1. Check it out! 1 3.2.3: Solving Radical Equations

2. A large canvas leaning against a wall forms a right triangle. The horizontal distance from the base of the canvas to the wall represents one leg of the triangle. The vertical distance from the floor to the point of contact between the canvas and the wall represents the other leg. The canvas itself represents the hypotenuse. The diagram at right illustrates the triangle created by the canvas, floor, and wall. Use the diagram and the Pythagorean Theorem, a 2 + b 2 = c 2, to solve the problems on the next slide. 2 3.2.3: Solving Radical Equations

3. 1.If the base of the canvas is 1.5 ft from the wall, and the top of the canvas touches the wall 3.6 ft from the floor, how long is the canvas? 2.Suppose the base of this same canvas were pulled away from the wall until the base was 2.34 ft from the wall. How high along the wall would the canvas now reach? 3.If a 6-foot-long canvas were set against the wall so that the angle formed by the canvas and the floor was 45°, how far from the wall would the base of the canvas be? 3 3.2.3: Solving Radical Equations

4. 1.If the base of the canvas is 1.5 ft from the wall, and the top of the canvas touches the wall 3.6 ft from the floor, how long is the canvas? Use the Pythagorean Theorem, a 2 + b 2 = c 2, to determine the length of the canvas. From the diagram, we are given that a = the horizontal leg, b = the vertical leg, and c = the hypotenuse. Solve for c, the length of the canvas. 4 3.2.3: Solving Radical Equations

5. a 2 + b 2 = c 2 Pythagorean Theorem (1.5) 2 + (3.6) 2 = c 2 Substitute 1.5 for a and 3.6 for b. 2.25 + 12.96 = c 2 Simplify. 15.21 = c 2 3.9 = c Take the square root of both sides. The canvas is 3.9 ft long. 5 3.2.3: Solving Radical Equations

6. 2.Suppose the base of this same canvas were pulled away from the wall until the base was 2.34 ft from the wall. How high along the wall would the canvas now reach? Again, use the Pythagorean Theorem to determine how high the canvas would reach. Solve for b, the vertical distance along the wall between the floor and the canvas. 6 3.2.3: Solving Radical Equations

7. a 2 + b 2 = c 2 Pythagorean Theorem (2.34) 2 + b 2 = (3.9) 2 Substitute 2.34 for a and 3.9 for c. 5.4756 + b 2 = 15.21 Simplify. b 2 = 9.7344 b = 3.12 Take the square root of both sides. The top of the canvas reaches 3.12 ft up the wall. 7 3.2.3: Solving Radical Equations

8. 3.If a 6-foot-long canvas were set against the wall so that the angle formed by the canvas and the floor was 45°, how far from the wall would the base of the canvas be? With the canvas forming a 45° angle with the floor, the right triangle is also an isosceles triangle, with both legs equal: a = b. Use the Pythagorean Theorem to determine how far from the wall the base of the canvas would be. 8 3.2.3: Solving Radical Equations

9. a 2 + b 2 = c 2 Pythagorean Theorem a 2 + (a) 2 = (6) 2 Substitute a for b and 6 for c. 2a 2 = 6 2 Simplify. 2a 2 = 36 Apply the exponent. a 2 = 18 Divide both sides by 2. a ≈ 4.24 Take the square root of both sides. The base of the canvas would be about 4.24 ft from the wall. 9 3.2.3: Solving Radical Equations