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1. find x evaluate independent practice lesson 12.4 homework complete p…

Question

  1. find x

evaluate independent practice
lesson 12.4 homework
complete problems 1-6 below for independent practice.
when you are finished, check the solutions with your teacher.
find the missing length using the geometric mean theorem.
1.
2.
3.
4.
5.
6.

Explanation:

Identify the geometric theorems

Using the Geometric Mean and Right Triangle Similarity knowledge points, we apply the geometric mean theorems for right triangles.

For a right triangle with an altitude drawn to the hypotenuse:

  1. Altitude Rule: The altitude \(h\) is the geometric mean of the two segments of the hypotenuse \(p_1\) and \(p_2\):
$$h^2 = p_1 \cdot p_2$$
  1. Leg Rule: Each leg \(a\) is the geometric mean of the adjacent hypotenuse segment \(p_1\) and the entire hypotenuse \(c\):
$$a^2 = p_1 \cdot c$$

Solve the top introductory problem

Using the Geometric Mean and Solving Linear Equations knowledge points:

  • The altitude is \(x\), and the two segments of the hypotenuse are \(5\) and \(7\).
  • Applying the Altitude Rule:
$$x^2 = 5 \cdot 7 = 35$$
$$x = \sqrt{35}$$

Solve Homework Problems 1 and 2

Using the Geometric Mean and Solving Linear Equations knowledge points:

  • Problem 1: The altitude is \(x\), and the segments of the hypotenuse are \(64\) and \(36\).
$$x^2 = 64 \cdot 36$$
$$x = \sqrt{64} \cdot \sqrt{36} = 8 \cdot 6 = 48$$
  • Problem 2: The altitude is \(x\), and the segments of the hypotenuse are \(144\) and \(25\).
$$x^2 = 144 \cdot 25$$
$$x = \sqrt{144} \cdot \sqrt{25} = 12 \cdot 5 = 60$$

Solve Homework Problems 3 and 4

Using the Geometric Mean and Solving Linear Equations knowledge points:

  • Problem 3: The altitude is \(x\). The entire hypotenuse is \(25\), and one segment is \(16\). The other segment is:
$$25 - 16 = 9$$

Applying the Altitude Rule:

$$x^2 = 16 \cdot 9$$
$$x = \sqrt{16} \cdot \sqrt{9} = 4 \cdot 3 = 12$$
  • Problem 4: The leg is \(580\). The adjacent segment of the hypotenuse is \(400\), and the entire hypotenuse is \(x\).

Applying the Leg Rule:

$$580^2 = 400 \cdot x$$
$$336400 = 400x$$
$$x = \frac{336400}{400} = 841$$

Solve Homework Problems 5 and 6

Using the Geometric Mean and Solving Linear Equations knowledge points:

  • Problem 5: The leg is \(1500\). The adjacent segment of the hypotenuse is \(900\), and the entire hypotenuse is \(x\).

Applying the Leg Rule:

$$1500^2 = 900 \cdot x$$
$$2250000 = 900x$$
$$x = \frac{2250000}{900} = 2500$$
  • Problem 6: The leg is \(960\). The adjacent segment of the hypotenuse is \(576\), and the entire hypotenuse is \(x\).

Applying the Leg Rule:

$$960^2 = 576 \cdot x$$
$$921600 = 576x$$
$$x = \frac{921600}{576} = 1600$$

Answer:

No.ProblemAnswer
1Find \(x\)\(x = 48\)
2Find \(x\)\(x = 60\)
3Find \(x\)\(x = 12\)
4Find \(x\)\(x = 841\)
5Find \(x\)\(x = 2500\)
6Find \(x\)\(x = 1600\)