Circles 4-6 continued

We continued our study of circles by examining irregular shapes and determined their perimeters.
To see each of the irregular shapes turned to page 131. The numbers we used in class were all different that those of 24-27, but use the shapes to help solve the following:

The circles in the diagrams are parts of circles and the angles are right angles. We found the perimeter of each figure.
The first figure was a semicircle ( see #24 with a diameter of 4).
We noticed that we needed to start with
C =∏d but then we only need half of that
so
∏d/2 or 4∏/2 = 2∏
Using ∏≈ 3.14
we found
≈3.14(2) = 6.26
BUT.. that only was the upper part we needed to add the diameter of 4 to make sure we had all we needed in our perimeter.
6.28 + 4 = 10.28 units... but then we needed to round to 3 digits-- according to our textbook so
10.3 units would be a good approximation for the first perimeter.

Our 2nd irregular shape was a quarter of a circle with a radius of 6
Again, look to our textbook, page 131 # 25 for the shape. Use 6 as the radius.
This time
C = 2∏r
BUT... we only need 1/4 so
(2⋅∏⋅6)/4 = 12∏/4 = 3∏
≈ 3.14(3) = 9.42
BUT... wait.. we aren't finished... we have to sides of this quarter circle that we need to include in our perimeter.
so the perimeter is approximately 9.42 + 6 + 6 = 9.42 + 12
≈ 21.42 which round to ≈ 21.4 units


The next irregular shape looks like something from Griffith Park observatory. Make sure to use the diagram for # 26 but use a radius of 2 as we did in class.

C = 2∏r
but you notice we only need half of the full circle so
2∏r/2 or just ∏r
Now substitute int he radius-- which is also 2
2∏ ≈ 2(3.14) = 6.28
But... we still need the bottom perimeter
so this shape ≈ 6.28 + 2 + 4 + 2 or 6.28 + 8
≈14.28
≈14.3 units

Period 7 said the next shape ( # 27 in our textbook) looks like a bandaid... What do you think?
(At least.. with the way I drew it.. having a radius of 10)
Again we need the formula
C = 2∏r
This time we realized we had two semicircles.. but that is one whole circles so we kept the formula and substituted in our radius of 10
C =2(10)∏
=20∏
≈20(3.14)= 62.8
Then we added the two sides of 10 and found the perimeter to be approximately
62.8 + 10 + 10
≈82.8 units


I described the last shape to be a teardrop. Make sure to check #28 in our textbook. I actually used the same radius as the book so that drawing is exactly what we did.

C =2∏r
C =2⋅6⋅∏ = 12∏
But... we only need 3/4 of the circle so what is 3/4 of 12... in class everyone knew it was 9 so
3/4 of 12∏ is 9∏
≈9(3.14) = 28.26
But then we need to make sure we include the two sides of 6 each
≈28.26 + 6 + 6 = 28.26 + 12
≈40.26 units
and rounding to three digits
≈ 40.3 units.