Different sizes of ribbon need to be cut to go around various shapes. All of the following sizes are in inches. π , √5 ,√8 ,2√5 (a)Without using your calculator, approximate the decimal equivalent of each number to the nearest tenth. (b)Order the ribbon sizes from least to greatest. Use the original numbers given.

(a) To approximate [tex] \pi [/tex], we are going to use a compass to draw a cirle of radius 2 and circumference 12.5. Remember that [tex] \pi [/tex] is the ratio between the circumference of a circle and its diameter; since the diameter of a circle is two times its radius, we can find [tex] \pi [/tex] using the equation: [tex] \pi = \frac{cricunference}{2*radius} [/tex]
[tex] \pi = \frac{12.5}{2(2)} [/tex]
[tex] \pi = \frac{12.5}{4} [/tex]
[tex] \pi =3.125[/tex] 
And to the nearest tenth:
[tex] \pi =3.1[/tex]

To approximate [tex] \sqrt{2} [/tex], we are going to use the Babylonian method:
Step 1. We are going to make a guess. Our guess is 2 because [tex] \sqrt{4} =2[/tex], so [tex] \sqrt{5} [/tex] must be close to 2.
Step 2. We are going to divide 5 by 2: 
[tex] \frac{5}{2} =2.5[/tex]
Step 3.  We are going to find the average between our initial guest and the result of our previous calculation:
[tex] \frac{2+2.5}{2} = \frac{4.5}{2} =2.25[/tex]
And to the nearest tenth: [tex] \sqrt{5} =2.2[/tex]

To approximate [tex] \sqrt{8} [/tex], we are going to use the Babylonian method one more time:
Step 1. Our guess is 3 because [tex] \sqrt{9} =3[/tex], so [tex] \sqrt{8} [/tex] must be close to 3.
Step 2. We are going to divide 8 by 3:
[tex] \frac{8}{3} =2.6[/tex]
Step 3.  We are going to find the average between our initial guest and the result of our previous calculation:
[tex] \frac{3+2.6}{2} = \frac{5.6}{2} =2.8[/tex]
[tex] \sqrt{8} =2.8[/tex]

To approximate [tex]2 \sqrt{5} [/tex] we are going to multiply our approximation of [tex] \sqrt{5} [/tex] and multiply it by 2:
[tex]2 \sqrt{5} =2*2.2[/tex]
[tex]2 \sqrt{5}=4.4[/tex]

(b) Since we have our irrational numbers expressed as decimal approximations, we can compare them to arrange them form least to greatest:
2.2,2.8,3.1,4.4
Now, we just need to replace the decimal approximations with the original numbers:
[tex] \sqrt{5} [/tex],[tex] \sqrt{8} [/tex],[tex] \pi [/tex],[tex]2 \sqrt{5} [/tex]

We can conclude that ordering the ribbon sizes from least to greatest we get:
[tex] \sqrt{5} [/tex],[tex] \sqrt{8} [/tex],[tex] \pi [/tex],[tex]2 \sqrt{5} [/tex]