7.1 Geometric Mean.  Find the geometric mean between two numbers  Solve problems involving relationships between parts of right triangles and the altitude.

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Presentation transcript:

7.1 Geometric Mean

 Find the geometric mean between two numbers  Solve problems involving relationships between parts of right triangles and the altitude to its hypotenuse

 The geometric mean between two numbers is the positive square root of their products.  In other words, given two positive numbers such as a and b, the geometric mean is the positive number x such that a : x = x : b We can also write this as fractions, a / x = x / b or as cross products, x 2 = ab.

Find the geometric mean between 2 and 50. Definition of geometric mean Let x represent the geometric mean. Cross products Take the positive square root of each side. Simplify. Answer: The geometric mean is 10. Example 1a:

Find the geometric mean between 25 and 7. Definition of geometric mean Let x represent the geometric mean. Cross products Take the positive square root of each side. Simplify. Answer: The geometric mean is about Use a calculator. Example 1b:

a. Find the geometric mean between 3 and 12. b. Find the geometric mean between 4 and 20. Answer: 6 Answer: 8.9 Your Turn:

 If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. ∆CBD ~ ∆ABC ∆ACD ~ ∆ABC ∆CBD ~ ∆ACD

 Theorem 7.2: In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments.  Theorem 7.3: In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. BD CD = AD AB CB = DB AB AC = AD

= = 6 x = x 3 18 = x 2 √18 = x √9 ∙ √2 = x 3 √2 = x y y 2 14 = y 2 7 y y 2 √14 = y

Example 2:

Cross products Take the positive square root of each side. Use a calculator. Answer: CD is about Example 2:

Answer: about 8.5 Your Turn:

KITES Ms. Turner is constructing a kite for her son. She has to arrange perpendicularly two support rods, the shorter of which is 27 inches long. If she has to place the short rod 7.25 inches from one end of the long rod in order to form two right triangles with the kite fabric, what is the length of the long rod? Example 3:

Draw a diagram of one of the right triangles formed. Let be the altitude drawn from the right angle of Example 3:

Cross products Divide each side by Answer: The length of the long rod is , or about 32.4 inches long. Example 3:

AIRPLANES A jetliner has a wingspan, BD, of 211 feet. The segment drawn from the front of the plane to the tail, intersects at point E. If AE is 163 feet, what is the length of the aircraft? Answer: about ft Your Turn:

Find c and d in Example 4:

is the altitude of right triangle JKL. Use Theorem 7.2 to write a proportion. Cross products Divide each side by 5. Example 4:

is the leg of right triangle JKL. Use the Theorem 7.3 to write a proportion. Answer: Cross products Take the square root. Simplify. Use a calculator. Example 4:

Find e and f. Answer: f Your Turn:

 Pre-AP Geometry Pg. 346 #13 – 38 & #44  Geometry Pg. 346 #13 – 32, #