The
Size of Shapes
In 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 32=9
and it cannot be 4 units since 42=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 sqrt10 and it means “the length of a side
of a square whose area is 10 square units.” All of these statements are true:
- sqrt9=3 because 32=9
- sqrt16=4 because 42=16
- sqrt10 is the side length of a square whose area is 10 square units, and (\srt10)2=10
Task
If each grid square represents 1
square unit, what is the side length of this titled square? Explain your
reasoning.
Solution
The side length is sqrt26 because the area of the square is 26 square units and the
square root of the area of a square is the side length.
Side
Lengths and Areas of Squares
This week your student will be
working with the Pythagorean Theorem, which is useful when working with
right triangles. 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, a2+b2=c2.
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:
We can first draw a right triangle
and determine the lengths of the two legs:
Next, since this is a right
triangle, we know 242+72=c2, which
means the length of the line segment, is 25 units.
Tasks
- 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.)
Solutions
- The length of the hypotenuse is 50−−√ units. With legs a and b both equal to 5 and an unknown value for the hypotenuse, c, we know the relationship 52+52=c2 is true. That means 50=c2, so c must be 50−−√ units.
- The length of p is 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 32+42=p2 is true. Since 32+42=25=p2, p must equal 25−−√ or 5 units.
Side
Lengths and Volumes of Cubes
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 it’s edge length is 4 units because sqrt16=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:
Even without the useful grid, we can
calculate that the edge length is 4 from the volume since 64−−√3=4.
Cube roots
that are not integers (like 4) are still numbers that we can plot on a number
line. If we have the three numbers 40−−√, 30−−√3, and 64−−√3, we can plot them on the number line by estimating what
integers they are near. For example, 40−−√ is between 6 and 7, since 40−−√>36−−√, which equals 6, and 40−−√<49−−√ and 49−−√=7. Similarly, 40−−√3 is between 3 and 4 because 30 is
between 27 and 64. Our number line will look like this:
Task
Plot the
given numbers on the number line. 28−−√;27−−√3;50−−√3
Solutions
Since 33=27 means 27−−√3=3,
we plot 27−−√3 at 3. The 50−−√3 is between 3 and 4 because 50−−√3>27−−√3, which equals 3, and 50−−√3<64−−√3, which equals 4. A similar argument
can be made for 28−−√
falling between 5 and 6.