Slide 1 8 1 Geometric Mean and Pythagorean

Slide #1

8. 1 Geometric Mean and Pythagorean Theorem Geometry

Objectives/Assignment l l Use Pythagorean theorem to solve problems Use Geometric Mean and Pythagorean Theorem to solve reallife problems

Geometric Mean l The geometric mean of two positive numbers a and b is the positive number x such that a x = x b If you solve this proportion for x, you find that x = √a ∙ b which is a positive number.

Book Example Pg 397

Geometric Mean Example l For example, the geometric mean of 8 and 18 is 12, because 8 12 = 12 18 and also because x = √ 8 ∙ 18 = x = √ 144 = 12

Practice “Geometric Mean”

Geometric mean l PAPER SIZES. International standard paper sizes are commonly used all over the world. The various sizes all have the same width-to-length ratios. Two sizes of paper are shown, called A 4 and A 3. The distance labeled x is the geometric mean of 210 mm and 420 mm. Find the value of x.

Solution: 210 x = x 420 X 2 = 210 ∙ 420 X = √ 210 ∙ 420 X = 297 mm Write proportion Cross product property Simplify

EXAMPLE 1 Find the length of a hypotenuse Find the length of the hypotenuse of the right triangle. (hypotenuse)2 = (leg)2 + (leg)2 Pythagorean Theorem x 2 = 62 + 82 Substitute. x 2 = 36 + 64 Square. x 2 = 100 Add. x = 10 Find the positive square root.

GUIDED PRACTICE for Example 1 1. Find the unknown side length of the right triangle. Write your answer in simplest radical form. ANSWER 4

GUIDED PRACTICE 2. for Example 1 Find the unknown side length of the right triangle. ANSWER 13

X = 3. 2 Y=X+5 Y = 3. 2 + 5 Y = 8. 2

Y² = 32 Y = 5. 7²+ 8² = x² 32. 49 + 64 = x² 96. 49 = x² X = 9. 8

x 4 x² = 20 x = 4. 5²+ 5² = y² 20 + 25 = y² 45 = y² X = 6. 7

x²+ 5² = 9² x² + 25 = 81 x² = 81 -25 X² = 56 X = 7. 5

30²+ 16² = x² 900+256 = x² x² = 1156 X = 34

x+ 60² = 65² x² +3600 = 4225 x² = 4225 -3600 x² = 625 X = 25