Question

geometric mean leg theorem
directions: find the value of x.
12.
13.
directions: find the values of x, y, and z.
14.
15.
© gina wilson (all things algebra®, llc), 2014 - 2018

Explanation:

Step1: Solve Q13 (find x)

Apply Geometric Mean Leg Theorem: If a leg of a right triangle is the geometric mean of the hypotenuse and the adjacent segment, then for the leg of length 12, $12^2 = 7 \times x$.
$x = \frac{12^2}{7} = \frac{144}{7} \approx 20.57$

Step2: Solve Q14 (find x, y, z)

Substep2.1: Find hypotenuse x

Use Pythagorean theorem on the large right triangle: $x = \sqrt{24^2 + 27^2}$
$x = \sqrt{576 + 729} = \sqrt{1305} = 3\sqrt{145} \approx 36.12$

Substep2.2: Find altitude y

Apply Geometric Mean Altitude Theorem: $y = \frac{24 \times 27}{x}$
$y = \frac{648}{3\sqrt{145}} = \frac{216}{\sqrt{145}} = \frac{216\sqrt{145}}{145} \approx 17.94$

Substep2.3: Find segment z

Apply Geometric Mean Leg Theorem: $z = \frac{27^2}{x}$
$z = \frac{729}{3\sqrt{145}} = \frac{243}{\sqrt{145}} = \frac{243\sqrt{145}}{145} \approx 20.13$

Step3: Solve Q15 (find x, y, z)

Substep3.1: Find segment x

Apply Geometric Mean Leg Theorem: $20^2 = 8 \times (8 + x)$
$400 = 64 + 8x \implies 8x = 336 \implies x = 42$

Substep3.2: Find hypotenuse y

Use Pythagorean theorem on the large right triangle: $y = \sqrt{(8+42)^2 + 20^2}$
$y = \sqrt{2500 + 400} = \sqrt{2900} = 10\sqrt{29} \approx 53.85$

Substep3.3: Find segment z

Apply Geometric Mean Leg Theorem: $z = \frac{20 \times 42}{y}$
$z = \frac{840}{10\sqrt{29}} = \frac{84}{\sqrt{29}} = \frac{84\sqrt{29}}{29} \approx 15.61$

Answer:

1. $x=\frac{144}{7}$
2. $x=3\sqrt{145}$, $y=\frac{216\sqrt{145}}{145}$, $z=\frac{243\sqrt{145}}{145}$
3. $x=42$, $y=10\sqrt{29}$, $z=\frac{84\sqrt{29}}{29}$