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CH 53  COMPOSITE SHAPES &
SEMICIRCLES
 Composite Shapes
Homework
Find the area and perimeter of each geometric shape:
1.
2.
3.
4.
5.
6.
30
2
4
3
6
6
4
12
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Semicircles
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 Semicircles
If we start with half of a circle, and then connect the endpoints with a
straight line (which is the diameter), we get the following shape, which
we’ll call a semicircle:
r
Remember that the area of a geometric figure is a measure of the size
of the region inside the figure, while the perimeter is a measure of the
distance all the way around the figure. For the examples and
homework on semicircles, let’s hone our skills with decimals by using
3.14 as our approximation for , and then round our answers to the
hundredths place.
EXAMPLE 1:
Find the area and the perimeter of a semicircle
with radius 5 cm.
Solution:
A semicircle of radius 5 cm would look like this:
5
PART 1: We’ll calculate the area first. It seems logical that we should first
find the area of the entire circle, and then cut that in half:
A = r 2 = 3.14(5 2 ) = 3.14(25) = 78.5 cm 2
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Dividing that by 2 gives the semicircle’s area:
39.25 cm 2
PART 2: As for the perimeter, the important fact is that the
perimeter consists of two parts: the semicircular part (the arc),
and the straight part (the diameter).
The semicircular part is exactly half the circumference of the full
circle; so we’ll calculate the circumference of the full circle with
radius 5 cm, and then divide that result by 2:
C = 2r = 2(3.14)(5) = 31.4 cm
Half of that would be 15.7 cm.
The straight part is clearly the diameter of the circle, which is
twice the radius, which is 2(5) = 10 cm.
So the circular part of the figure is 15.7 and the straight part is
10, for a total perimeter of
25.7 cm
Homework
7.
Find the area and perimeter of the semicircle with the given
radius:
a. r = 10
b. r = 6
c. r = 5
d. r = 1
e. r = 22
f. r = 3.4
g. r = 8.7
h. r = 7.1
i. r = 0.8
j. r = 0.1
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 The Norman Window
Developed by the Normans around the 12th century
A.D, and based on Roman architecture, the
Norman window has the shape of a rectangle with
a semicircle on top. The area of the window will, of
course, be the area of the semicircle plus the area of
the rectangle.
EXAMPLE 2:
Find the area of the Norman window with a
length of 7.5 m and a width of 6 m.
Solution:
Let’s draw a figure to represent the information:
7.5
6
It seems that the total area of the window would be the sum of
the areas of the rectangle and the semicircle, so let’s work these
areas separately and then add them together.
For the rectangular part, the area is the product of its
dimensions: A = lw = 7.5  6 = 45 m 2 .
The key to finding the area of the semicircle is to note that the
radius of the semicircle is half of 6, or 3. Be sure you understand
why. As in the earlier discussion of semicircles, we’ll find the
area of the full circle, and then cut that in half:
A = r 2 = 3.14(3 2 ) = 3.14(9) = 28.26 m 2 ,
half of which is 14.13 m 2 .
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Summing the rectangle’s area, 45 m 2 , with that of the
semicircle, 14.13 m 2 , the Norman window has a total area of
59.13 m 2
Homework
8.
Find the area of the Norman window with the given length
and width:
a. l = 6; w = 4
b. l = 8; w = 6
c. l = 10; w = 2
d. l = 1.2; w = 10
e. l = 2.3; w = 12
f. l = 3; w = 2
g. l = 2.3; w = 5
h. l = 20; w = 9
i. l = 10.6; w = 6.8
 The Inscribed Circle
Our final shape is a circle inside a square, where four points of the
circle touch the four sides of the square (thus giving the largest circle
that can fit inside the square.) Each side of the square has length s.
s
Suppose we are asked to find the area of the shaded region, which can
be described as the region that is outside the circle and inside the
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square. It’s a funny-shaped region, but its area is actually quite easy to
calculate. If you think you see a method, start the homework problems
now. If not, there’s a hint in the solutions.
Homework
9.
Find the area of the shaded region in the figure on the
previous page (the region inside the square but outside the
circle), given the side of the square:
a. s = 6
b. s = 10
c. s = 20
d. s = 30
e. s = 7
f. s = 11
g. s = 1
h. s = 21
i. s = 4.2
j. s = 3.4
k. s = 7.8
l. s = 5.6
Practice
Problems
10.
The radius of a semicircle is 6.8. Find the area and the perimeter.
11.
A Norman window has a length of 20 and a width of 8.6. Find the area
of the window.
12.
A circle is inscribed in a square each of whose sides is 8.2. Find the
area of the region inside the square but outside the circle.
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Solutions
1.
A = 38; P = 30
2.
A = 35; P = 36
3.
A = 9; P = 16
4.
A = 24; P = 24
5.
A = 30; P = 26
6.
A = 192; P = 80
7.
a.
d.
g.
j.
157; 51.4
1.57; 5.14
118.83; 44.72
0.02; 0.51
b.
e.
h.
56.52; 30.84
759.88; 113.08
79.14; 36.49
8.
a.
d.
g.
30.28
51.25
21.31
b.
e.
h.
62.13
84.12
211.79
9.
Hints:
c.
f.
i.
c.
f.
i.
39.25; 25.7
18.15; 17.48
1; 4.11
21.57
7.57
90.23
1.
The radius of the circle is half the side of the square.
2.
The shaded region is what would remain if you were to remove the
circle from the square.
a.
7.74
b.
21.5
c.
86
d.
193.5
e.
10.54
f.
26.02
g.
0.22
h.
94.82
i.
3.79
j.
2.49
k.
13.08
l.
6.74
10. A = 72.60; P = 34.95
11. A = 201.03
12. A = 14.46
Ch 53  Composite Shapes &
Semicircles
300
“You
can’t
direct
the
wind,
but you can adjust
the sails.”
Anonymous
Ch 53  Composite Shapes &
Semicircles