The image shows a series of right triangle problems. We need to solve for the following: Problem 1: Simplify $\sqrt{125}$. Problem 2: Simplify $\sqrt{100}$. Problem 3: Find the length of the hypotenuse of a right triangle with legs of length 12 and 16. Problem 4: Find the length of one leg of a right triangle given the other leg is 15 and the hypotenuse is 25. Problem 5: Find the length of the hypotenuse of a right triangle with legs of length 5 cm and 2 cm. Problem 6: If a 39-foot ladder is placed against a building such that the base of the ladder is 33 feet from the building, find the height of the building. Problem 7: Which of the following sets of numbers could represent the three sides of a right triangle?

1. Problem Description

The image shows a series of right triangle problems. We need to solve for the following:

Problem 1: Simplify $\sqrt{125}$.

Problem 2: Simplify $\sqrt{100}$.

Problem 3: Find the length of the hypotenuse of a right triangle with legs of length 12 and

1

6. Problem 4: Find the length of one leg of a right triangle given the other leg is 15 and the hypotenuse is

2

5. Problem 5: Find the length of the hypotenuse of a right triangle with legs of length 5 cm and 2 cm.

Problem 6: If a 39-foot ladder is placed against a building such that the base of the ladder is 33 feet from the building, find the height of the building.

Problem 7: Which of the following sets of numbers could represent the three sides of a right triangle?

2. Solution Steps

Problem 1: Simplify $\sqrt{125}$.

$\sqrt{125} = \sqrt{25 \times 5} = \sqrt{25} \times \sqrt{5} = 5\sqrt{5}$

Problem 2: Simplify $\sqrt{100}$.

$\sqrt{100} = 10$

Problem 3: Find the length of the hypotenuse.

Using the Pythagorean theorem:

$a^2 + b^2 = c^2$

$12^2 + 16^2 = c^2$

$144 + 256 = c^2$

$400 = c^2$

$c = \sqrt{400} = 20$

Problem 4: Find the length of one leg of a right triangle.

Using the Pythagorean theorem:

$a^2 + b^2 = c^2$

$a^2 + 15^2 = 25^2$

$a^2 + 225 = 625$

$a^2 = 625 - 225$

$a^2 = 400$

$a = \sqrt{400} = 20$

Problem 5: Find the hypotenuse of a right triangle with legs 5 cm and 2 cm.

Using the Pythagorean theorem:

$a^2 + b^2 = c^2$

$5^2 + 2^2 = c^2$

$25 + 4 = c^2$

$29 = c^2$

$c = \sqrt{29} \approx 5.4$

Problem 6: Find the height of the building.

Using the Pythagorean theorem:

$a^2 + b^2 = c^2$

where $a$ is the height of the building, $b$ is the distance from the base of the building to the ladder (33 feet), and $c$ is the length of the ladder (39 feet).

$a^2 + 33^2 = 39^2$

$a^2 + 1089 = 1521$

$a^2 = 1521 - 1089$

$a^2 = 432$

$a = \sqrt{432} \approx 20.8$

Problem 7: Which of the following sets could represent the sides of a right triangle?

We need to check if $a^2 + b^2 = c^2$ for each set, where $c$ is the largest number in the set.

A. {20, 21, 28}: $20^2 + 21^2 = 400 + 441 = 841$. $28^2 = 784$. $841 \neq 784$. No.

B. {24, 32, 39}: $24^2 + 32^2 = 576 + 1024 = 1600$. $39^2 = 1521$. $1600 \neq 1521$. No.

C. {31, 60, 68}: $31^2 + 60^2 = 961 + 3600 = 4561$. $68^2 = 4624$. $4561 \neq 4624$. No.

D. {6, 8, 10}: $6^2 + 8^2 = 36 + 64 = 100$. $10^2 = 100$. $100 = 100$. Yes.

3. Final Answer

Problem 1: $5\sqrt{5}$

Problem 2: 10

Problem 3: 20

Problem 4: 20

Problem 5: 5.4

Problem 6: 20.8

Problem 7: D. {6, 8, 10}