1. § 3.3 Problem Solving in Geometry

2. Geometry Blitzer, Introductory Algebra, 5e – Slide #2 Section 3.3 Geometry is about the space you live in and the shapes that surround you. For thousands of years, people have studied geometry in some form to obtain a better understanding of the world in which they live. In this section, we will look at some basic geometric formulas for perimeter, area, and volume. Remember that perimeter is a linear measure, that is, a measure of length. It will be measured in units such as feet, or meters. Area is a measure of square units and it will be measured in units such as square feet, square miles, or square meters. How many floor tiles are in your kitchen? That’s an area measure. Volume is a measure of cubic units. It measures the number of cubic units in a three dimensional object or rather the amount of space occupied by the figure. A liter is a measure of volume. Cups and gallons are also measures of volume. The volume of a solid is the number of cubic units that can be contained in the solid.

3. Common Formulas for Perimeter and Area Square Rectangle s l s w A = lw P = 4sP = 2l + 2w Perimeter and Area of Rectangle Blitzer, Introductory Algebra, 5e – Slide #3 Section 3.3

4. Common Formulas for Area Triangle Trapezoid b h b a Area of Triangle and Trapezoid Blitzer, Introductory Algebra, 5e – Slide #4 Section 3.3

5. 1.Let l represent the unknown length. l = length of the rectangle. 2.Represent other unknowns in terms of l. there are no other unknowns 3.Write an equation that describes l the condition. 2w + 2l = P 2(10) + 2l = 125 20 + 2l =125 Example : A rectangle has a width of 10 inches and a perimeter of 125 inches. Find the length Perimeter of Rectangle Blitzer, Introductory Algebra, 5e – Slide #5 Section 3.3

6. 4.Solve the equation and answer the question. 20 + 2l = 125 2l = 105 l = 52.5 The length of the rectangle is 52.5 inches. 5.Check the solution in the original problem. The perimeter of a rectangle that is 10 inches by 52.5 inches is 125 inches. 2(10) + 2(52.5) = 125 Example Continued: A rectangle has a width of 10 inches and a perimeter of 125 inches. Find the length. Perimeter of a Rectangle Blitzer, Introductory Algebra, 5e – Slide #6 Section 3.3

7. Formulas for Circles r The Circle Blitzer, Introductory Algebra, 5e – Slide #7 Section 3.3 The circumference of a circle is a measure of length. The circumference of a circle is the distance around it, or its perimeter.

8. Example : Find the area and circumference of a circle with a radius of 5 inches. Area and Circumference of Circle Blitzer, Introductory Algebra, 5e – Slide #8 Section 3.3

9. Common Formulas for Volume Cube Rectangular Solid Volume of a Cube and a Rectangular Solid Blitzer, Introductory Algebra, 5e – Slide #9 Section 3.3

10. Common Formulas for Volume Circular Cylinder Sphere Volume of a Cylinder and a Sphere Blitzer, Introductory Algebra, 5e – Slide #10 Section 3.3

11. Cone Volume of a Cone Blitzer, Introductory Algebra, 5e – Slide #11 Section 3.3

12. Example : Find the volume of a cylinder that has a height of 4 feet and a radius of 5 feet. Volume of a Cylinder Blitzer, Introductory Algebra, 5e – Slide #12 Section 3.3

13. Angles of a Triangle The sum of the measures of the three angles of a triangle is 180º. Important Fact about Triangles Blitzer, Introductory Algebra, 5e – Slide #13 Section 3.3

14. Example : One angle of a triangle is three times another. The measure of the third angle is 10º more than the smallest angle. Find the angles. 1.Let x represent one of the quantities. x = measure of the first angle 2.Represent other quantities in terms of x. 3x = measure of the second angle x + 10 = measure of the third angle 3.Write an equation in x that describes the conditions. x + 3x + x + 10 = 180 (Sum of the angles of a triangle is 180.) Angles of a Triangle Blitzer, Introductory Algebra, 5e – Slide #14 Section 3.3

15. Example continued : One angle of a triangle is three times another. The measure of the third angle is 10º more than the smallest angle. Find the angles. 4.Solve the equation and answer the question. x + 3x + x + 10 = 180 5x + 10 = 180 Combine like terms. 5x =170 Subtract 10 from both sides. x = 34 Divide both sides by 5. 3x = 102 measure of the second angle x + 10 = 44 measure of the third angle The angles are 34º, 102º and 44º. 5.Check the solution. 102º is three times 34º, thus one angle is three times another. The measure of the third, 44º is 10º more than the smallest angle. The angles sum to 180º Angles of a Triangle Blitzer, Introductory Algebra, 5e – Slide #15 Section 3.3

16. Complementary and Supplementary Complementary Angles: Two angles whose sum is 90º. Supplementary Angles: Two angles whose sum is 180º. Special Angles Blitzer, Introductory Algebra, 5e – Slide #16 Section 3.3

17. Algebraic Expressions for Angle Measures Measure of an angle: x Measure of the angle’s complement: 90 – x Measure of the angle’s supplement: 180 – x Complements and Supplements Blitzer, Introductory Algebra, 5e – Slide #17 Section 3.3

18. Example : Find the measure of an angle that is 12 º less than its complement. 1.Let x represent one of the quantities. x = measure of the first angle 2.Represent other quantities in terms of x. 90 - x = measure of the complement angle 3.Write an equation in x that describes the conditions. x = (90 – x) – 12 The angle is 12 less than the complement. Complement of an Angle Blitzer, Introductory Algebra, 5e – Slide #18 Section 3.3

19. Example continued: Find the measure of an angle that is 12 º less than its complement. 4.Solve the equation and answer the question. x = (90 – x) – 12 x = 90 – x – 12 Remove parentheses. x = 78 – x Combine like terms. 2x = 78 Add x to both sides. x = 39 Divide both sides by 2. x = 39º measure of the first angle 90 – x = 51 measure of the complement angle The angles are 39º and 51º. 5.Check the solution. 39º is 12º less than 51º and the angles sum to 90º Complement of an Angle Blitzer, Introductory Algebra, 5e – Slide #19 Section 3.3

20. Blitzer, Introductory Algebra, 5e – Slide #20 Section 3.3 Area of a Circle, An ApplicationEXAMPLE SOLUTION Which one of the following is a better buy: a large pizza with a 16 inch diameter for $12.00 or two small pizzas, each with a 10 inch diameter, for $12.00? STEP 1: Since the cost is the same, we must just determine which has greatest area – the one big pizza or the two small ones. STEP 2: Find the area of the big pizza. The diameter is 16, so the radius is 8. The area of the big pizza is then: A = which is approximately 201 square inches.

21. Blitzer, Introductory Algebra, 5e – Slide #21 Section 3.3 Application, continuedCONTINUED Now we must find the area of each of the two smaller pizzas. A = which is approximately 78.5 square inches for each smaller pizza. We would have two of these, so the total area would be around 157 square inches. But still… the best buy is the one large. And we know something now that the average person doesn’t consider when purchasing pizza. And that is, when you are considering how big a pizza is – don’t just consider the radius. Consider the square of the radius! And that’s why one sixteen inch pizza would be bigger than two eight inch ones. It’s even bigger than two ten inch ones as we discovered above. Hungry yet?