 900 Pentagon 5 3 x1800 = 5400 5400/5 = 1080 Hexagon 6 4 x1800 = 7200 7200/6 = 1200 Heptagon 7 5 x1800 = 9000 9000/7 = 128.30 “n” sided polygon n Association with no. of sides Association with no. of triangles Association with sum of interior angles

Sum of the interior angles Regular Polygon No. of sides No. of diagonals No. of Sum of the interior angles Each interior angle Triangle 3 1 1800 1800/3 = 600 Quadrilateral 4 2 2 x1800 = 3600 3600/4 = 900 Pentagon 5 3 x1800 = 5400 5400/5 = 1080 Hexagon 6 4 x1800 = 7200 7200/6 = 1200 Heptagon 7 5 x1800 = 9000 9000/7 = 128.30 “n” sided polygon n n - 3 n - 2 (n - 2) x1800 (n - 2) x1800 / n

Polygons with 3 sides… Triangles Polygons with 4 sides… Quadrilaterals Polygons with 5 sides.. Pentagons But wait we have more polygons Polygons with 6 sides… Hexagons Polygons with 7 sides… Heptagons Polygons with 8 sides… Octagons But still we have more polygons Polygons with 9 sides… Nonagons Polygons with 10 sides… Decagons Polygons with 12 sides… Dodecagons And now we have our polygons

1 Calculate the Sum of Interior Angles and each interior angle of each of these regular polygons. 7 sides Septagon/Heptagon Sum of Int. Angles 900o Interior Angle 128.6o 2 3 4 9 sides 10 sides 11 sides Nonagon Decagon Hendecagon Sum 1260o I.A. 140o Sum 1440o I.A. 144o Sum 1620o I.A. 147.3o

Find the unknown angles below. Diagrams not drawn accurately. Find the unknown angles below. x 115o 110o 75o 95o 75o 100o 70o w 2 x 180o = 360o 3 x 180o = 540o 360 – 245 = 115o 540 – 395 = 145o y 117o 121o 100o 125o 140o z 133o 137o 138o 125o 105o 4 x 180o = 720o 5 x 180o = 900o 720 – 603 = 117o 900 – 776 = 124o

EXTERIOR ANGLES OF A POLYGON

An exterior angle of a regular polygon is formed by extending one side of the polygon. Angle CDY is an exterior angle to angle CDE D E Y B C A F 1 2 Exterior Angle + Interior Angle of a regular polygon =1800

Sum of exterior angles = 360º No matter what type of polygon we have, the sum of the exterior angles is ALWAYS equal to 360º.   Sum of exterior angles = 360º Polygons

In a regular polygon with ‘n’ sides Sum of interior angles = (n -2) x 1800 i.e. 2(n – 2) x right angles Exterior Angle + Interior Angle =1800 Each exterior angle = 3600/n No. of sides = 3600/exterior angle                Polygons

Let us explore few more problems Find the measure of each interior angle of a polygon with 9 sides. Ans : 1400 Find the measure of each exterior angle of a regular decagon. Ans : 360 How many sides are there in a regular polygon if each interior angle measures 1650? Ans : 24 sides Is it possible to have a regular polygon with an exterior angle equal to 400 ? Ans : Yes Polygons

Review

Important Terms CONSECUTIVE VERTICES are two endpoints of any side. A VERTEX is the point of intersection of two sides CONSECUTIVE VERTICES are two endpoints of any side. F A B C D E A segment whose endpoints are two nonconsecutive vertices is called a DIAGONAL. Sides that share a vertex are called CONSECUTIVE SIDES.

Polygons are named by listing its vertices consecutively. F E D

More Important Terms EQUILATERAL - All sides are congruent EQUIANGULAR - All angles are congruent REGULAR - All sides and angles are congruent

Three more important terms Interior Angles Exterior Angles SUPPLEMENTARY the SUM of an interior angle and it’s corresponding exterior angle = 180o

Polygons can be CONCAVE or CONVEX

Classify each polygon as convex or concave.

It’s what’s inside that counts  Finding and using the interior angle measures of polygons

What is the sum of the measures of the interior angles of a triangle? 180° 180° 180° What is the sum of the measures of the interior angles of any quadrilateral? 360°

Sum of measures of interior angles # of triangles Sum of measures of interior angles # of sides 1(180) = 180 3 1 2(180) = 360 4 2 3 3(180) = 540 5 6 4 4(180) = 720 n-2 (n-2) 180 n

If a convex polygon has n sides, then the sum of the measure of the interior angles is (n – 2)(180°)

If you know the sum of the angles of a regular polygon, how can you find the measure of one of the congruent angles? Regular Polygon Interior angle sum Measure of one angle Triangle 180o Quadrilateral 360o Pentagon Hexagon Heptagon Octagon Decagon Dodecagon n-gon

If a regular convex polygon has n sides, then the measure of one of the interior angles is

Perimeter and Area

Area Definition: The number of square units needed to cover a surface. (INSIDE) Length x width

Perimeter Definition: The sum of the length of all the sides of a polygon.

Base Definition: A side or face of a figure on which the figure stands.

Height Definition: The distance from the bottom to the top of a figure.

Width Definition: How wide a figure is from side to side.

Length Definition: The measure of the distance across an figure.

Warm Up Graph the line segment for each set of ordered pairs. Then find the length of the line segment. 1. (–7, 0), (0, 0) 2. (0, 3), (0, 6) 3. (–4, –2), (1, –2) 4. (–5, 4), (–5, –2) 7 units 3 units 5 units 6 units

Vocabulary perimeter area base height composite figure

Perimeter is the distance around a polygon Perimeter is the distance around a polygon. To find the perimeter of any polygon, you add the lengths of all its sides. Since opposite sides of a parallelogram are equal in length, you can find a formula for the perimeter of a parallelogram. P = w + l + w + l = w + w + l + l = 2w + 2l w l

Additional Example 1: Finding the Perimeter of Parallelograms A. Find the perimeter of the figure. 5 14 P = 2w + 2l Perimeter of a parallelogram. = 2(5) + 2(14) Substitute 5 for w and 14 for l. = 10 + 28 = 38 units

Additional Example 1: Finding the Perimeter of Parallelograms B. Find the perimeter of the figure. 20 16 P = 2w + 2l Perimeter of a parallelogram. = 2(16) + 2(20) Substitute 16 for w and 20 for l. = 32 + 40 = 72 units

Check It Out! Example 1 A. Find the perimeter of the figure. 6 11 P = 2w + 2l Perimeter of a parallelogram. = 2(6) + 2(11) Substitute 6 for w and 11 for l. = 12 + 22 = 34 units

Check It Out! Example 1 B. Find the perimeter of the figure. 5 13 P = 2w + 2l Perimeter of a parallelogram. = 2(5) + 2(13) Substitute 5 for w and 13 for l. = 10 + 26 = 36 units

The area of a plane figure is the number of unit squares needed to cover the figure. The base of a parallelogram is the length of one side. The height is the perpendicular distance from the base to the opposite side. Height Side Base

While perimeter is expressed in linear units, such as inches (in While perimeter is expressed in linear units, such as inches (in.) or meters (m), area is expressed in square units, such as square feet (ft2). You can cut a parallelogram and shift the cut piece to form a rectangle whose base and height are the same as those of the original parallelogram. The same number of unit squares are needed to cover the two figures. So a parallelogram and a rectangle that have the same base and height have the same area.

Since the base and height of a rectangle are the same as its length and width, the formula for the area of a rectangle can also be written as A = lw. Helpful Hint

Additional Example 2: Using a Graph to Find Area Graph and find the area of the figure with the given vertices. A. (–1, –2), (2, –2), (2, 3), (–1, 3) Area of a rectangle. A = bh Substitute 3 for b and 5 for h. A = 3 • 5 A = 15 units2

The height of a parallelogram is not the length of its slanted side The height of a parallelogram is not the length of its slanted side. The height of a figure is always perpendicular to the base. Caution!

Additional Example 2: Using a Graph to Find Area Graph and find the area of the figure with the given vertices. B. (0, 0), (5, 0), (6, 4), (1, 4) Area of a parallelogram. A = bh Substitute 5 for b and 4 for h. A = 5 • 4 A = 20 units2

Graph and find the area of the figure with the given vertices. Check It Out! Example 2 Graph and find the area of the figure with the given vertices. A. (–3, –2), (1, –2), (1, 3), (–3, 3) x y (–3, –2) (1, –2) (1, 3) (–3, 3) 4 5 Area of a rectangle. A = bh Substitute 4 for b and 5 for h. A = 4 • 5 A = 20 units2

Area of a parallelogram. Check It Out! Example 2 Graph the figure with the given vertices. Then find the area of the figure. B. (–1, –1), (3, –1), (5, 3), (1, 3) (5, 3) x y (–1, –1) (3, –1) (1, 3) 4 Area of a parallelogram. A = bh Substitute 4 for b and 4 for h. A = 4 • 4 A = 16 units2

A composite figure is made up of basic geometric shapes such as rectangles, triangles, trapezoids, and circles. To find the area of a composite figure, find the areas of the geometric shapes and then add the areas.

Additional Example 3: Finding Area and Perimeter of a Composite Figure Find the perimeter and area of the figure. 6 6 3 3 6 5 5 The length of the side that is not labeled is the same as the sum of the lengths of the sides opposite, 18 units. P = 5 + 6 + 3 + 6 + 3 + 6 + 5 + 18 = 52 units

Additional Example 3 Continued 6 6 3 3 6 5 5 A = 6 • 5 + 6 • 2 + 6 • 5 Add the areas together. = 30 + 12 + 30 = 72 units2

Find the perimeter of the figure. Check It Out! Example 3 Find the perimeter of the figure. The length of the side that is not labeled is 2. 2 4 6 7 7 2 6 2 P = 6 + 2 + 4 + 7 + 6 + 4 + 2 + 2 + 2 + 7 ? = 42 units 4

Check It Out! Example 3 Continued 2 Find the area of the figure. 4 6 7 Add the areas together. A = 2 • 6 + 7 • 2 + 2 • 2 + 4 • 2 7 2 2 6 = 12 + 14 + 4 + 8 2 2 = 38 units2 2 6 4 2 4 7 2 2 + + +

Lesson Quiz: Part I 1. Find the perimeter of the figure. 44 ft 2. Find the area of the figure. 108 ft2

Lesson Quiz: Part II Graph and find the area of each figure with the given vertices. 3. (–4, 2), (6, 2), (6, –3), (–4, –3) 50 units2

Lesson Quiz: Part III Graph and find the area of each figure with the given vertices. 4. (4, –2), (–2, –2), (–3, 5), (3, 5) 42 units2

Warm Up A rectangle has sides lengths of 12 ft and 20 ft. 1. Find the perimeter. 64 ft 2. Find the area. 240 ft2

Additional Example 1: Using Perimeter Find the missing measurement when the perimeter is 71 in. 18 in. 15 in. d 22 in. P = 22 + 15 + 18 + d 71 = 55 + d Substitute 71 for P. -55 -55 Subtract 55 from both sides. 16 = d d = 16 in.

Check It Out! Example 1 Find the missing measurement when the perimeter is 58 in. 14 in. 7 in. d 28 in. P = 28 + 7 + 14 + d 58 = 49 + d Substitute 58 for P. -49 -49 Subtract 49 from both sides. 9 = d d = 9 in.

Additional Example 2: Multi-Step Application A homeowner wants to plant a border of shrubs around her yard that is in the shape of a right triangle. She knows that the length of the shortest side of the yard is 12 feet and the length of the longest side is 20 feet. How long will the border be? Step 1: Find the length of the third side. a2 + b2 = c2 Use the Pythagorean Theorem. 122 + b2 = 202 Substitute 12 for a and 20 for c. 144 + b2 = 400 b2 = 256 b = 16 √256 = 16.

Additional Example 3 Continued Step 2: Find the perimeter of the yard. P = a + b + c = 12 + 20 + 16 Add all sides. = 48 The border will be 48 feet long.

Check It Out! Example 2 A gardener wants to plant a border of flowers around the building that is in the shape of a right triangle. He knows that the length of the shortest sides of the building are 38 feet and 32 feet. How long will the border be? Step 1: Find the length of the third side. a2 + b2 = c2 Use the Pythagorean Theorem. 382 + 322 = c2 Substitute 38 for a and 32 for b. 1444 + 1024 = c2 2468 = c2 c ≈ 49.68 √2468  49.68.

Check It Out! Example 2 Continued Step 2: Find the perimeter of the yard. P = a + b + c  38 + 32 + 49.68 Add all sides.  119.68 The border will be about 119.68 feet long.

area of a triangle or trapezoid have as a factor. A triangle or trapezoid can be thought of as half of a parallelogram. Therefore, the formulas for the area of a triangle or trapezoid have as a factor. 1 2

Additional Example 3: Finding the Area of Triangles and Trapezoids Graph and find the area of the figure with the given vertices. A. (–2, 2), (4, 2), (0, 5) Area of a triangle y A = bh 1 2 (0, 5) Substitute for b and h. = • 6 • 3 1 2 3 (–2, 2) (4, 2) 6 = 9 units2 x

Additional Example 3: Finding the Area of Triangles and Trapezoids Graph and find the area of the figure with the given vertices. B. (–1, 1), (4, 1), (4, 4), (0, 4) y Area of a trapezoid A = h(b1 + b2) 1 2 (0, 4) 4 (4, 4) 3 Substitute for h, b1, and b2. (–1, 1) (4, 1) x = • 3(5 + 4) 1 2 5 = 13.5 units2

1 A = h(b1 + b2) 2 1 = • 6(8 + 4) 2 Check It Out! Example 3 Graph and find the area of the figure with the given vertices. A. (–1, –2), (5, –2), (5, 2), (–1, 6) y (–1, 6) Area of a trapezoid A = h(b1 + b2) 1 2 (5, 2) Substitute for h, b1, and b2. 8 6 4 x = • 6(8 + 4) 1 2 (–1, –2) (5, –2) = 36 units2

1 A = bh 2 1 = • 6 • 4 2 Check It Out! Example 3 Graph and find the area of the figure with the given vertices. B. (–1, 1), (5, 1), (1, 5) Area of a triangle y A = bh 1 2 (1, 5) Substitute for b and h. = • 6 • 4 1 2 4 (–1, 1) (5, 1) = 12 units2 x 6

1. the perimeter of the triangle 36 cm Lesson Quiz Use the figure to find the following measurements. 1. the perimeter of the triangle 36 cm 2. the perimeter of the trapezoid 44 cm 3. the perimeter of the combined figure 64 cm 4. the area of the triangle 54 cm2 5. the area of the trapezoid 104 cm2

Circles

Warm Up 1. Two angles are supplementary. One angle measures 61°. Find the measure of the other angle. 2. Two angles are complementary. One angle measures 19°. Find the measure of the other angle. 3. Two angles are supplementary and congruent. Find the measure of each angle. 119° 71° 90°

Vocabulary circle center of a circle arc radius diameter chord central angle sector

A circle is the set of all points in a plane that are the same distance from a given point, called the center of a circle. This distance is called the radius of the circle. A circle is named by its center. For example, if point A is the center of a circle, then the name of the circle is circle A. There are special names for the different parts of a circle.

Diameter Arc Part of a circle named by its endpoints Line segment that passes through the center of a circle, and whose endpoints lie on the circle Radius Line segment whose endpoints are the center of a circle and any point on the circle Chord Line segment whose endpoints are any two points on a circle

Additional Example 1: Identifying Parts of Circles Name the parts of circle M. N A. radii: MN, MR, MQ, MO O P B. diameters: NR, QO M Q C. chords: NR, QO, QN, NP R Radii is the plural form of radius. Reading Math

Name the parts of circle M. Check It Out! Example 1 Name the parts of circle M. B A C A. radii: GB, GA, GF, GD G B. diameters: BF, AD D H F C. chords: AH, AB, CE, BF, AD E

A central angle of a circle is an angle formed by two radii. A sector of a circle is the part of the circle enclosed by two radii and an arc connecting them. Sector ) The sum of the measures of all of the central angles in a circle is 360°. We say that there are 360° in a circle. Central angle

Additional Example 2: Problem Solving Application The circle graph shows the results of a survey about favorite types of muffins. Find the central angle measure of the sector that shows the percent of people whose favorite type of muffin is blueberry.

Additional Example 2 Continued 1 Understand the Problem List the important information: • The percent of people whose favorite muffin is blueberry is 40%.

Additional Example 2 Continued Make a Plan The central angle measure of the sector that represents this group is 40% of the angle measure of the whole circle. The angle measure of a circle is 360°. Since the sector is 40% of the circle graph, the central angle is 40% of the 360° in the circle. 40% of 360° = 0.40 · 360° Solve 3 0.40 · 360° = 144° Multiply. The central angle of the sector measures 144°.

Additional Example 2 Continued 4 Look Back The 40% sector is less than half the graph, and 144° is less than half of 360°. Since 144° is close to 180°, the answer is reasonable.

Check It Out! Example 2 The circle graph shows the results of a survey about favorite types of muffins. Find the central angle measure of the sector that shows the percent of people whose favorite type of muffin is banana nut.

Check It Out! Example 2 Continued 1 Understand the Problem List the important information: • The percent of people whose favorite muffin is banana nut is 35%.

Check It Out! Example 2 Continued Make a Plan The central angle measure of the sector that represents this group is 35% of the angle measure of the whole circle. The angle measure of a circle is 360°. Since the sector is 35% of the circle graph, the central angle measure is 35% of the 360° in the circle. 35% of 360° = 0.35 · 360° Solve 3 0.35 · 360° = 126° Multiply. The central angle of the sector measures 126°.

Check It Out! Example 2 Continued 4 Look Back The 35% sector is about one-third the graph, and 126° is about one-third of 360°. Since 126° is close to 120°, the answer is reasonable.

Lesson Quiz Name the parts of circle B. 1. radii 2. diameter(s) 3. chord(s) 4. BA, BC AC DE, FE, AC A pie is cut into 8 equal sectors. Find the measure of the central angle of one slice of pie. 45°

Vocabulary circumference

d = 2r Radius Center The diameter d is twice the radius r. Diameter Circumference The circumference of a circle is the distance around the circle.

What is pi? Pi=3.14159… Pi is the mathematical constant whose value is the ratio of a circle’s circumference to its diameter. Circumference = The distance around a circle Diameter = The width of a circle.

Remember! Pi () is an irrational number that is often approximated by the rational numbers 3.14 and . 22 7

Additional Example 1: Finding the Circumference of a Circle Find the circumference of each circle, both in terms of  and to the nearest tenth. Use 3.14 for . A. circle with a radius of 4 m C = 2r = 2(4) = 8 m  25.1 m B. circle with a diameter of 3.3 ft C = d =  (3.3) = 3.3 ft  10.4 ft

Check It Out! Example 1 Find the circumference of each circle, both in terms of  and to the nearest tenth. Use 3.14 for . A. circle with a radius of 8 cm C = 2r = 2(8) = 16 cm  50.2 cm B. circle with a diameter of 4.25 in. C = d = (4.25) = 4.25 in.  13.3 in.

Additional Example 2: Finding the Area of a Circle Find the area of each circle, both in terms of  and to the nearest tenth. Use 3.14 for . A. circle with a radius of 4 in. A = r2 = (42) = 16 in2  50.2 in2 B. circle with a diameter of 3.3 m d 2 = 1.65 A = r2 = (1.652) = 2.7225 m2  8.5 m2

Check It Out! Example 2 Find the area of each circle, both in terms of  and to the nearest tenth. Use 3.14 for . A. circle with a radius of 8 cm A = r2 =  (82) = 64 cm2  201.0 cm2 B. circle with a diameter of 2.2 ft d 2 = 1.1 A = r2 =  (1.12) = 1.21 ft2  3.8 ft2

Additional Example 3: Finding the Area and Circumference on a Coordinate Plane Graph the circle with center (–2, 1) that passes through (1, 1). Find the area and circumference, both in terms of  and to the nearest tenth. Use 3.14 for . A = r2 C = d = (32) = (6) = 9 units2 = 6 units  28.3 units2  18.8 units

Check It Out! Example 3 Graph the circle with center (–2, 1) that passes through (–2, 5). Find the area and circumference, both in terms of  and to the nearest tenth. Use 3.14 for . y A = r2 C = d (–2, 5) = (42) = (8) 4 = 16 units2 = 8 units  50.2 units2  25.1 units x (–2, 1)

Additional Example 4: Measurement Application A Ferris wheel has a diameter of 56 feet and makes 15 revolutions per ride. How far would someone travel during a ride? Use for . 22 7 C = d = (56) Find the circumference.  (56)  22 7 56 1 22 7  176 ft The distance is the circumference of the wheel times the number of revolutions, or about 176  15 = 2640 ft.

22 7 22 22 14  (14)   44 in. 7 7 1 Check It Out! Example 4 A second hand on a clock is 7 in. long. What is the distance it travels in one hour? Use for . 22 7 C = d =  (14) Find the circumference. 14 1 22 7 12 3 6 9  (14)  22 7  44 in. The distance is the circumference of the clock times the number of revolutions, or about 44  60 = 2640 in.

Lesson Quiz Find the circumference of each circle, both in terms of  and to the nearest tenth. Use 3.14 for . 1. radius 5.6 m 11.2 m; 35.2 m 2. diameter 113 m 113 mm; 354.8 mm Find the area of each circle, both in terms of  and to the nearest tenth. Use 3.14 for . 3. radius 3 in. 9 in2; 28.3 in2 4. diameter 1 ft 0.25 ft2; 0.8 ft2

Warm Up Find each area. Round to the nearest tenth, if necessary. 1. a rectangle with length 10 cm and width 4 cm 2. a parallelogram with base 18 ft and height 12 ft 3. a triangle with base 16 cm and height 8 cm 4. a circle with radius 5 in. 40 cm2 216 ft2 64 cm2 78.5 in2

Additional Example 1A: Finding the Area of Composite Figures by Adding Find the shaded area. Round to the nearest tenth, if necessary. 2 m Divide the figure into a rectangle and a trapezoid. 8 m 2 m 4 m 8 m 6 m 6 m area of the rectangle: 12 m A = bh A = 12 • 6 A = 72 m2

Additional Example 1A Continued Find the shaded area. Round to the nearest tenth, if necessary. area of the trapezoid: 2 m A = h(b1 + b2) 1 2 __ 8 m 2 m 4 m A = 2(4 + 2) 1 2 __ 6 m 6 m A = (12) 1 2 __ A = 6 m2 Add the area of the rectangle and the area of the trapezoid. total area: A = 72 + 6 = 78 m2.

Additional Example 1B: Finding the Area of Composite Figures by Adding Find the shaded area. Round to the nearest tenth, if necessary. Divide the figure into a rectangle and a semicircle. 12 in. 8 in. 8 in. 8 in. area of the rectangle: 20 in. A = bh A = 20 • 8 A = 160 in2

Additional Example 1B Continued Find the shaded area. Round to the nearest tenth, if necessary. area of the semicircle: 12 in. 8 in. A = (pr2) 1 2 __ 8 in. A  (3.14 • 42) 1 2 __ 20 in. A  (50.24) 1 2 __ A  25.1 in2 Add the area of the rectangle and the area of the semicircle. total area: A = 160 + 25.1 = 185.1 in2.

Find the shaded area. Round to the nearest tenth, if necessary. Check It Out! Example 1A Find the shaded area. Round to the nearest tenth, if necessary. 10 yd 9 yd 8 yd Divide the figure into a rectangle and a triangle. area of the rectangle: A = bh A = 8 • 9 A = 72 yd2

Check It Out! Example 1A Continued Find the shaded area. Round to the nearest tenth, if necessary. area of the triangle: A = bh 1 2 __ 10 yd 9 yd 8 yd A = (2 • 9) 1 2 __ A = (18) 1 2 __ A = 9 yd2 Add the area of the rectangle and the area of the triangle. total area: A = 72 + 9 = 81 yd2

Find the shaded area. Round to the nearest tenth, if necessary. Check It Out! Example 1B Find the shaded area. Round to the nearest tenth, if necessary. Divide the figure into a rectangle and a semicircle. 12 m 10 m 10 m area of the rectangle: 22 m A = bh A = 22 • 10 A = 220 m2

Check It Out! Example 1B Continued Find the shaded area. Round to the nearest tenth, if necessary. area of the semicircle: A = (pr2) 1 2 __ 12 m 10 m A  (3.14 • 52) 1 2 __ 10 m A  (78.5) 1 2 __ 22 m A  39.3 m2 Add the area of the rectangle and the area of the semicircle. total area: A = 220 + 39.3 = 259.3 m2

Subtract the area of the triangle from the area of the rectangle. Additional Example 2: Finding the Area of Composite Figures by Subtracting 5 ft Find the shaded area. 6 ft 9 ft Subtract the area of the triangle from the area of the rectangle. 12 ft Area of the rectangle: Area of the triangle: A = bh 1 2 __ A = bh A = 12 • 9 = 108 ft2 A = • (6)(7) 1 2 __ A = • 42 = 21 ft2 1 2 __ Shaded area: A = 108 – 21 = 87 ft2

Subtract the area of the triangle from the area of the rectangle. Check It Out! Example 2 9 in. Find the shaded area. 5 in. 8 in. Subtract the area of the triangle from the area of the rectangle. 16 in. Area of the rectangle: Area of the triangle: A = bh 1 2 __ A = bh A = 16 • 8 = 128 in2 A = • (5)(7) 1 2 __ A = • 35 = 17.5 in2 1 2 __ Shaded area: A = 128 – 17.5 = 110.5 in2

Additional Example 3: Landscaping Application 6 ft What is the area of the room floor shown in the figure? Round to the nearest tenth. 12 ft 12 ft 6 ft 18 ft To find the area, divide the composite figure into a square, a rectangle, and a semicircle.

Additional Example 3 Continued 6 ft Area of the square: 12 ft 12 ft A = s2 6 ft A = 62 = 36 ft2 18 ft Area of the rectangle: Area of the semicircle: A = pr2 1 2 __ A = bh A = 18 • 6 = 108 ft2 A  • 3.14 • (9)2 1 2 __ A  (254.34)  127.17 ft2 1 2 __

Additional Example 3 Continued 6 ft What is the area of the room floor shown in the figure? Round to the nearest tenth. 12 ft 12 ft 6 ft 18 ft Area of the room: A = 36 + 108 + 127.17 = 271.17 ft2 The area of the room is approximately 271.2 ft2.

Check It Out! Example 3 5 ft What is the area of the stage floor shown in the figure? Round to the nearest tenth. 10 ft 10 ft 5 ft 15 ft To find the area, divide the composite figure into a square, a rectangle, and a semicircle.

Check It Out! Example 3 Continued 5 ft Area of the square: 10 ft 10 ft A = s2 5 ft A = 52 = 25 ft2 15 ft Area of the rectangle: Area of the semicircle: A = pr2 1 2 __ A = bh A = 15 • 5 = 75 ft2 A  • 3.14 • (7.52) 1 2 __ A  (314)  88.3 ft2 1 2 __

Check It Out! Example 3 Continued 5 ft What is the area of the room floor shown in the figure? Round to the nearest tenth. 10 ft 10 ft 5 ft 15 ft Area of the room: A = 25 + 75 + 88.3 = 188.3 ft2 The area of the room is 188.3 ft2.

Lesson Quiz: Part l Find the perimeter and area of each figure. Round to the nearest tenth, if necessary. 1. 2. 3. 111 ft2 5 ft 45.9 in2 188.9 mm2

Lesson Quiz: Part ll 4. The figure gives the dimensions of a backdrop for a school play. Find the area of the backdrop. 97 ft2

Warm Up Find the area of each figure. Round to the nearest tenth, if necessary. 1. a trapezoid with bases 10 cm and 12 cm and height 5 m 2. a parallelogram with base 9 in. and height 8 in. 3. a semicircle with radius 5 m 4. a semicircle with diameter 8 cm 55 cm2 72 in2 39.3 m2 25.1 cm2

One method of estimating the area of an irregular figure is to count the number of squares the figure covers.

Additional Example 1: Finding Area by Counting Find the area of each figure. A. Count the full squares: 10 Count the half-full squares: 4 Add the number of full squares plus half the number of half-full squares: 10 + ( • 4) = 10 + 2 =12 1 2 The area of the figure is 12 square units.

For shapes such as the one in Example 1B you can only estimate the area. Helpful Hint

Additional Example 1: Finding Area by Counting Find the area of each figure. B. Count the full and almost-full squares: 11 Count the half-full squares: 4 Add the number of full squares plus half the number of half-full squares: 11 + ( • 4) = 11 + 2 =13 1 2 The area of the figure is approximately 13 square units.

Check It Out! Example 1 Find the area of each figure. A. Count the full squares: 11 Count the half-full squares: 8 Add the number of full squares plus half the number of half-full squares: 11 + ( • 8) = 11 + 4 = 15 1 2 The area of the figure is 15 square units.

Check It Out! Example 1 Find the area of each figure. B. Count the full and almost-full squares: 11 Count the half-full squares: 6 Add the number of full squares plus half the number of half-full squares: 11 + ( • 6) = 11 + 3 =14 1 2 The area of the figure is about 14 square units.

Additional Example 2A: Estimating Area Using Composite Figures Use a composite figure to estimate the shaded area. Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

Additional Example 2A Continued Use a composite figure to estimate the shaded area. area of the trapezoid: A = h(b1 + b2) 1 2 = 1(4 + 2) = 3 1 2 area of the triangle: A = bh 1 2 __ = (4 • 2) = 4 1 2 __ The shaded area is approximately 7 square units.

Additional Example 2B: Estimating Area Using Composite Figures Use a composite figure to estimate the shaded area. Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

Additional Example 2B Continued Use a composite figure to estimate the shaded area. area of the rectangle: A = bh = 1 • 4 = 4 area of the semicircle: A = pr2 1 2 __  • 3.14(22)  6.28 1 2 __ The shaded area is approximately 10.3 square units.

Check It Out! Example 2A Use a composite figure to estimate the shaded area. Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

Check It Out! Example 2A Continued Use a composite figure to estimate the shaded area. area of the rectangle: A = bh = 3 • 2 = 6 area of the triangle: A = bh 1 2 __ = (3 • 2) = 3 1 2 __ The shaded area is approximately 9 square units.

Check It Out! Example 2B Use a composite figure to estimate the shaded area. Draw a composite figure that approximates the irregular shape. Divide the composite figure into simple shapes.

Check It Out! Example 2B Continued Use a composite figure to estimate the shaded area. area of the square: A = s2 = 2 • 2 = 4 area of the semicircle: A = pr2 1 2 __  • 3.14(12)  1.57 1 2 __ The shaded area is approximately 5.5 square units.

Lesson Quiz: Part l Find the area of each figure. 1. 2. 8 square units approximately 12 square units

Lesson Quiz: Part ll Use a composite figure to estimate the shaded area. 3. 4. 1 2 approximately 4 square units approximately 6 square units

Competition Problems

The length of a rectangle is 4 inches longer than the width The length of a rectangle is 4 inches longer than the width. If the perimeter is 28 inches, what is the width?

Answer: x + x + (x+4) + (x+4) = 28 4x + 8 = 28 4x = 20 x = 5

A regular pentagon has a side measure of 1 ft 4 in A regular pentagon has a side measure of 1 ft 4 in. What is its perimeter?

16in 16in Answer: 16in + 16in + 16in + 16in + 16in = 5·16in = 80 inches = 6ft 8in 16in 1ft 4in = 12in + 4in = 16in 16in

Andrew and Ted built a snowman 98 inches tall by stacking three balls of snow. The largest ball had a diameter that was twice as long as the middle ball’s diameter. The smallest ball had a diameter that was half of the middle ball’s diameter. What was the circumference of the middle ball of snow? Use 3.14 for π.

Answer: 4d + 2d + d = 98 7d = 98 d = 14 = 87.92 inches d Middle Circle diameter 2d = 2(14) = 28 inches Circumference = π· d = π (28) = 3.14 (28) = 87.92 inches 2d Answer: 4d + 2d + d = 98 7d = 98 d = 14 98” 4d

Which of the following points would be in Quadrant III. A. (1,3) B Which of the following points would be in Quadrant III?   A.(1,3) B.(-1,4) C.(4,-3) D.(-4,-1) E.NOTA

Answer: D. (-4,-1)

If the perimeter of a square is 56 inches, what is the area of the square?  

Answer: P = 4S = 56 P = 56/4 P = 14 inches A = S·S = 14in·14in = 196 in2

Carter and Alex are building a circular garden Carter and Alex are building a circular garden. A four-foot high fence will encircle the garden, except for a three-foot opening for a gate. If the radius of the garden is 15 feet, how much fencing will they need? Use 3.14 for pi.  

Answer: radius = 15 diameter = 30 circumference = πd = 3. 14(30) = 94 Answer: radius = 15 diameter = 30 circumference = πd = 3.14(30) = 94.2 less 3’ opening for gate =94.2 – 3 =91.2ft

A rectangle has an area of 48 sq. cm. and a width of 3 cm A rectangle has an area of 48 sq. cm. and a width of 3 cm. If its length is reduced by 5 cm., and its width is increased by 5 cm., what is the resulting area?  

Answer: A = L·W = 11(8) = 88 cm2 A = 48 W = 3cm A = L·W 48=L (3) Reduced Length = 16-5 = 11cm A = 48 W = 3cm Increase Width = 3 + 5 = 8cm A = L·W 48=L (3) 16cm=L A = L·W = 11(8) = 88 cm2