Unit IV D Question Liberal Arts Math

MAT 1301, Liberal Arts Math 1

Course Learning Outcomes for Unit IV Upon completion of this unit, students should be able to:

1. Apply mathematical principles used in real-world situations. 1.1 Solve problems involving the relationships among angles, arcs, and circles. 1.2 Solve problems involving angle relationships of polygons and similar polygons. 1.3 Calculate the perimeter and area of geometric objects. 1.4 Solve problems involving right triangles using the Pythagorean Theorem. 1.5 Calculate the circumference and area of circles.

4. Apply geometric principles and formulas to solve problems.

4.1 Apply the basic properties of geometric objects. 4.2 Utilize the fundamental properties of angles to solve problems. 4.3 Apply the basic terminology and properties of polygons. 4.4 Determine how formulas for geometric figures are related. 4.5 Demonstrate conversions between metric measurements. 4.6 Demonstrate conversions between measurement systems using dimensional analysis.

Reading Assignment Chapter 9: Geometry: Ancient and Modern Mathematics Embrace

 Section 9.1: Lines, Angles, and Circles, pp. 437-445

 Section 9.2: Polygons, pp. 446-455

 Section 9.3: Perimeter and Area, pp. 456-467

 Section 9.5: The Metric System and Dimensional Analysis, pp. 477-487

Unit Lesson

Look at any page of the textbook. That page is a plane. The sentences, shapes, and lines on the page are all in the plane. Now take another sheet of paper and hover it above the page. Is the paper in the same plane as the textbook page? No. In fact, the paper is in a completely different plane!

Chapter 9: Geometry Geometry is the study of points, lines, shapes, and space. While reading and practicing several math geometry problems, one might find that geometry concepts are used all around. While watching a basketball game, look at the arcs painted on the court. Those arcs had to be painted perfectly to provide a direct relationship to the hoop. The players on the court also use geometry by angling their bodies to the hoop and creating an arc with the ball as it is shot through the air into the basketball hoop. Section 9.1, Lines, Angles, and Circles: Before exploring other areas where geometry may be used, it is important to begin by introducing the basic properties of geometry. In this section, points, lines, planes, angles, and circles will be discussed. Planes A plane is a flat surface. It is two-dimensional and can be any shape or size.

UNIT IV STUDY GUIDE

Geometry

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For the purposes of this course, points, lines, and angles will be drawn in one plane. The gray rectangle in the figure below represents a plane. This is drawn to show that both lines lie within the same plane.

Points and Lines Basic properties for points and lines are as follows:

 Label points with capital letters, such as A, B, and C.  Label lines with lowercase letters such as l and m, and subscripts such as l1 and l2 may be included.

The following figures represent different types of lines: (a) line, (b) half line, (c) ray, and (d) line segment (e) parallel lines, and (f) intersecting lines. (a) Line

 A line does not end. The arrows on each end of the line signify that the line goes on forever in the same plane.

 l1 and l2 are lines.

 Line l2 is divided by point A into three parts, a point and two half lines.

(b) Half line

 A half line has an open point at one end. This means that the point is not included in the line.

(c) Ray

 A ray has a closed point at one end of a half line. This means that the point is included in the line.

(d) Line segment AB

 A line segment has two closed points on each end of the line.

 This line is a segment because it does not continue.

Parallel Lines (Pirnot, 2014, p. 438)

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 We state that this is line segment AB, because the end points of the line are identified as point A and point B.

(e) Parallel lines

 Parallel lines do not touch or intersect.

 The “||” symbol is used to state that two lines are parallel. Therefore, l1 || l2 could be written. (f) Intersecting lines

 Intersecting lines are lines that cross at one point.

 Lines l3 and l4 are intersecting lines and have point D in common.

Angle

 Angles are formed by two rays that share an endpoint.

 Angles are denoted by the symbol, “∠”.

 Angles may be labelled using the vertex or by using all three points that create the angle. Therefore, the angle below may be called ∠B or ∠ABC. Note: To label an angle using three points, name the point on the initial side, name the point of the vertex, and then the point on the terminal side.

Parallel Lines (Pirnot, 2014, p. 438)

Intersecting lines (Pirnot, 2014, p. 438)

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 Angles are measured in degrees. The figure below shows that ∠ABC = 45°. 

Additional Terms and Definitions

Term Definition Example

Right Angle An angle whose measure is 90°.

Acute Angle An angle whose measure is

between 0° and 90°.

Obtuse Angle An angle whose measure is between 90° and 180°.

Straight Angle An angle whose measure is

180°.

Vertical Angles Pair of opposite angles made by intersecting lines.

Think of the terminal side of an angle as the ending side or the finishing position.

Think of the initial side of an angle as the starting side. Imagine that this ray was copied and moved following the direction of the arc. This created an angle.

The vertex of an angle is the common endpoint between the two rays.

Angle formed by rotating a ray about point B (Pirnot, 2014, p. 438)

45°

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Property of Vertical Angles

Vertical angles have equal measures.

Perpendicular Lines Two lines that intersect and form a right angle.

Complementary Angles

A pair of angles whose sum is 90°.

Supplementary Angles A pair of angles whose sum is 180°.

The figure below represents two parallel lines cut by another line called a transversal. These angles have special names and unique properties.

Right angle, acute angle, obtuse angle, straight angle, vertical angles, property of vertical angles, perpendicular lines, complementary angles, supplementary angles (Pirnot, 2014, p. 439)

Transversal (Pirnot, 2014, p. 440)

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Example:

Use the figure above to identify the following angles. Lines l and m are parallel.

1. Right Angle 2. Acute Angle 3. Obtuse Angle 4. Vertical Angles 5. Complementary Angles 6. Supplementary Angles 7. Corresponding Angles 8. Alternate Interior Angles 9. Alternate Exterior Angles

Solution: The solutions to this problem are found using the definitions and properties of angles. Figure 9.10 and table 9.1 on page 440 of the textbook provide guidance for this question. Solutions for questions 1-9 are given below. Additionally, explanations are provided for questions 1, 2, and 5.

1. ∠1, ∠6, ∠11

Explanation:

∠11 has a square at the vertex of the angle. Therefore, ∠11 is known to be a 90° or right angle. ∠ 1 and ∠11 are corresponding angles. Therefore, ∠1 is also a 90° or right angle. ∠6 and ∠1 are vertical angles. Therefore, ∠6 is a 90° or right angle.

2. ∠2, ∠3, ∠4, ∠5, ∠7, ∠9

Properties of pairs of angles formed by cutting parallel lines with a transversal (Pirnot, 2014, p. 440)

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Explanation: ∠1 + ∠2 + ∠3 = 180°. This is known because the total angle measurement of a straight line equals 180°. It is also known from part (1) that ∠1 = 90°, so that value can be substituted as follows: 90° + ∠2 + ∠3 = 180° Subtracting 90° from both sides gives: ∠2 + ∠3 = 90° From this, it can be seen that both ∠2 and ∠3 must each be less than 90° and added together must also be equal to 90°. Therefore, ∠2 and ∠3 are both acute angles. The same principle can be used for ∠6, ∠5, and ∠4. Therefore, ∠5 and ∠4 must each be less than 90°, so ∠5 and ∠4 are acute angles.

3. ∠8, ∠10

4. ∠7 and ∠9, ∠8 and ∠10

Explanation: When two lines intersect, the opposite angles formed are equal to each other. These angles are called vertical angles. Thus, ∠7 and ∠9 are vertical angles, and ∠8 and ∠10 are also vertical angles.

5. ∠2 and ∠3, ∠4 and ∠5

6. ∠9 and ∠10, ∠7 and ∠10, ∠7 and ∠8, ∠8 and ∠9, ∠3 and ∠10, ∠4 and ∠10, ∠4 and ∠8, ∠3 an ∠8

7. ∠3 and ∠9, ∠4 and ∠7

8. ∠3 and ∠7

9. ∠9 and ∠4 Example: What is the complementary and supplementary angle for each angle?

1. 30° 2. 110.4°

Solution:

1. 30°

Two angles are complementary if the sum of their angles is 90°. Therefore, the complementary angle of 30° is 90 – 30 = 60°. Two angles are supplementary if the sum of their angles is 180°. Therefore, the supplementary angle of 30° is 180 – 30 = 150°.

1. 110.4°

Two angles are complementary if the sum of their angles is 90°. Therefore, this angle does NOT have a complementary angle, because 110.4° is greater than 90°.

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Two angles are supplementary if the sum of their angles is 180°. Therefore, the supplementary angle of 110.4° is 180 – 110.4 = 69.6°.

Example: In problems 1-2, find the measures of angles a, b, and c. Lines l and m are parallel.

1.

Solution: The property of vertical angles will be used to start solving this problem.

 ∠c = 36° because they are vertical angles.

 ∠a + 36° = 180° because these angles form a straight angle. Straight angles are 180°. Therefore, ∠a + 36° = 180° ∠a + 36° - 36° = 180° - 36° ∠a = 144°

 ∠a and ∠b are vertical angles. Therefore, ∠b = ∠a. It was found that ∠a = 144°, so ∠b = 144°.

1. Let x = 40°. Solve for angles a, b, and c.

Solution:

 ∠x and ∠a are corresponding angles and are equal. Since ∠x = 40°, then ∠a = 40°.

 ∠a + ∠b are corresponding angles to the perpendicular angle in the upper left corner of the figure. So, ∠a + ∠b = 90° 40° + ∠b = 90° 40°- 40° + ∠b = 90° - 40° ∠b = 50°

 ∠a and ∠c are supplementary angles. So, ∠a + ∠c = 180° 40° + ∠c = 180° 40°- 40° + ∠c = 180° - 40° ∠c = 140°

Circles Circles are common geometric symbols. The bottom of a coffee cup, the tires of a car, and the face of a watch all resemble the shape of a circle.

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A circle is the set of all points lying on a plane that are located at a fixed distance called the radius, from a given point called the center. The diameter, radius, and circumference are important properties of a circle. See below.

 Diameter – A diameter of a circle is a line segment passing through the center with both endpoints lying on the circle.

 Radius – A line segment from the center of the circle to a point on the circle is a radius. The radius is half the length of the diameter.

 Circumference – The distance around the circle is the circumference.

 Central Angle – A central angle is an angle with the vertex at the center of the circle.

The length of the arc AB is proportional to the measure of ∠ACB. We can use the following ratio to find missing pieces of information, such as the length of the arc.

𝑇ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴𝐶𝐵

360° =

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐ℎ 𝐴𝐵

𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒

Arc AB is the distance around the circle that is created by the central angle.

Angle ACB is a central angle (Pirnot, 2014, p. 441)

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Example: Find the length of arc AB is the circumference is 12 meters and the measure of angle ACB (we write this as m∠ACB).

Solution:

𝑇ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴𝐶𝐵

360° =

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐ℎ 𝐴𝐵

𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒

120°

360° =

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐 𝐴𝐵

12 Cross Multiply.

360𝑥 = 1440 Divide each side of the equation by 360.

𝑥 = 4 𝑚𝑒𝑡𝑒𝑟𝑠 The length of arc AB is 4 meters.

Example: Two of the following three pieces of information will be given: the circumference of the circle, the measure of the central angle ACB, and the length of arc AB. The third piece of information will be found.

1. Circumference = 150 centimeters; m<ACB = 72o

Solution:

𝑇ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴𝐶𝐵

360° =

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐ℎ 𝐴𝐵

𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒

x = 30 The length of arc AB is 30 cm.

Circle containing a 120-degree angle (Pirnot, 2014, p. 441)

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2. Circumference = 240 inches; length of arc AB = 40 inches Solution:

𝑇ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴𝐶𝐵

360° =

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐ℎ 𝐴𝐵

𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒

𝑥

360° =

40

240 Cross Multiply

240𝑥 = 14400 Divide each side of the equation by 240.

𝑋 = 60° m∠ACB = 60°

Section 9.2: Polygons: Before you can learn about polygons, some vocabulary must be learned. Two important terms that are used in the definition of a polygon are closed and simple.

 Closed – Figure is closed if it can be drawn without lifting the pencil and if the starting and ending points are the same.

 Simple – Figure is simple if one can draw it without lifting the pencil, and in drawing it, the same point is not passed twice.

Here are some examples of closed and simple figures.

 Characteristics of a polygon:

o Must be closed. o Must be simple. o Must only consist of line segments called edges.

Closed and simple plane figures (Pirnot, 2014, p. 446)

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Here are some examples of polygons. Polygons are named according to the number of their sides. The following table provides some names of common polygons.

Number of Sides Name of Polygon

3 Triangle

4 Quadrilateral

5 Pentagon

6 Hexagon

7 Heptagon

8 Octagon

9 Nonagon

10 Decagon Polygons can also be convex or nonconvex. A convex polygon can be thought of as having all outward edges, while a nonconvex polygon will have at least one inward edge. Examples of convex and non-convex polygons are shown below.

Examples of polygons (Pirnot, 2014, p. 447)

Convex and nonconvex polygons (Pirnot, 2014, p. 447)

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Polygons and Angles Some questions will ask you to find the sum of all interior angles within a polygon. To do this, the following formula will be used: Angle Sum of a Polygon = (𝑛 − 2) x 180 Note: n is the number of sides of a polygon. Example: What is the interior angle sum of a hexagon? Solution: A hexagon has 6 sides. Therefore, n=6 in the equation. Interior angle sum = (6-2) x 180 Interior angle sum = 4 x 180 Interior angle sum = 720° Sometimes, you will need to know the measure of each interior angle of a regular polygon. A regular polygon is a polygon where sides and angles are the same length or measure.

Interior Angles of Regular Polygon = (n−2) × 180

n

Example: What is the measure of an interior angle of a regular 20-sided polygon? Solution: Since the polygon is a regular polygon, the measure of each interior angle can be found by using the following formula:

Interior angles of regular polygon = (n−2) × 180

n

Interior angles of regular polygon = (20−2) × 180

20

Interior angles of regular polygon = (18) × 180

20

Interior angles of regular polygon = 3240

20

Interior angles of regular polygon = 162° Similar Polygons Two polygons are similar if their corresponding sides are proportional and their corresponding angles are equal.

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In other words, polygons are similar if they are the same shape but different sizes. The shapes below are similar polygons.

Example: Each pair of triangles in the figure below is similar. Find x.

Solution: The triangles are similar and have corresponding sides.

The blue sides of each triangle are corresponding sides and are proportional. The red sides of each triangle are corresponding sides are also proportional.

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The following proportion can be set up:

6

3 =

10

𝑥

Cross multiply.

6𝑥 = 30 Divide both sides of the equation by 6.

𝐱 = 𝟓

9.3 Perimeter and Area: The words perimeter and area are probably very familiar. These words may have been heard from a contractor or a roofer, and some may have even used the formulas for area and perimeter. This unit will teach you the concept of area and perimeter and will introduce formulas that can be used for identifying the perimeter and area of triangles, parallelograms, trapezoids, and other shapes.

 Perimeter The perimeter of a polygon is the sum of the lengths of its sides. For example, if you wanted to find the perimeter of the shape below, you would add 3 + 4 + 4.75 + 2.25 + 7 + 4.25 = 25.25 cm.

 Area The area of a polygon is the measurement of its surface. For example, a large square is created by four square tiles in the figure below. If each square tile is one square foot, the area of the larger square would equal 4.

Perimeter and Area Formulas The following are general formulas for some basic shapes. These formulas will be used to derive perimeter and area formulas of irregular shapes.

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Example: Find the area of the figure:

Using the table given above, we know that the area of a triangle is 𝑨 = 𝟏

𝟐 𝒉 ∙ 𝒃. Since we are given the height,

we can label our diagram and plug the given values into our formula.

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𝐴 = 1

2 ∙ 6 ∙ 15 Plug in values for b and h.

𝐴 = 3 ∙ 15 Multiply from left to right: 1

2 ∙ 6 = 3.

𝐴 = 45𝑖𝑛2 Multiply 3 ∙ 15 to find A.

Heron’s Formula Heron’s formula can be used to find the area of triangle. This formula can be helpful if the height of triangle is not given.

Heron’s formula is 𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) where 𝑠 = 1

2 (𝑎 + 𝑏 + 𝑐)

Example: Use Heron’s formula to find the area of each triangle.

Solution: Label the legs of triangle (𝑎, 𝑏, 𝑐) in ascending order.

Heron’s formula is 𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) where 𝑠 = 1

2 (𝑎 + 𝑏 + 𝑐)

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First, solve for s:

𝑠 = 1

2 (18 + 40 + 52)

𝑠 = 1

2 (110)

𝑠 = 55 in Next plug in the value of s in Heron;s formula.

𝑠 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

𝐴

= √55(55 − 18)(55 − 40)(55 − 52) Plug in values for s, a, b, and c.

𝐴 = √55(37)(15)(3) Subtract the numbers in each set or parentheses.

𝐴 = √91575 Multiply from left to right.

𝐴 = 302.61 𝑖𝑛2 Take the square root.

By using Heron’s formula, the area of the triangle was found to be 302.61 in2. Area of Trapezoid A trapezoid is created by placing two triangles on either side of a rectangle. The formula for finding the area of a trapezoid can be found by using the formula for the area of a triangle and a rectangle.

The formula of a trapezoid is 𝐴 = 1

2 (𝑏1 + 𝑏2) × ℎ.

Figure 9.39 on page 459 shows the properties of a trapezoid. The bases of the trapezoid are labeled b1 and b2. The height of a trapezoid is labeled h.

Trapezoids (Pirnot, 2014, p. 459)

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Example: Find the area of the figure.

Solution:

Recall that the formula for the area of a trapezoid is 𝐴 = 1

2 (𝑏1 + 𝑏2) × ℎ. The diagram can then be labeled as

shown.

Pythagorean Theorem The Pythagorean Theorem is a well-known theorem that is used to find the length of a side in a right triangle.

The Pythagorean Theorem states that 𝑎2 + 𝑏2 = 𝑐2 where a and b are the legs of a triangle and c is the hypotenuse of the right triangle.

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Recall that the right triangle has the following features: Example: Find the length of side x for each triangle. Solution:

Since this is a right triangle, the Pythagorean Theorem 𝑎2 + 𝑏2 = 𝑐2 can be used to find the length of the missing side. Start by labeling the sides of the triangle.

𝑥2 = 82 + 142 Plug in the values for a and b.

𝑥 = √82 + 142 Take the square root of both sides of the equation.

𝑥 = √64 + 196 Evaluate the exponents under the radical; 82 = 64 and 142 = 196.

A right triangle has an angle

that equals 90°.

The Pythagorean Theorem (Pirnot, 2014, p. 461)

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𝑥 = √260 Add the numbers under the radical: 64 = 196 = 260.

𝑥 = 16.12 𝑓𝑡 Take the square root: √260 = 16.12

Therefore, the length of x or the hypotenuse, is 16.12 feet. Example: Find the length of side x for each triangle. Solution:

Since this is a right triangle, the Pythagorean Theorem 𝑎2 + 𝑏2 = 𝑐2 can be used to find the length of the missing side.

92 + 𝑥2 = 192 Plug in the values for a and c.

92 − 92 + 𝑥2 = 192 − 92 Subtract 92 from both sides of the equation to isolate x2 on the left-hand side.

𝑥 = √192 − 92 Take the square root of both sides.

𝑥 = √361 − 81 Evaluate the exponents under the radical: 192 = 361 and 92 = 81

𝑥 = √280 Subtract the numbers under the radical: 361 – 81 = 280

𝑥 = 16.73 𝑖𝑛 Take the square root to find

𝑥 − √280 = 16.73

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Circles As stated previously, circles are not polygons because a circle does not contain line segments called edges. However, circles are common shapes, and it is important to know the formulas that are associated with circles. Instead of finding the perimeter, the circumference can be found. The circumference is the distance around the circle. The area of a circle can also be found. Circumference of a circle: C = 2𝝅r Area of a circle: A = 𝝅r2 The symbol 𝝅 can be found on a calculator. In most cases, the number 3.14 can be substituted for 𝜋. Example: Find the circumference and area of a circle that has a radius of 7 cm. Substitute 3.14 for 𝜋. Solution: First, the circumference of the circle is found by using the formula C = 2𝜋r. C = 2𝜋(7) Plug in the value given for the radius. C = 14𝜋 Multiply 7 and 2. C = 14(3.14) Substitute 3.14 for 𝜋. C = 43.96 cm Multiply 14 and 3.14. Next, the area of the circle is found by using the formula, A = 𝜋r2. A = 𝜋(7)2 Plug in the value given for the radius. A = 𝜋(49) Evaluate the exponential: (7)2 = 7•7 = 49.

A = 3.14(49) Substitute 3.14 for 𝜋. A = 153.86 cm2 Multiply 3.14 and 49. After becoming familiar with the concepts of area, you can start finding the areas of different portions of shapes. For example, look at the following diagram.

To find the area of the shaded region, you can subtract the area of the circle from the area of the square.

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First, find the area of the square. The sides of the square are twice the length of the area of the circle. Therefore, s = 10 in. A = s2

A = 102

A = 100 in2 Next, find the area of the circle. The radius of the circle is given. Therefore, r = 5 in. Substitute 3.14 for 𝜋. A = 𝜋(5)2

A = 𝜋(25) A = 3.14(25) A = 78.5 in2

Lastly, subtract the area of the circle from the area of the square to find the shaded region. area of the shaded region = area of the square – area of the circle area of the shaded region = 100 – 78.5 area of the shaded region = 21.5 in2

The next two examples use all geometric properties that were discussed. Example: Find the area of the shaded region.

The shape can be imagined as a square (dashed lined) with two semi circles inside.

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Solution:

The formula for the area of the square is 𝐴 = 𝑠2, and the area of a semi-circle is 𝐴 = 1

2 𝜋𝑟2. So, the formula for

the shaded region is the area of the square minus twice the area of the semi-circle.

𝐴 = 𝑠2 − 2 [ 1

2 𝜋𝑟2]

Since the diameter of the circle is given, divide by 2 to find 𝑟 = 2

2 = 1. The diameter of the circle is also the

length of the side of the square, thus 𝑠 = 2. The number can now be plugged in and solved for 𝐴.

𝐴 = 22 − 2 [ 1

2 𝜋 ∙ 12] Assume 𝜋 = 3.14 but leave the symbol in place until the end.

𝐴 = 4 − 2 [ 1

2 𝜋 ∙ 1] Evaluate the exponents: 22 = 4 and 12 = 1

𝐴 = 4 − 2 [ 1

2 𝜋] Multiply within the brackets: 2 [

1

2 𝜋] = 𝜋

= 4 − 𝜋 Distribute 2 to the inside of the bracket: 2 [ 1

2 𝜋] = 𝜋

𝐴 = 4 − 3.14 Now replace 𝜋 with 3.14 and subtract to find 𝐴.

𝑨 = 𝟎. 𝟖𝟔 𝒎𝟐 Example:

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Find the area of the shaded region:

Solution: Here the shaded region is a circle with a square taken from the middle. The radius of the circle is given as 1 meter, but it is necessary to find the length of a side of the square. The side of the square 𝑠 is actually the hypotenuse of a triangle created by drawing lines from the center of the square to the two right corners.

The Pythagorean Theorem 𝑎2 + 𝑏2 = 𝑐2 now applies to the red triangle above, so

𝑠2 = 12 + 12 Both sides are length 1

𝑠 = √12 + 12 Take the square root of both sides

𝑠 = √1 + 1 Square the ones: 12 = 1

𝑠 = √2 ≈ 1.41 Add the numbers under the radical and use a calculator to find the approximate value.

The area of shaded region is the area of the circle 𝐴 = 𝜋𝑟2 minus the area of the square 𝐴 = 𝑠2. So the whole formula can be written as 𝐴 = 𝜋𝑟2 − 𝑠2 𝐴 = 𝜋 ∙ 12 − 1.41 Since 𝑟 = 1 is given, use 𝑠 = 1.41 as found above.

𝐴 = 𝜋 − √2 Evaluate the exponent: 12 = 1. 𝐴 = 3.14 − 1.41 Use 3.14 as an approximation for 𝜋 and then subtract to find 𝐴.

𝑨 = 𝟏. 𝟕𝟑 𝒎𝟐

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To view more information about what has been covered so far, please view the following interactive lesson: https://media.pearsoncmg.com/pcp/pls/pls_mycoursetools/fufillment/mct_1256689785_csu/beginning_algebra /redirect_index_10.html 9.5 The Metric System and Dimensional Analysis: The metric system is comprised of basic units of measurement and is used throughout the world. Although the United States uses a system of measurement called the U.S. customary system, most of the world uses the metric system. In order to understand the metric system, the basic units that are used to measure the length, weight, or volume of an object will be discussed. The table presented below represents some units of measurement that are included in the metric system.

Type of Measurement Unit

Length Meter

Volume Liter

Weight Gram Prefixes may be placed on the units to change the amount of the units. By adding prefixes, much heavier objects can be described. For example, one gram is about the weight of a paperclip. Suppose a picture frame weighed one thousand grams. Instead of writing 1,000 grams, 1 kilogram can be written because the prefix kilo means 1,000. The table below explains the meaning of prefixes.

Prefix Scale

mega- million 1,000,000

kilo- thousand 1,000

hecto- hundred 100

deca- ten 10

base unit one 1

deci- tenth 1 10

centi- hundredth 1 100

milli- thousandth 1 1,000

micro- millionth 1 1,000,000

Metric Conversions Units can be converted to other units by using the prefix definitions listed in the previous table. Example: Use the table to covert 28 decimeters to millimeters.

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Solution: Decimeters are being converted to millimeters. According to the chart, two conversions are needed to get to the final answer in millimeters.

28 decimeters = 28 × (10 centimeters) = 280 centimeters 280 centimeters = 280 × (10 millimeters) = 2,800 millimeters Dimensional Analysis Dimensional analysis is used to convert customary units to metric units. Customary units are used by the United States. Feet, inches, yards, and miles are all customary units that are used when measuring the length. Review tables 9.6, 9.7, 9.8, and 9.9 to see some basic relationships between customary and metric units of length. The last four examples explain how to use dimensional analysis when converting from customary units to metric units and vice versa. Example: Convert 7,600 centimeters to meters. Solution: Dimensional analysis is used to solve this problem:

7,600 𝑐𝑚 × 𝟏 𝒎

𝟏𝟎𝟎 𝒄𝒎 1 𝑚𝑒𝑡𝑒𝑟 = 100 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑒𝑟𝑠 is the unit

fraction.

= (7,600 𝑐𝑚)(1 𝑚)

100 𝑐𝑚 Rewrite the problem as a single fraction

= (7,600)(1 𝑚)

100 Cancel the centimeters in the numerator

and denominator.

= 𝟕𝟔 𝒎 Multiply the numerator values and then divide by the denominator: 7,600 × 1 = 7,600 and 7,600 ÷ 100 = 76

Example: Convert 507,820 milligrams to pounds.

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Solution: Dimensional analysis is used to solve this problem:

507,820 𝑚𝑔 × 𝟏 𝒌𝒈

𝟏, 𝟎𝟎𝟎, 𝟎𝟎𝟎 𝒎𝒈 1 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚 = 1,000,000 𝑔𝑟𝑎𝑚𝑠 is our

unit fraction.

= (507,820 𝑚𝑔)(1 𝑘𝑔)

1,000,000 𝑚𝑔 Rewrite the problem as a single fraction

= 507,820 𝑘𝑔

1,000,000 ×

𝟐. 𝟐 𝒍𝒃

𝟏 𝒌𝒈 2.2 𝑝𝑜𝑢𝑛𝑑𝑠 = 1 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚

= (507,820 𝑘𝑔)(2.2 𝑙𝑏)

1,000,000 𝑘𝑔 Rewrite the problem as a single fraction

= 1,117,204 𝑙𝑏

1,000,000 Cancel the kilograms in the numerator

and denominator. Then multiply the numberator values: 507,820 × 2.2 = 1,117,204

= 𝟏. 𝟏𝟐 𝒍𝒃 Divide: 1,117,204 ÷ 1,000,000

Example:

Eric is driving in Switzerland at a speed of 55 𝑚𝑖

ℎ𝑟 . What is his speed in kilometers per hour?

Solution: Dimensional analysis is used to solve this problem:

55 𝑚𝑖

1 ℎ𝑟 ×

𝟏. 𝟔𝟎𝟗 𝒌𝒎

𝟏 𝒎𝒊 1.609 𝑘𝑖𝑙𝑜𝑚𝑒𝑡𝑒𝑟 = 1 𝑚𝑖𝑙𝑒 is our unit

fraction

= (55 𝑚𝑖)(1.609 𝑘𝑚)

(1 𝑚𝑖)(1 ℎ𝑟) Rewrite the problem as a single fraction

= 88.5 𝑘𝑚

1 ℎ𝑟 Multiply the numbers in the numerator

55 × 1.609 = 88.5 then the numbers in the denominator 1 × 1 = 1

= 𝟖𝟖. 𝟓 𝒌𝒎

𝒉𝒓 Any number over one is itself:

88.5 𝑘𝑚

1 ℎ𝑟 =

88.5 𝑘𝑚

ℎ𝑟

Example: Adrian’s rectangular swimming pool is 20 𝑓𝑡 wide, 40 𝑓𝑡 long, and averages 6 𝑓𝑡 in depth. How many kiloliters of water are needed to fill the pool? Solution:

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First, we find the volume in 𝑓𝑡 3.

𝑣 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡

𝑣 = 40 𝑓𝑡 × 20 𝑓𝑡 × 6 𝑓𝑡 Plug in the values that were given.

= 4,800 𝑓𝑡3 Multiply the values and units: 40 × 20 × 6 = 4800 and 𝑓𝑡 × 𝑓𝑡 × 𝑓𝑡 = 𝑓𝑡3

Dimensional analysis I used to convert to 𝑘𝐿:

4,800 𝑓𝑡3 × 𝟏 𝒎𝟑

𝟑𝟓. 𝟑𝟏𝟒𝟕 𝒇𝒕𝟑

The unit conversion is 1 𝑐𝑢𝑏𝑖𝑐 𝑚𝑒𝑡𝑒𝑟 = 35.3147 𝑐𝑢𝑏𝑖𝑐 𝑓𝑒𝑒𝑡.

= (4,800 𝑓𝑡3 )(1 𝑚3)

35.3147 𝑓𝑡3

Rewrite the problem as a single fraction.

= 4,800 𝑚3

35.3147

Multiply the values in the numerator: 4,800 × 1 = 4,800

= 135.92 𝑚3 Divide: 4,800 ÷ 35.3147 = 135.92

= 135.92 𝑚3 × 𝟏 𝒌𝑳

𝟏 𝒎𝟑 = 135.92 𝑘𝐿 Use unit conversion: 1 𝑐𝑢𝑏𝑖𝑐 𝑚𝑒𝑡𝑒𝑟 =

1 𝑘𝑖𝑙𝑜𝑙𝑖𝑡𝑒𝑟

For more information regarding measurement, please view the following interactive lesson and video: https://media.pearsoncmg.com/pcp/pls/pls_mycoursetools/fufillment/mct_1256689785_csu/beginning_algebra /redirect_index_03.html

Reference

Pirnot, T. L. (2014). Mathematics all around (5th ed.). Boston, MA: Pearson.