The image presents several problems related to the Pythagorean theorem. We will solve problems 1, 2, 4 and 5. Problem 1: Determine if a right triangle can be formed from straws of lengths 5cm, 7cm, and 12cm. Problem 2: Find the missing side length, $x$, of a right triangle with hypotenuse 24m and one side 22m, rounding to the nearest tenth. Problem 4: Find the length of FG. Problem 5: A ladder of 8 feet is placed against a house. The base of the ladder is 6.5 feet away from the house. How high up the wall is the ladder, to the nearest tenth?

1. Problem Description

The image presents several problems related to the Pythagorean theorem. We will solve problems 1, 2, 4 and

5.

Problem 1: Determine if a right triangle can be formed from straws of lengths 5cm, 7cm, and 12cm.

Problem 2: Find the missing side length, $x$, of a right triangle with hypotenuse 24m and one side 22m, rounding to the nearest tenth.

Problem 4: Find the length of FG.

Problem 5: A ladder of 8 feet is placed against a house. The base of the ladder is 6.5 feet away from the house. How high up the wall is the ladder, to the nearest tenth?

2. Solution Steps

Problem 1:

To determine if the straws can form a right triangle, we need to check if the Pythagorean theorem holds.

Let $a=5$, $b=7$, and $c=12$.

We need to check if $a^2 + b^2 = c^2$.

$5^2 + 7^2 = 25 + 49 = 74$

$12^2 = 144$

Since $74 \neq 144$, the straws cannot form a right triangle.

Problem 2:

Using the Pythagorean theorem, we have:

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

where $a$ and $b$ are the sides and $c$ is the hypotenuse.

In this case, $22^2 + x^2 = 24^2$.

$484 + x^2 = 576$

$x^2 = 576 - 484$

$x^2 = 92$

$x = \sqrt{92}$

$x \approx 9.59166$

Rounding to the nearest tenth, $x \approx 9.6$

Problem 4:

We need to find the coordinates of points F and G on the grid, then use the distance formula.

From the image, F = (1,5) and G = (5,2).

The distance formula is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$FG = \sqrt{(5 - 1)^2 + (2 - 5)^2}$

$FG = \sqrt{(4)^2 + (-3)^2}$

$FG = \sqrt{16 + 9}$

$FG = \sqrt{25}$

$FG = 5$

Problem 5:

Let the height the ladder reaches on the wall be $h$. We have a right triangle with the ladder as the hypotenuse (8 feet), the distance from the house as one leg (6.5 feet), and the height on the wall as the other leg ($h$).

Using the Pythagorean theorem:

$h^2 + 6.5^2 = 8^2$

$h^2 + 42.25 = 64$

$h^2 = 64 - 42.25$

$h^2 = 21.75$

$h = \sqrt{21.75}$

$h \approx 4.66369$

Rounding to the nearest tenth, $h \approx 4.7$

3. Final Answer

Problem 1: No, the straws cannot form a right triangle because $5^2 + 7^2 \neq 12^2$.

Problem 2: $x \approx 9.6$ m

Problem 4: $FG = 5$

Problem 5: $4.7$ feet