Question

extra practice
in exercises 1 and 2, identify the similar triangles.
1.
2.
hkj~kij~jhi
npm~pom~mno
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.
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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 \(b=6\), we have \(\sqrt{2\times6}=\sqrt{12}=2\sqrt{3}\).

Step3: Calculate geometric mean for 5 and 45

For \(a = 5\) and \(b = 45\), we have \(\sqrt{5\times45}=\sqrt{225}=15\).

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

In a right - triangle, if the altitude \(x\) is drawn to the hypotenuse of a right - triangle, then \(x^{2}=9\times16\). So \(x=\sqrt{9\times16}=\sqrt{144}=12\).

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

Using the geometric - mean relationships in right - triangles, if the segments of the hypotenuse are \(2\) and \(9 - 2=7\), and the altitude is drawn to the hypotenuse, then \(y^{2}=2\times9 = 18\), so \(y=\sqrt{18}=3\sqrt{2}\).

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

If we assume similar right - triangles or geometric - mean relationships, if the segments of the hypotenuse are \(7\) and \(49\), then \(t^{2}=7\times49\), so \(t=\sqrt{7\times49}=7\sqrt{7}\).

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

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

Answer:

1. \(2\sqrt{3}\)
2. \(15\)
3. \(12\)
4. \(3\sqrt{2}\)
5. \(7\sqrt{7}\)
6. \(8\)