Unit 8 Pythagorean Theorem and Irrational Numbers (Family Materials)

Section A Side Lengths and Areas of Squares

This week your student will be working with the relationship between the side length and area of squares. We know two main ways to find the area of a square:

Multiply the square’s side length by itself.
Decompose and rearrange the square so that we can see how many square units are inside. For example, if we decompose and rearrange the tilted square in the diagram, we can see that its area is 10 square units.

But what is the side length of this tilted square? It cannot be 3 units since $3^{2} = 9$ and it cannot be 4 units since $4^{2} = 16$ . In order to write “the side length of a square whose area is 10 square units,” we use notation called a square root. We write “the square root of 10” as $\sqrt{10}$ and it means “the length of a side of a square whose area is 10 square units.” All of these statements are true:

$\sqrt{9} = 3$ because $3^{2} = 9$
$\sqrt{16} = 4$ because $4^{2} = 16$
$\sqrt{10}$ is the side length of a square whose area is 10 square units, and ${(\sqrt{10})}^{2} = 10$

There are 3 squares on a square grid, arranged in order of area, from smallest, on the left, to largest, on the right. The left most square is aligned to the grid and has side lengths of 3 with an area of 9. The middle square is tilted on the grid so that its sides are diagonal to the grid. The square is labeled with a side length of square root of 10 and an area of 10. The right most square is aligned to the grid and has side lengths of 4 with an area of 16.

Here is a task to try with your student:

If each grid square represents 1 square unit, what is the side length of this tilted square? Explain your reasoning.

A square, not aligned to the horizontal or vertical gridlines, is on a square grid. The square is drawn such that the first vertex is on the left. The second vertex is 1 grid square down and 5 grid squares right from the first vertex. The third vertex is 5 grid squares down and 1 grid square left from the second vertex. The fourth vertex is 1 grid square up and 5 grid squares left from the third vertex. The first vertex is 5 grid squares up and 1 grid square right from the fourth vertex.

Solution:

The side length is $\sqrt{26}$ because the area of the square is 26 square units and the square root of the area of a square is the side length.

Section B The Pythagorean Theorem

This week your student will work with the Pythagorean Theorem, which describes the relationship between the sides of any right triangle. A right triangle is any triangle with a right angle. The side opposite the right angle is called the hypotenuse, and the two other sides are called the legs. Here we have a triangle with hypotenuse $c$ and legs $a$ and $b$ . The Pythagorean Theorem states that for any right triangle, the sum of the squares of the legs are equal to the square of the hypotenuse. In other words, $a^{2} + b^{2} = c^{2}$ .

A right triangle with legs labeled “a” and “b.” The hypotenuse is labeled “c.”

We can use the Pythagorean Theorem to tell if a triangle is a right triangle or not, to find the value of one side length of a right triangle if we know the other two, and to answer questions about situations that can be modeled with right triangles. For example, let’s say we wanted to find the length of this line segment:

A line segment slanted downward and to the right on a square grid. The bottom endpoint is 7 units down and 24 units to the right from the top endpoint.

We can first draw a right triangle and determine the lengths of the two legs:

A right triangle on a square grid. The horizontal side has a length of 24 and the vertical side has a length of 7. The hypotenuse is labeled c.

Next, since this is a right triangle, we know that $24^{2} + 7^{2} = c^{2}$ , which means the length of the line segment is 25 units.

Here is a task to try with your student:

Find the length of the hypotenuse as an exact answer using a square root.
What is the length of line segment $p$ ? Explain or show your reasoning. (Each grid square represents 1 square unit.)

Solution:

The length of the hypotenuse is $\sqrt{50}$ units. With legs $a$ and $b$ both equal to 5 and an unknown value for the hypotenuse, $c$ , we know the relationship $5^{2} + 5^{2} = c^{2}$ is true. That means $50 = c^{2}$ , so $c$ must be $\sqrt{50}$ units.
The length of $p$ is $\sqrt{25}$ or 5 units. If we draw in the right triangle, we have legs of length 3 and 4 and hypotenuse $p$ , so the relationship $3^{2} + 4^{2} = p^{2}$ is true. Since $3^{2} + 4^{2} = 25 = p^{2}$ , $p$ must equal $\sqrt{25}$ or 5 units.

Section C Decimal Representation of Rational and Irrational Numbers

This week your student will learn about cube roots. We previously learned that a square root is the side length of a square with a certain area. For example, if a square has an area of 16 square units then its edge length is 4 units because $\sqrt{16} = 4$ . Now, think about a solid cube. The cube has a volume, and the edge length of the cube is called the cube root of its volume. In this diagram, the cube has a volume of 64 cubic units:

A solid cube composed of 64 unit cubes. Each edge length is 4 unit cubes.

Even without the useful grid, we can calculate that the edge length is 4 from the volume since $\sqrt[3]{64} = 4$ .

Cube roots that are not integers are still numbers that we can plot on a number line. If we have the three numbers $\sqrt{40}$ , $\sqrt[3]{30}$ , and $\sqrt[3]{64}$ , we can plot them on the number line by estimating what integers they are near. For example, $\sqrt{40}$ is between 6 and 7, since $\sqrt{36} < \sqrt{40} < \sqrt{49}$ and $\sqrt{36} = 6$ while $\sqrt{49} = 7$ . Similarly, $\sqrt[3]{30}$ is between 3 and 4 because 30 is between 27 and 64. Our number line will look like this:

Here is a task to try with your student:

Plot the given numbers on the number line: $\sqrt{28}$ , $\sqrt[3]{27}$ , $\sqrt[3]{50}$

Solution:

Since $3^{3} = 27$ means $\sqrt[3]{27} = 3$ , we can plot $\sqrt[3]{27}$ at 3. $\sqrt[3]{50}$ is between 3 and 4 because 50 is between $3^{3} = 27$ and $4^{3} = 64$ . $\sqrt{28}$ is between 5 and 6 because 28 is between $5^{2} = 25$ and $6^{2} = 36$ .