8-1 Geometric Mean
altitude drawn to the hypotenuse of a right triangle
separates the hypotenuse into two segments. The
length of this altitude is the geometric mean between
the lengths of these two segments. Solve for y. Write a similarity statement identifying the
three similar triangles in the figure.
15. SOLUTION: 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.
By the Geometric Mean (Leg) Theorem the altitude
drawn to the hypotenuse of a right triangle separates
the hypotenuse into two segments. The length of a
leg of this triangle is the geometric mean between the
length of the hypotenuse and the segment of the
hypotenuse adjacent to that leg. So,
is the altitude to the hypotenuse of the right triangle XYW. Therefore,
19. Solve for x. SOLUTION: By the Geometric Mean (Leg) Theorem the altitude
drawn to the hypotenuse of a right triangle separates
the hypotenuse into two segments. The length of a
leg of this triangle is the geometric mean between the
length of the hypotenuse and the segment of the
hypotenuse adjacent to that leg. Solve for z.
By the Geometric Mean (Altitude) Theorem the
altitude drawn to the hypotenuse of a right triangle
separates the hypotenuse into two segments. The
length of this altitude is the geometric mean between
the lengths of these two segments. Solve for y. eSolutions Manual - Powered by Cognero
Page 1
By the Geometric Mean (Leg) Theorem the altitude
drawn to the hypotenuse of a right triangle separates
the hypotenuse into two segments and the length of a
leg of this triangle is the geometric mean between the
length of the hypotenuse and the segment of the
hypotenuse adjacent to that leg.
drawn to the hypotenuse of a right triangle separates
the hypotenuse into two segments. The length of a
leg of this triangle is the geometric mean between the
length of theMean
hypotenuse and the segment of the
8-1 Geometric
hypotenuse adjacent to that leg. Solve for z.
Solve for y. By the Geometric Mean (Leg) Theorem the altitude
drawn to the hypotenuse of a right triangle separates
the hypotenuse into two segments and the length of a
leg of this triangle is the geometric mean between the
length of the hypotenuse and the segment of the
hypotenuse adjacent to that leg.
22. SOLUTION: Solve for z. By the Geometric Mean (Altitude) Theorem the
altitude drawn to the hypotenuse of a right triangle
separates the hypotenuse into two segments and the
length of this altitude is the geometric mean between
the lengths of these two segments. Solve for x. Find the geometric mean between each pair of
numbers.
27. By the Geometric Mean (Leg) Theorem the altitude
drawn to the hypotenuse of a right triangle separates
the hypotenuse into two segments and the length of a
leg of this triangle is the geometric mean between the
length of the hypotenuse and the segment of the
hypotenuse adjacent to that leg.
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and
SOLUTION: By the definition, the geometric mean x of any two
numbers a and b is given by
is
Therefore, the geometric mean of
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Solve for y. ALGEBRA Find the value(s) of the variable.
8-1 Geometric Mean
Find the geometric mean between each pair of
numbers.
27. and
SOLUTION: By the definition, the geometric mean x of any two
numbers a and b is given by
Therefore, the geometric mean of
is
36. SOLUTION: By the Geometric Mean (Altitude) Theorem the
altitude drawn to the hypotenuse of a right triangle
separates the hypotenuse into two segments and the
length of this altitude is the geometric mean between
the lengths of these two segments. ALGEBRA Find the value(s) of the variable.
35. SOLUTION: By the Geometric Mean (Altitude) Theorem the
altitude drawn to the hypotenuse of a right triangle
separates the hypotenuse into two segments and the
length of this altitude is the geometric mean between
the lengths of these two segments. Use the quadratic formula to find the roots of the
quadratic equation.
If w = –16, the length of the altitude will be –16 + 4 =
–12 which is not possible, as a length cannot be
negative. Therefore, w = 8.
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36. SOLUTION: Page 3