59
CH 10  INTRO TO GEOMETRY
 Introduction
G
eo: Greek for earth, and metros: Greek for
measure. These roots are the origin of the
word “geometry,” which literally means “earth
measurement.” The study of geometry has gone
way beyond the notions of triangles and circles -from the shapes of molecules to the structure of the
4-dimensional space-time continuum.
Consider the following figure:
First question: How many little squares are there in the figure? If you
simply count them up, you’ll get a total of 30 squares.
Second question: Find the total distance around the figure. The total
distance around consists of four pieces: the top, with length 6; the
right, with length 5; the bottom, with length 6; and the left, with length
5. Adding the lengths of all four sides gives a total distance of 22.
Notice that these two questions are very different. The first counts the
number of squares inside the figure, while the second measures the
distance all the way around the figure. Since the answers are different,
we see that these two concepts are not the same.
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60
Area
The area of a geometric figure is a measure of how big a region is
enclosed inside the figure.
For example, the area of the region below is 18 square units.
Just count the number of little squares, and you’ve got the area.
Perimeter
The distance around a geometric figure is called its perimeter.
For example, the perimeter of the region above is 20 units. Start
at a corner (or anywhere you like), and march along the edge of
the region until you reach your starting point. The distance
you’ve traveled (not the number of squares you walk by) is the
perimeter. To help you remember this, note that peri means
“around,” as in periscope or peripheral vision.
Homework
Find the area and perimeter of each geometric shape:
1.
2.
3.
4.
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61
 Rectangles and Squares
In Manhattan, Fifth Avenue (which runs north-south) meets 42nd
Street (which runs east-west) at a 90 angle. The floor and the wall
also meet at a 90 angle.
A rectangle is a four-sided figure with all inside angles
90
90
equal to 90. This implies that adjacent sides (sides
90
next to each other) are perpendicular and opposite sides 90
are parallel. Notice that a square (where all four sides have the same
length) is also a four-sided figure with all four inside angles equal to
90. Therefore, by definition, a square is a special kind of rectangle.
We can conclude that every square is a rectangle, but certainly not
every rectangle is a square.
We know that the distance around the rectangle (the sum of all four of
its sides) is called its perimeter. The area of a rectangle is a measure
of the size of the region enclosed by the rectangle.
From the first page of this chapter, we counted the
little squares in the rectangle and got a total of 30.
But notice that the rectangle could be viewed as
having 5 rows with 6 little squares in each row.
Multiplying 5  6 gives 30, the area.
So for any rectangle (and therefore any square),
Perimeter: Add the four sides
Area: Multiply the number of rows by the
number of little squares in each row
(or, multiply the length by the width).
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62
Homework
5.
The length of a rectangle is 12 and its width is 5. Find the
perimeter and the area of the rectangle.
6.
Each side of a square is 10. Find the perimeter and the area of
the square.
 Challenging Examples
1.
The perimeter of a square is 44. What is the area of the square?
To find the area of a square, we need to know the length of one
side of the square, since all sides are equal. But the side isn’t
given; however, we can find it from the perimeter of 44. Since the
perimeter is 44, it follows that each side must be 11.
Now we can compute the area of the square by multiplying its
length by its width (which are both 11 in this case). Since 11  11
is 121, we conclude that the area of the square is 121.
2.
The area of a square is 144. What is the square’s perimeter?
The perimeter of a square depends upon its side. We don’t have
the side, but we do have the area. Since the area is 144, and
since the area is found by multiplying the length by the width
(which must be the same in a square), we ask ourselves, What
number times itself is 144? The answer is 12, so each side of the
square is 12.
Now, the perimeter of a square is the distance all the way around
-- that is, all four sides added together. Adding four 12’s together
gives a perimeter of 48.
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63
3.
The length of a rectangle is 5 more than its width. If the perimeter
of the rectangle is 42, find the rectangle’s area.
This is a hard problem compared to the two previous ones, so
we’ll attack it by taking some educated guesses until all the
conditions of the problem are met.
The first requirement is that the length must be 5 more than the
width. So consider the following rectangle:
11
6
I chose 6 for the width off the top of my
head. Since 5 more than 6 is 11, the
length would have to be 11.
First, check that the length is actually 5 more than the width. It
is, so this rectangle is possibly useful to us. Now see if the
perimeter is 42, as it’s supposed to be. Adding the four sides:
11 + 11 + 6 + 6 = 34 
We failed, so we’ll try another (larger) rectangle. How about:
13
8
The length is still 5 more than the width. So we’ll check this
rectangle against the requirement that the perimeter be 42:
13 + 13 + 8 + 8 = 42 
This time we hit the perimeter right on. So we know that the
dimensions of the rectangle we’ve been seeking are 13 by 8. Since
the problem asked for the rectangle’s area, we multiply the
length by the width to get an area of 104.
Ch 10  Intro to Geometry
64
4.
The length of a rectangle is 5 times its width. If the area of the
rectangle is 80, find the rectangle’s perimeter.
This problem will also be done by taking some educated guesses
until we land on the right answer. Let’s randomly choose a
rectangle whose width is 3. Since the length must be 5 times the
width, we are considering the rectangle
15
3
This rectangle has an area of 45, nowhere near
the 80 it’s supposed to be.
We’ll increase the width to 5 and see what happens:
25
5
This rectangle has an area of 125, way too big.
Let’s compromise; since a width of 3 produced a rectangle too
small, and a width of 5 produced one too big, let’s hope that a
width of 4 will do the trick.
20
4
The area of this rectangle is 80, exactly what was
specified -- we’ve found our rectangle. 
It should now be clear that the perimeter of this rectangle is 48.
Homework
7.
If the perimeter of a square is 8, find the area of the square.
8.
If the area of a square is 25, find the perimeter of the square.
9.
The length of a rectangle is 8 more than its width. If the perimeter of
the rectangle is 40, what is the area of the rectangle?
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10.
The length of a rectangle is 2 times its width. If the perimeter of the
rectangle is 30, what is the area of the rectangle?
11.
The length of a rectangle is 2 times its width. If the area of the
rectangle is 18, what is the perimeter of the rectangle?
12.
The length of a rectangle is 6 more than its width. If the area of the
rectangle is 72, what is the perimeter of the rectangle?
13.
The length of a rectangle is 2 times its
width. If the perimeter of the
rectangle is 24, what is the area of
the rectangle?
14.
The length of a rectangle is 1 more
than its width. If the area of the rectangle is 6, what is the perimeter
of the rectangle?
15.
The length of a rectangle is 4 times its width. If the area of the
rectangle is 16, what is the perimeter of the rectangle?
16.
The length of a rectangle is 8 more than its width. If the perimeter of
the rectangle is 20, what is the area of the rectangle?
17.
If the area of a square is 100, find the perimeter of the square.
18.
The length of a rectangle is 5 times its width. If the area of the
rectangle is 5, what is the perimeter of the rectangle?
19.
The length of a rectangle is 1 more than its width. If the perimeter of
the rectangle is 30, what is the area of the rectangle?
20.
The length of a rectangle is 3 times its width. If the perimeter of the
rectangle is 24, what is the area of the rectangle?
21.
If the perimeter of a square is 28, find the area of the square.
22.
The length of a rectangle is 8 more than its width. If the area of the
rectangle is 84, what is the perimeter of the rectangle?
23.
If the area of a square is 64, find the perimeter of the square.
24.
The length of a rectangle is 4 times its width. If the perimeter of the
rectangle is 40, what is the area of the rectangle?
25.
If the area of a square is 144, find the perimeter of the square.
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26.
The length of a rectangle is 6 more than its width. If the perimeter of
the rectangle is 28, what is the area of the rectangle?
27.
The length of a rectangle is 2 times its width. If the area of the
rectangle is 32, what is the perimeter of the rectangle?
28.
If the perimeter of a square is 48, find the area of the square.
29.
The length of a rectangle is 12 more than its width. If the area of the
rectangle is 45, what is the perimeter of the rectangle?
30.
If the perimeter of a square is 40, find the area of the square.
31.
The length of a rectangle is 3 times its width. If the area of the
rectangle is 27, what is the perimeter of the rectangle?
32.
The length of a rectangle is 4 more than its width. If the perimeter of
the rectangle is 20, what is the area of the rectangle?
33.
The length of a rectangle is 10 more than its width. If
the area of the rectangle is 39, what is the perimeter of
the rectangle?
34.
The length of a rectangle is 3 times its width. If the
perimeter of the rectangle is 8, what is the area of the
rectangle?
Solutions
1.
A = 38; P = 30
2.
A = 35; P = 36
3.
A = 9; P = 16
4.
A = 9; P = 22
5.
To find the perimeter, add the 4 sides: 12 + 5 + 12 + 5 = 34
For the area, multiply the length by the width: 12(5) = 60
6.
To find the perimeter, add the 4 sides: 10 + 10 + 10 + 10 = 40
For the area, multiply the length by the width: 10(10) = 100
Ch 10  Intro to Geometry
67
7.
4
8.
20
9.
84
10.
50
11.
18
12.
36
13.
32
14.
10
15.
20
16.
9
17.
40
18.
12
19.
56
20.
27
21.
49
22.
40
23.
32
24.
64
25.
48
26.
40
27.
24
28.
144
29.
36
30.
100
31.
24
32.
21
33.
32
34.
3
Ch 10  Intro to Geometry
68
“Each problem that I solved
became a rule which served
afterwards to solve other
problems.”
Rene Descartes (1596-1650), Discourse de la Methode
Ch 10  Intro to Geometry