Use the following Media4Math resources with this Illustrative Math lesson.
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with algebraic expressions. Two rectangles are shown, with the smaller one having sides of 5 and 12, and the larger one having sides of x + 5 and 3x + 6. The problem requires setting up a proportion: 5 / 12 = (x + 5) / (3x + 6). Solving this equation leads to x = 10, which then allows us to find the side lengths of the larger rectangle as 15 and 36. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with algebraic expressions. Two rectangles are shown, with the smaller one having sides of 5 and 12, and the larger one having sides of x + 5 and 3x + 6. The problem requires setting up a proportion: 5 / 12 = (x + 5) / (3x + 6). Solving this equation leads to x = 10, which then allows us to find the side lengths of the larger rectangle as 15 and 36. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms. Two parallelograms are shown, with the smaller one having sides of 8 and 18, and the larger one having sides of 20 and x. The problem requires finding the length of side x by setting up a proportion: 8 / 18 = 20 / x. Solving this equation leads to x = 45. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms. Two parallelograms are shown, with the smaller one having sides of 8 and 18, and the larger one having sides of 20 and x. The problem requires finding the length of side x by setting up a proportion: 8 / 18 = 20 / x. Solving this equation leads to x = 45. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, with the smaller one having sides of 9 and 21, and the larger one having sides of x + 12 and 4x + 3. The problem requires setting up a proportion: 9 / 21 = (x + 12) / (4x + 3). Solving this equation leads to x = 15, which then allows us to find the side lengths of the larger parallelogram as 27 and 63. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, with the smaller one having sides of 9 and 21, and the larger one having sides of x + 12 and 4x + 3. The problem requires setting up a proportion: 9 / 21 = (x + 12) / (4x + 3). Solving this equation leads to x = 15, which then allows us to find the side lengths of the larger parallelogram as 27 and 63. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two triangles are similar using proportions. Two triangles are shown, both with a 70° angle. The first triangle has sides of 12 and 10, while the second has sides of 18 and 15. The problem requires setting up a proportion to check for similarity: 12 / 10 = 18 / 15. After simplifying, both ratios are equal (6 / 5 = 6 / 5), confirming that the triangles are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two triangles are similar using proportions. Two triangles are shown, both with a 70° angle. The first triangle has sides of 12 and 10, while the second has sides of 18 and 15. The problem requires setting up a proportion to check for similarity: 12 / 10 = 18 / 15. After simplifying, both ratios are equal (6 / 5 = 6 / 5), confirming that the triangles are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for b in a proportion where a = 8, c = 4, and d = 3. We set up the proportion 8 / b = 4 / 3 and solve for b, resulting in b = 6. This problem shows how to find an unknown value in the denominator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for b in a proportion where a = 8, c = 4, and d = 3. We set up the proportion 8 / b = 4 / 3 and solve for b, resulting in b = 6. This problem shows how to find an unknown value in the denominator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two triangles are not similar using proportions. Two triangles are shown, both with a 75° angle. The first triangle has sides of 15 and 9, while the second has sides of 28 and 18. The problem requires setting up a proportion to check for similarity: 15 / 9 = 28 / 18. After simplifying, the ratios are not equal (5 / 3 ≠ 14 / 9), concluding that the triangles are not similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two triangles are not similar using proportions. Two triangles are shown, both with a 75° angle. The first triangle has sides of 15 and 9, while the second has sides of 28 and 18. The problem requires setting up a proportion to check for similarity: 15 / 9 = 28 / 18. After simplifying, the ratios are not equal (5 / 3 ≠ 14 / 9), concluding that the triangles are not similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two right triangles are similar using proportions. Two right triangles are shown, one with legs of length 4 and 3, and the other with legs of length 10 and 7.5. The problem requires setting up a proportion to check for similarity: 4 / 3 = 10 / 7.5. After simplifying, both ratios are equal (4 / 3 = 4 / 3), confirming that the triangles are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two right triangles are similar using proportions. Two right triangles are shown, one with legs of length 4 and 3, and the other with legs of length 10 and 7.5. The problem requires setting up a proportion to check for similarity: 4 / 3 = 10 / 7.5. After simplifying, both ratios are equal (4 / 3 = 4 / 3), confirming that the triangles are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two right triangles are not similar using proportions. Two right triangles are shown, one with legs of length 12 and 5, and the other with legs of length 35 and 15. The problem requires setting up a proportion to check for similarity: 12 / 5 = 35 / 15. After simplifying, the ratios are not equal (12 / 5 ≠ 7 / 3), concluding that the triangles are not similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two right triangles are not similar using proportions. Two right triangles are shown, one with legs of length 12 and 5, and the other with legs of length 35 and 15. The problem requires setting up a proportion to check for similarity: 12 / 5 = 35 / 15. After simplifying, the ratios are not equal (12 / 5 ≠ 7 / 3), concluding that the triangles are not similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two rectangles are similar using proportions. Two rectangles are shown, with the first having dimensions 3 and 8, and the second having dimensions 9 and 24. The problem requires setting up a proportion to check for similarity: 3 / 8 = 9 / 24. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two rectangles are similar using proportions. Two rectangles are shown, with the first having dimensions 3 and 8, and the second having dimensions 9 and 24. The problem requires setting up a proportion to check for similarity: 3 / 8 = 9 / 24. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two rectangles are similar using proportions with algebraic expressions. Two rectangles are shown, with the first having dimensions 11.5 and 23, and the second having dimensions 23x and 46x. The problem requires setting up a proportion to check for similarity: 11.5 / 23 = 23x / 46x. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar for any value of x. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two rectangles are similar using proportions with algebraic expressions. Two rectangles are shown, with the first having dimensions 11.5 and 23, and the second having dimensions 23x and 46x. The problem requires setting up a proportion to check for similarity: 11.5 / 23 = 23x / 46x. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar for any value of x. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are similar using proportions. Two parallelograms are shown, with the first having dimensions 6 and 9, and the second having dimensions 15 and 22.5. The problem requires setting up a proportion to check for similarity: 6 / 9 = 15 / 22.5. After simplifying, the ratios are equal, confirming that the parallelograms are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are similar using proportions. Two parallelograms are shown, with the first having dimensions 6 and 9, and the second having dimensions 15 and 22.5. The problem requires setting up a proportion to check for similarity: 6 / 9 = 15 / 22.5. After simplifying, the ratios are equal, confirming that the parallelograms are indeed similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are not similar using proportions. Two parallelograms are shown, with the first having dimensions 9 and 28, and the second having dimensions 18 and 54. The problem requires setting up a proportion to check for similarity: 9 / 18 ≠ 28 / 54. After simplifying, the ratios are not equal, concluding that the parallelograms are not similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are not similar using proportions. Two parallelograms are shown, with the first having dimensions 9 and 28, and the second having dimensions 18 and 54. The problem requires setting up a proportion to check for similarity: 9 / 18 ≠ 28 / 54. After simplifying, the ratios are not equal, concluding that the parallelograms are not similar. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar triangles with algebraic expressions. Two triangles are shown, one with side lengths 9 and 18, and the other with expressions 6x and 10x + 6. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 9 / 18 = 6x / (10x + 6). Solving this equation leads to x = 3. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar triangles with algebraic expressions. Two triangles are shown, one with side lengths 9 and 18, and the other with expressions 6x and 10x + 6. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 9 / 18 = 6x / (10x + 6). Solving this equation leads to x = 3. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, one with side lengths 4 and 3, and the other with expressions 3x and 2x + 1. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 4 / 3 = 3x / (2x + 1). Solving this equation leads to x = 4. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, one with side lengths 4 and 3, and the other with expressions 3x and 2x + 1. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 4 / 3 = 3x / (2x + 1). Solving this equation leads to x = 4. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with an unknown side length. Two rectangles are shown, one with side lengths of 4 and x, and the other with side lengths of 15 and 4. The problem requires setting up a proportion to determine the value of x for which the rectangles are similar: 4 / 15 = x / 4. Solving this equation leads to x = 16/15. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with an unknown side length. Two rectangles are shown, one with side lengths of 4 and x, and the other with side lengths of 15 and 4. The problem requires setting up a proportion to determine the value of x for which the rectangles are similar: 4 / 15 = x / 4. Solving this equation leads to x = 16/15. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for b in a proportion where a is expressed as x + 2, and c and d are given constants (c = 5, d = 2). We set up the equation (x + 2) / b = 5 / 2 and solve for b, resulting in the expression b = (2(x + 2)) / 5. This problem demonstrates how to handle algebraic expressions in proportions. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for b in a proportion where a is expressed as x + 2, and c and d are given constants (c = 5, d = 2). We set up the equation (x + 2) / b = 5 / 2 and solve for b, resulting in the expression b = (2(x + 2)) / 5. This problem demonstrates how to handle algebraic expressions in proportions. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, one with side lengths of 6 and x, and the other with expressions of 3x + 2 and 22. The problem requires setting up a proportion to determine the value of x for which the parallelograms are similar: 6 / 22 = x / (3x + 2). Solving this equation leads to x = 3. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, one with side lengths of 6 and x, and the other with expressions of 3x + 2 and 22. The problem requires setting up a proportion to determine the value of x for which the parallelograms are similar: 6 / 22 = x / (3x + 2). Solving this equation leads to x = 3. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for c in a proportion where a = 9, b = 4, and d = 12. We set up the proportion 9 / 4 = c / 12 and solve for c, resulting in c = 27. This problem shows how to find an unknown value in the numerator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for c in a proportion where a = 9, b = 4, and d = 12. We set up the proportion 9 / 4 = c / 12 and solve for c, resulting in c = 27. This problem shows how to find an unknown value in the numerator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for c in a proportion where a = 8, b = 3, and d is expressed as x + 3. We set up the equation 8 / 3 = c / (x + 3) and solve for c, resulting in the expression c = (8(x + 3)) / 3. This problem demonstrates how to handle algebraic expressions in the denominator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for c in a proportion where a = 8, b = 3, and d is expressed as x + 3. We set up the equation 8 / 3 = c / (x + 3) and solve for c, resulting in the expression c = (8(x + 3)) / 3. This problem demonstrates how to handle algebraic expressions in the denominator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for d in a proportion where a = 12, b = 5, and c = 48. We set up the proportion 12 / 5 = 48 / d and solve for d, resulting in d = 20. This problem shows how to find an unknown value in the denominator of a proportion when all other values are known. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for d in a proportion where a = 12, b = 5, and c = 48. We set up the proportion 12 / 5 = 48 / d and solve for d, resulting in d = 20. This problem shows how to find an unknown value in the denominator of a proportion when all other values are known. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for d in a proportion where a = 11, b = 5, and c is expressed as x - 4. We set up the equation 11 / 5 = (x - 4) / d and solve for d, resulting in the expression d = (5(x - 4)) / 11. This problem demonstrates how to handle algebraic expressions in the numerator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for d in a proportion where a = 11, b = 5, and c is expressed as x - 4. We set up the equation 11 / 5 = (x - 4) / d and solve for d, resulting in the expression d = (5(x - 4)) / 11. This problem demonstrates how to handle algebraic expressions in the numerator of a proportion. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar triangles. Two triangles are shown, with one having sides of 3 and 4, and the other having sides of 6 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 3 / 4 = 6 / x. Solving this equation leads to x = 8. |
Proportions | |
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar triangles. Two triangles are shown, with one having sides of 3 and 4, and the other having sides of 6 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 3 / 4 = 6 / x. Solving this equation leads to x = 8. |
Proportions | |
Math Example--Solving Equations--Solving Equations Using Angle Properties: Example 1 | Solving Equations Using Angle Properties: Example 1TopicEquations DescriptionThis example demonstrates solving equations using angle properties, specifically focusing on supplementary angles. Supplementary angles are two angles that add up to 180 degrees. In this case, we have one known angle of 135° and an unknown angle x. To solve such equations, we use the fundamental property of supplementary angles: their sum equals 180°. |
Applications of Equations and Inequalities and Definition of an Angle | |
Math Example--Solving Equations--Solving Equations Using Angle Properties: Example 10 | Solving Equations Using Angle Properties: Example 10TopicEquations DescriptionThis example illustrates solving equations using angle properties, focusing on parallel lines cut by a transversal and supplementary angles. When parallel lines are cut by a transversal, pairs of supplementary angles are formed, meaning they sum to 180°. In this scenario, we have one known angle of 118° and an unknown angle x. However, angle y and the 118° angle are alternate exterior angles, which are congruent. This means we can set up this equation using the property of supplementary angles: 118 + x = 180 |
Applications of Equations and Inequalities and Definition of an Angle | |
Math Example--Solving Equations--Solving Equations Using Angle Properties: Example 2 | Solving Equations Using Angle Properties: Example 2TopicEquations DescriptionThis example illustrates solving equations using angle properties, focusing on supplementary angles. Supplementary angles are two angles that sum to 180 degrees. In this scenario, we have one known angle of 75° and an unknown angle x. The equation for supplementary angles is always in the form: angle1 + angle2 = 180°. Here, we can write 75 + x = 180. To solve for x, we subtract 75 from both sides: x = 180 - 75, giving us x = 105°. |
Applications of Equations and Inequalities and Definition of an Angle | |
Math Example--Solving Equations--Solving Equations Using Angle Properties: Example 3 | Solving Equations Using Angle Properties: Example 3TopicEquations DescriptionThis example demonstrates solving equations using angle properties, specifically focusing on straight angles and vertical angles. A straight angle measures 180°, and vertical angles are always congruent. In this scenario, we have two known angles (36° and 72°) and an unknown angle x. This unknown angle x is vertical (and therefore congruent) to angle z. The angles 36, 72, and z form a straight, but since z is congruent to x, we can write the sum of this straight angle: x + 36 + 72 = 180 x = 72° |
Applications of Equations and Inequalities and Definition of an Angle | |
Math Example--Solving Equations--Solving Equations Using Angle Properties: Example 4 | Solving Equations Using Angle Properties: Example 4TopicEquations DescriptionThis example illustrates solving equations using angle properties, focusing on straight angles and vertical angles. A straight angle measures 180°, and vertical angles are always congruent. In this scenario, we have two known angles (42° and 55°) and an unknown angle x. However, angle x is vertical (and therefore congruent) to angle y. The equation can be set up based on the fact that the sum of angles on a straight line is 180°. Thus, we have: 42 + 55 + x = 180 |
Applications of Equations and Inequalities and Definition of an Angle | |
Math Example--Solving Equations--Solving Equations Using Angle Properties: Example 5 | Solving Equations Using Angle Properties: Example 5TopicEquations DescriptionThis example demonstrates solving equations using angle properties, specifically focusing on complementary angles and the exterior angle theorem. Complementary angles are two angles that sum to 90°, while the exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. In this scenario, we have a right angle (90°), an exterior angle (125°), and an unknown angle x. Using the exterior angle theorem, we can set up the equation: x + 90 = 125. To solve for x, we subtract 90 from both sides: x = 35°. |
Applications of Equations and Inequalities and Definition of an Angle |