Question

t 12.4 i can use geometric means theorems to solve problems. /10
leave all answers in radical form (if necessary).

1. what is the geometric mean of 2 and 32? (2 pts)
2. ____________
3. what is the geometric mean of 4 and 75? (2 pts)
4. ____________
5. use the figure to find the value of x. (3 pts)
6. x = ________
7. use the figure to find the value of x. (3 pts)
8. x = ________

extra credit:
solve for x.

Explanation:

Step1: Geometric mean formula

The geometric mean of two numbers $a$ and $b$ is $\sqrt{a \times b}$.

Q12: Calculate geometric mean

$\sqrt{2 \times 32} = \sqrt{64} = 8$

Q13: Calculate geometric mean

$\sqrt{4 \times 75} = \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3}$

Step2: Apply geometric mean (Q14)

For a right triangle, the leg is the geometric mean of the adjacent segment and the hypotenuse.
$x = \sqrt{9 \times 25} = \sqrt{225} = 15$

Step3: Apply geometric mean (Q15)

The altitude to the hypotenuse is the geometric mean of the two hypotenuse segments.
First find the other segment: $100 - 36 = 64$
$x = \sqrt{36 \times 64} = \sqrt{2304} = 48$

Step4: Proportional segments (Extra Credit)

Use the triangle proportionality theorem: $\frac{4}{2x-3} = \frac{12}{10}$
Cross multiply: $12(2x-3) = 4 \times 10$
$24x - 36 = 40$
$24x = 76$
$x = \frac{76}{24} = \frac{19}{6}$

Answer:

1. $8$
2. $10\sqrt{3}$
3. $x=15$
4. $x=48$

Extra Credit: $x=\frac{19}{6}$