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1. Solve x2 = 100. ANSWER 10, –10 2. Solve x2 + 9 = 25. ANSWER 4, –4

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Presentation on theme: "1. Solve x2 = 100. ANSWER 10, –10 2. Solve x2 + 9 = 25. ANSWER 4, –4"— Presentation transcript:

1 1. Solve x2 = 100. ANSWER 10, –10 2. Solve x2 + 9 = 25. ANSWER 4, –4 ANSWER 2 5 3. Simplify 20. ANSWER 8 3 4. Simplify 5. Find x2 - 4x – 2 = 0 . ANSWER 6. Find 3x2 = 4x + 4 . ANSWER 2, – 2 3

2 Use Pythagorean’s Theorem and its converse.
Target Use Pythagorean’s Theorem and its converse. You will… Find side lengths of right triangles.

3 VOCABULARY Pythagorean Theorem 7.1 In a right triangle, the square of the lengths of the hypotenuse is equal to the sum of the squares of the lengths of the legs. Pythagorean Triples (page 429) 3, 4, 5 5, 12, , 15, , 24, 25 6, 8, 10 10, 24, , 30, , 48, 50 9, 12, 15 15, 36, , 45, , 72, 75 30, 40, , 120, , 150, , 240, 250 3x, 4x, 5x 5x, 12x, 13x x, 15x, 17x 7x, 24x, 25x

4 Find the length of a hypotenuse
EXAMPLE 1 Find the length of a hypotenuse Find the length of the hypotenuse of the right triangle. SOLUTION (hypotenuse)2 = (leg)2 + (leg)2 Pythagorean Theorem x2 = Substitute. x2 = Multiply. x2 = 100 Add. x = 10 Find the positive square root.

5 GUIDED PRACTICE for Example 1 Identify the unknown side as a leg or hypotenuse. Then, find the unknown side length of the right triangle. Write your answer in simplest radical form. 1. 2. ANSWER Leg; 4 hypotenuse; 2 13 ANSWER

6 EXAMPLE 2 Standardized Test Practice SOLUTION = +

7 Standardized Test Practice
EXAMPLE 2 Standardized Test Practice = + 162 = 42 + x2 Substitute. 256 = 16 + x2 Multiply. 240 = x2 Subtract 16 from each side. 240 = x Find positive square root. ≈ x Approximate with a calculator. ANSWER The ladder is resting against the house at about 15.5 feet above the ground. The correct answer is D.

8 GUIDED PRACTICE for Example 2 The top of a ladder rests against a wall, 23 feet above the ground. The base of the ladder is 6 feet away from the wall. What is the length of the ladder? 3. about 23.8 ft ANSWER The Pythagorean Theorem is only true for what type of triangle? 4. right triangle ANSWER

9 EXAMPLE 3 Find the area of an isosceles triangle Find the area of the isosceles triangle with side lengths 10 meters, 13 meters, and 13 meters. SOLUTION STEP 1 Draw a sketch. By definition, the length of an altitude is the height of a triangle. In an isosceles triangle, the altitude to the base is also a perpendicular bisector. So, the altitude divides the triangle into two right triangles with the dimensions shown.

10 Find the area of an isosceles triangle
EXAMPLE 3 Find the area of an isosceles triangle Use the Pythagorean Theorem to find the height of the triangle. STEP 2 c2 = a2 + b2 Pythagorean Thm. 132 = 52 + h2 Substitute. 169 = 25 + h2 Multiply. 144 = h2 Subtract 25 from each side. 12 = h Find the positive square root.

11 EXAMPLE 3 Find the area of an isosceles triangle STEP 3 Find the area. 1 2 (base) (height) = (10) (12) = 60 m2 1 2 Area = ANSWER The area of the triangle is 60 square meters.

12 GUIDED PRACTICE for Example 3 Find the area of each triangle. 5. 6. ANSWER about ft2. ANSWER 240 m2.

13 Find the length of the hypotenuse of the right triangle.
EXAMPLE 4 Find the length of a hypotenuse using two methods Find the length of the hypotenuse of the right triangle. SOLUTION Method 1: Use a Pythagorean triple. A common Pythagorean triple is 5, 12, 13. Notice that if you multiply the lengths of the legs of the Pythagorean triple by 2, you get the lengths of the legs of this triangle: 5 2 = 10 and = 24. So, the length of the hypotenuse is = 26.

14 Method 2: Use the Pythagorean Theorem.
EXAMPLE 4 Find the length of a hypotenuse using two methods SOLUTION Method 2: Use the Pythagorean Theorem. x2 = Pythagorean Theorem x2 = Multiply. x2 = 676 Add. x = 26 Find the positive square root.

15 GUIDED PRACTICE for Example 4 Find the unknown side length of the right triangle using the Pythagorean Theorem. Then use a Pythagorean triple. 7. 8. ANSWER 15 in. ANSWER 50 cm. Pythagorean triple: 3, 4, 5 Pythagorean triple: 7, 24, 25 3(3) = 9, 4(3) = 12, so 5(3) = 15 7(2) = 14, 24(2) = 48, so 25(2) = 50


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