1 / 19

2.5 Formulas and Additional Applications from Geometry

2.5 Formulas and Additional Applications from Geometry. Formulas. Definition.

tansy
Download Presentation

2.5 Formulas and Additional Applications from Geometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 2.5Formulas and Additional Applications from Geometry

  2. Formulas Definition. Many applied problems can be solved with formulas. A formula is an equation in which variables are used to describe a relationship. For example, formulas exist for geometric figures such as squares and circles, for distance, for money earned on bank savings, and for converting English measurements to metric measurements. Slide 2.5-3

  3. Objective 1 Solve a formula for one variable, given values of the other variables. Slide 2.5-4

  4. Solve a formula for one variable, given values of the other variables. Given the values of all but one of the variables in a formula, we can find the value of the remaining variable. In Example 1, we use the idea of area. The areaof a plane (two-dimensional) geometric figure is a measure of the surface covered by the figure. In the following formula A is the area of a rectangle with length L and width W. Slide 2.5-5

  5. Using Formulas to Evaluate Variables CLASSROOM EXAMPLE 1 Find the value of the remaining variable. Solution: The length of the rectangle is 38. Once the values of the variables are substituted, the resulting equation is linear in one variable and is solved as in previous sections. Slide 2.5-6

  6. Objective 2 Use a formula to solve an applied problem. Slide 2.5-7

  7. Use a formula to solve an applied problem. It is a good idea to draw a sketch when solving an applied problem that involves a geometric.Examples 2 and 3 use the idea of perimeter. Theperimeter of a plane (two-dimensional) geometric figure is the distance around the figure. For a polygon (e.g., a rectangle, square, or triangle), it is the sum of the lengths of its sides. In the following formula P is the perimeter of a rectangle with length L and width W. Slide 2.5-8

  8. Finding the Dimensions of a Rectangular Yard CLASSROOM EXAMPLE 2 A farmer has 800 m of fencing material to enclose a rectangular field. The width of the field is 50 m less than the length. Find the dimensions of the field. Solution: Let L = the length of the field, then L − 50 = the width of the field. The length of the field is 225 m and the width is 175 m. Slide 2.5-9

  9. Finding the Dimensions of a Triangle CLASSROOM EXAMPLE 3 The longest side of a triangle is 1 in. longer than the medium side. The medium side is 5 in. longer than the shortest side. If the perimeter is 32 in., what are the lengths of the three sides? Solution: Let x− 5 = the length of the shortest side, then x =the length of the medium side, and x + 1 = the length of the longest side. The shortest side of the triangle is 7 in., the medium side is 12 in., and the longest side is 13 in. Slide 2.5-10

  10. Finding the Height of a Triangular Sail CLASSROOM EXAMPLE 4 The area of a triangle is 120 m2. The height is 24 m. Find the length of the base of the triangle. Solution: Let b = the base of the triangle. The base of the triangle is 10 m. Slide 2.5-11

  11. Objective 3 Solve problems involving vertical angles and straight angles. Slide 2.5-12

  12. The figure below shows two intersecting lines forming angles that are numbered , , , and . Angles and lie “opposite” each other. They are called vertical angles. Another pair of vertical angles is and . In geometry, it is known that vertical angles have equal measures. Solve problems involving vertical angles and straight angles. Now look at angles and . When their measures are added, we get 180°, the measure of a straight angle. There are three other such pairs of angles: and , and , and and . Slide 2.5-13

  13. Finding Angle Measures CLASSROOM EXAMPLE 5 Find the measure of each marked angle in the figure. Solution: The two angle measures are 149° and 31°. The answer is not the value of x. Remember to substitute the value of the variable into the expression given for each angle. Slide 2.5-14

  14. Objective 4 Solve a formula for a specified variable. Slide 2.5-15

  15. Solve a formula for a specified variable. Sometimes it is necessary to solve a number of problems that use the same formula. For example, a surveying class might need to solve the formula for the area of a rectangle, A = LW.Suppose that in each problem area (A) and the length (L) of a rectangle are given, and the width (W) must be found. Rather than solving for W each time the formula is used, it would be simpler to rewrite the formula so that it is solved for W. This process is calledsolving for a specified variable,orsolving for a literal equation. When solving a formula for a specified variable, we treat the specified variable as if it were the ONLY variable in the equation and treat the other variables as if they were numbers. We use the same steps to solve the equation for specified variables that we use to solve equations with just one variable. Slide 2.5-16

  16. Solve for t. or Solving for a Specified Variable CLASSROOM EXAMPLE 6 Solution: Slide 2.5-17

  17. Solve for h. or Solving for a Specified Variable CLASSROOM EXAMPLE 7 Solution: Slide 2.5-18

  18. Solve for t. Solving for a Specified Variable CLASSROOM EXAMPLE 8 Solution: or Slide 2.5-19

  19. Solve for h. or Solving for a Specified Variable CLASSROOM EXAMPLE 9 Solution: Slide 2.5-20

More Related