Question

geometric mean - leg theorem
complete problems 1 - 6. show your work:

1. $x = \underline{15.9}$

$\frac{x}{14}=\frac{18}{x}$
$\sqrt{x^2}=\sqrt{252}$
$x=15.9$

1. $x = \underline{10}$

$\frac{x}{5}=\frac{20}{x}$
$\sqrt{x^2}=\sqrt{100}$
$x=10$

1. $x = \underline{21.3}$

$\frac{16}{12}=\frac{x}{16}$
$\frac{12x}{12}=\frac{256}{12}$
$x=21.3$

1. $x = \underline{\quad\quad}$
2. $x = \underline{\quad\quad}$
3. $x = \underline{\quad\quad}$ $y = \underline{\quad\quad}$ $z = \underline{\quad\quad}$

directions: solve for x, y, and z.

Explanation:

Step1: Solve problem 4 (Leg Theorem)

By the Geometric Mean (Leg) Theorem, the leg of the right triangle is the geometric mean of the hypotenuse and the adjacent segment. For the leg of length 20:
$\frac{x}{20} = \frac{20}{8}$
$x = \frac{20^2}{8} = \frac{400}{8} = 50$

Step2: Solve problem 5 (Leg Theorem)

Apply the Geometric Mean (Leg) Theorem: the leg of length 55 is the geometric mean of the hypotenuse segment $x$ and the full hypotenuse 63.
$\frac{63}{55} = \frac{55}{x}$
$x = \frac{55^2}{63} = \frac{3025}{63} \approx 48.02$

Step3: Solve problem 6 (Find x first)

Use the Geometric Mean (Altitude) Theorem first for the altitude 6: $\frac{15}{x} = \frac{x}{15-6}$
$\frac{15}{x} = \frac{x}{9}$
$x^2 = 15 \times 9 = 135$
$x = \sqrt{135} = 3\sqrt{15} \approx 11.62$

Step4: Solve problem 6 (Find z)

Use the Geometric Mean (Leg) Theorem for leg z: $\frac{15}{z} = \frac{z}{6}$
$z^2 = 15 \times 6 = 90$
$z = \sqrt{90} = 3\sqrt{10} \approx 9.49$

Step5: Solve problem 6 (Find y)

Use the Geometric Mean (Leg) Theorem for leg y: $\frac{15}{y} = \frac{y}{9}$
$y^2 = 15 \times 9 = 135$
$y = \sqrt{135} = 3\sqrt{15} \approx 11.62$
(Or use Pythagoras: $y = \sqrt{x^2 - 6^2} = \sqrt{135 - 36} = \sqrt{99} = 3\sqrt{11} \approx 9.95$ correction: correct leg-hypotenuse pairing: hypotenuse is 15, segment adjacent to y is 9, so $y = \sqrt{15 \times 9} = 3\sqrt{15}$ is wrong. Correct: $y = \sqrt{9 \times 15}$ is for x, correct for y: $\frac{15}{y} = \frac{y}{9}$ is incorrect. Correct: the altitude splits hypotenuse into 9 and 6. So $y = \sqrt{15 \times 6} = \sqrt{90} = 3\sqrt{10}$? No, correction: Leg Theorem: leg = $\sqrt{\text{hypotenuse} \times \text{adjacent segment}}$. For leg x: adjacent segment is 9, so $x = \sqrt{15 \times 9} = 3\sqrt{15}$. For leg y: adjacent segment is 6, so $y = \sqrt{15 \times 6} = 3\sqrt{10}$. For z (altitude): $z = \sqrt{9 \times 6} = \sqrt{54} = 3\sqrt{6} \approx 7.35$. Final correction for step 4 and 5:
Altitude z: $z = \sqrt{9 \times 6} = 3\sqrt{6} \approx 7.35$
Leg x: $x = \sqrt{15 \times 9} = 3\sqrt{15} \approx 11.62$
Leg y: $y = \sqrt{15 \times 6} = 3\sqrt{10} \approx 9.49$

Answer:

1. $x = 50$
2. $x \approx 48.02$ (or $x = \frac{3025}{63}$)
3. $x = 3\sqrt{15} \approx 11.62$, $y = 3\sqrt{10} \approx 9.49$, $z = 3\sqrt{6} \approx 7.35$