Question

in exercises 3 and 4, find the geometric mean of the two numbers.

1. 2 and 6
2. 5 and 45

in exercises 5 - 8, find the value of the variable.
5.
6.
7.
8.

Explanation:

Step1: Recall geometric - mean formula

The geometric mean of two positive numbers \(a\) and \(b\) is \(\sqrt{ab}\).

Step2: Calculate geometric mean for 2 and 6

For \(a = 2\) and \(b=6\), the geometric mean \(G=\sqrt{2\times6}=\sqrt{12} = 2\sqrt{3}\).

Step3: Calculate geometric mean for 5 and 45

For \(a = 5\) and \(b = 45\), the geometric mean \(G=\sqrt{5\times45}=\sqrt{225}=15\).

Step4: Solve for \(x\) in Exercise 5

Using the geometric - mean theorem in a right - triangle (\(CD^{2}=AD\times DB\)), we have \(x^{2}=9\times16\), so \(x=\sqrt{9\times16}=12\).

Step5: Solve for \(y\) in Exercise 6

Using the geometric - mean theorem in a right - triangle, if the segments of the hypotenuse are \(2\) and \(9 - 2=7\), then \(y^{2}=2\times7\), so \(y=\sqrt{14}\).

Step6: Solve for \(t\) in Exercise 7

Using the geometric - mean theorem in a right - triangle, if the segments of the hypotenuse are \(7\) and \(49+7 = 56\), then \(t^{2}=7\times56\), \(t=\sqrt{7\times56}=\sqrt{7\times7\times8}=14\sqrt{2}\).

Step7: Solve for \(a\) in Exercise 8

Using the geometric - mean theorem in a right - triangle (\(6^{2}=3\times(a + 4)\)). First, expand the right - side: \(36=3a+12\). Then, subtract 12 from both sides: \(3a=36 - 12=24\). Divide both sides by 3: \(a = 8\).

Answer:

1. Geometric mean of 2 and 6: \(2\sqrt{3}\)
2. Geometric mean of 5 and 45: 15
3. Value of \(x\) in Exercise 5: 12
4. Value of \(y\) in Exercise 6: \(\sqrt{14}\)
5. Value of \(t\) in Exercise 7: \(14\sqrt{2}\)
6. Value of \(a\) in Exercise 8: 8