51 basic shapes and formulas

Basic Geometrical Shapes and Formulas

If we connect the two
ends of a rope that’s
resting flat in a plane,
we obtain a loop.

we obtain a loop.
The loop forms a
perimeter or border
that encloses a flat area, or a plane-shape.

we obtain a loop.
The loop forms a
perimeter or border
The length of the border, i.e. the length of the rope,
is also referred to as the perimeter of the area.

we obtain a loop.
The loop forms a
perimeter or border
All the areas above are enclosed by the same rope,
so they have equal perimeters.

we obtain a loop.
The loop forms a
perimeter or border
Following shapes are polygons:
A plane-shape is a polygon if it is formed by straight lines.

we obtain a loop.
The loop forms a
perimeter or border
Following shapes are polygons: These are not polygons:
A plane-shape is a polygon if it is formed by straight lines.

Three sided polygons
are triangles.

are triangles.
If the sides of a triangle are labeled as
a, b, and c, then the perimeter is
P = a + b + c.
a
b
c

Triangles with three equal sides are
equilateral triangles.
are triangles.
P = a + b + c.
a
b
c

are triangles.
P = a + b + c.
a
b
c
The perimeter of an equilateral triangle is P = 3s.

are triangles.
P = a + b + c.
a
b
c
Rectangles are 4-sided
polygons where the sides
are joint at a right angle as
shown.

are triangles.
P = a + b + c.
a
b
c
shown. s
s
ss
A square
Rectangle with four equal sides are squares.

are triangles.
P = a + b + c.
a
b
c
shown. s
s
ss
A square
Rectangle with four equal sides are squares.
The perimeter of a squares is
P = s + s + s + s = 4s

Example A. We have to fence in an area with two squares
and an equilateral triangle as shown.
How many feet of fences
do we need?
20 ft

do we need?
20 ft
The area consists of two squares and an equilateral triangle
so all the sides, measured from corner to corner, are equal.

do we need?
20 ft
There are 9 sections where each is 20 ft hence we need
9 x 20 = 180 ft of fence.

do we need?
20 ft
If we know two adjacent sides of a rectangle, then we know
all four sides because their opposites sides are identical.

do we need?
20 ft
We will use the word “height” for
the vertical side and “width” for the
horizontal side.
width (w)
height (h)

do we need?
20 ft
We will use the word “height” for
the vertical side and “width” for the
horizontal side. The perimeter of a
rectangle is h + h + w + w or that
width (w)
height (h)
P = 2h + 2w

Example B. a. We want to rope off a 50-meter by 70-meter
rectangular area and also rope off sections of area as shown.
How many meters of rope do we need?
50 m
70 m

50 m
70 m
We have three heights where each
requires 50 meters of rope,

and three widths where each
requires 70 meters of rope.
50 m
70 m

Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m

Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
b. What is the perimeter of the following step-shape
if all the short segments are 2 feet?
2 ft
The perimeter of the step-shape is the
same as the perimeter of the rectangle
that boxes it in as shown.

Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
2 ft

Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
2 ft
2 ft

Hence it requires
3(50) + 3(70) = 150 + 210 = 360 meters of rope.
50 m
70 m
2 ft
2 ft
The height of the rectangle is 6 ft and the width is 10 ft, so the
perimeter P = 2(6) +2(10) = 32 ft.

Area
the rope form a loop
that encloses an area.

Area
The word “area”
also denotes the amount of surface enclosed.

Area
The word “area”
If each side of a square is 1 unit, we define the area of
the square to be 1 unit x 1 unit = 1 unit2, i.e. 1 square-unit.

Area
The word “area”
1 in
1 in
1 in2
Hence the areas of the following squares are:
1 square-inch

Area
The word “area”
1 in
1 in
1 in2
1 m
1 m
1 m2
1 square-inch 1 square-meter

Area
The word “area”
1 in
1 in
1 in2
1 m
1 m
1 mi
1 mi
1 m2 1 mi2
1 square-inch 1 square-meter 1 square-mile

Area
The word “area”
1 in
1 in
1 in2
1 m
1 m
1 mi
1 mi
1 m2 1 mi2
1 square-inch 1 square-meter 1 square-mile
We find the area of rectangles by cutting them into squares.

2 mi
3 miArea
A 2 mi x 3 mi rectangle may be cut into
six 1 x 1 squares so it covers an area of
2 x 3 = 6 mi2 (square miles).
w
= 6 mi2
2 x 3

2 mi
3 miArea
In general, given the rectangle with
height = h (units)
width* = w (units), h
w
= 6 mi2
2 x 3
A = h x w (unit2)
then its area A = h x w (unit2).

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
The area of a square is s*s = s2.

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
By cutting and pasting, we may find areas of other shapes.

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
Example C. a. Find the area of R as shown.
Assume the unit is meter.
4
4
R
12 12

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
4
4
There are two basic approaches.
R
12 12

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
4
4
R
I. We may view R as a 12 x 12 square
with a 4 x 8 corner removed.
12
8
12
4
4
R
12 12

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
4
4
R
12
8
12
4
4
R
12 12
Hence the area of R is
12 x 12 – 4 x 8

2 mi
3 miArea
height = h (units)
w
= 6 mi2
2 x 3
A = h x w (unit2)
4
4
R
12
8
12
4
4
R
12 12
Hence the area of R is
12 x 12 – 4 x 8 = 144 – 32 = 112 m2.

Area
Il. We may dissect R into two
rectangles labeled I and II.
12
12
4 4
I II

Area
12
12
4 4
8
I II
Area of I is 12 x 8 = 96,

Area
12
12
4 4
8
I II
area of II is 4 x 4 = 16.

Area
12
12
4 4
8
I II
The area of R is the sum of the
two or 96 + 16 = 112 m2.

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
b. Find the area of the following shape R where all the short
segments are 2 ft.
2 ft

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
segments are 2 ft.
Let’s cut R into three rectangles
as shown.
2 ft

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
segments are 2 ft.
as shown.
I
II
III 2 ft

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
segments are 2 ft.
as shown.
area of II is 2 x 6 = 12,
I
II
III 2 ft

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
segments are 2 ft.
as shown.
and area of III is 2 x 5 = 10.
I
II
III 2 ft

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
segments are 2 ft.
as shown.
Hence the total area is 4 + 12 + 10 = 16 ft2.
I
II
III 2 ft

Area
12
12
4 4
8
I II
two or 96 + 16 = 112 m2.
segments are 2 ft.
as shown.
Hence the total area is 4 + 12 + 10 = 16 ft2.
I
II
III 2 ft
By cutting and pasting we obtain the following area formulas.

Area
A parallelogram is a shape enclosed by two sets of parallel lines.
h
b

Area
By cutting and pasting, we may arrange a parallelogram into a
rectangle.
h
b

Area
rectangle.
h
b
h
b

Area
rectangle.
Hence the area of the parallelogram is A = h x b where
h = height and b = base.
h
b
h
b

Area
rectangle.
h
b
h
b
For example, the area of all
the parallelograms shown
here is 8 x 12 = 96 ft2,
so they are the same size.
12 ft
8 ft

Area
rectangle.
h
b
h
b
here is 8 x 12 = 96 ft2,
12 ft
8 ft 12 ft8 ft

Area
rectangle.
h
b
h
b
here is 8 x 12 = 96 ft2,
12 ft
8 ft
8 ft8 ft
12 ft12 ft
12 ft8 ft

Area
A triangle is half of a parallelogram.

Area
Given a triangle, we may copy it
and paste to itself to make a
parallelogram,
h
b
h
b

Area
parallelogram, and the area of the
triangle is half of the area of the
parallelogram formed.
h
b
h
b

Area
h
b
h
b
Therefore the area of a triangle is
h x b
2
A = (h x b) ÷ 2 or A =
where h = height and b = base.

Area
h
b
h
b
h x b
2
A = (h x b) ÷ 2 or A =
the triangles shown here is
(8 x 12) ÷ 2 = 48 ft2,
i.e. they are the same size.12 ft
8 ft 8 ft

Area
h
b
h
b
h x b
2
A = (h x b) ÷ 2 or A =
the triangles shown here is
(8 x 12) ÷ 2 = 48 ft2,
i.e. they are the same size.12 ft
8 ft
8 ft8 ft
12 ft12 ft
12 ft
8 ft

Area
A trapezoid is a 4-sided figure with
one set of opposite sides parallel.

Area
Example D. Find the area of the
following trapezoid R.
12
5
8
R

Area
By cutting R parallel to one side as shown, we split R into two
areas, one parallelogram and one triangle.
12
8
84
5
R

Area
12
8
84
The parallelogram has base = 8 and height = 5,
hence its area is 8 x 5 = 40 m2.
5
R

Area
12
8
84
The triangle has base = 4 and height = 5,
hence its area is (4 x 5) ÷ 2 = 10 m2.
5
R

Area
12
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
8
84
5
R

Area
12
Therefore the area of the trapezoid is 40 + 10 = 50 m2.
8
84
We may find the area of any trapezoid by cutting it into one
parallelogram and one triangle.
5
R

Circumference and Area of Circles
A circle has a center x,
x

A circle has a center x, and the distance
from any location on the circle to C
is a fixed number r,
r is called the radius of the circle.
r
r
x

r
r
x
The diameter d of the circle is the length
any straight line that goes through the
center x connecting two opposite points.

r
r
x
rr x
d (diameter)

r
r
x
rr x
d (diameter)
Hence d = 2r.

r
r
x
rr x
d (diameter)
Hence d = 2r.
The perimeter C of a circle is called
the circumference and
C = πd or C = 2πr where π ≈ 3.14…

r
r
x
rr x
d (diameter)
Hence d = 2r.
We may use 3 as an under–estimation for π.

r
r
x
rr x
d (diameter)
Hence d = 2r.
Example D. Is 25 feet of rope enough
to mark off a circle of radius r = 9 ft on the ground?

r
r
x
rr x
d (diameter)
Hence d = 2r.
Example D. Is 25 feet of rope enough
to mark off a circle of radius r = 9 ft on the ground?
No, 25 ft is not enough since the circumference C is at least
3 x 9 = 27 ft.

r
x
The area A (enclosed by) of a circle is
A = πr2 where π ≈ 3.14…
A

r
x
A = πr2 where π ≈ 3.14…
Example E. a. Approximate the area of
the circle with a 5–meter radius
using 3 as the estimated value of π.
A

r
x
A = πr2 where π ≈ 3.14…
Estimating using 3 in stead of 3.14 …
we have the area A to be at least
3 x 52
A

r
x
A = πr2 where π ≈ 3.14…
3 x 52 = 3 x 25 = 75 m2.
A

r
x
A = πr2 where π ≈ 3.14…
3 x 52 = 3 x 25 = 75 m2.
b. Approximate the area of the circle with a 5–meter radius
using π = 3.14 as the estimated value of π.
A

r
x
A = πr2 where π ≈ 3.14…
The better approximate answer using π = 3.14
is 3.14 x 52 = 3.14 x 75 = 78.5 m2
3 x 52 = 3 x 25 = 75 m2.
b. Approximate the area of the circle with a 5–meter radius
using π = 3.14 as the estimated value of π.
A

Volume
The volume of a solid is the measurement of the amount of
“room” or “space” the solid occupies.

Volume
s
A cube is a square–box,
i.e. a box whose edges are the same. s
s
A cube

Volume
s
We define the volume of a cube whose sides are 1 unit to be
1 unit x 1 unit x 1 unit = 1 unit3, i.e. 1 cubic unit.
s
A cube

Volume
s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in

Volume
s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in
1 m 1 m
1 m3
1 cubic meter
1 m

Volume
s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in
1 m 1 m
1 m3
1 cubic meter
1 m
1 mi 1 mi
1 mi31 mi
1 cubic mile

Volume
s
s
A cube
1 in 1 in
1 in3
1 cubic inch
1 in
1 m 1 m
1 m3
1 cubic meter
1 m
1 mi 1 mi
1 mi31 mi
1 cubic mile
We can cut larger cubes into smaller cubes to calculate their
volume.

Volume
A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,
a 3 x 3 x 3 cube has volume 33 = 27,
a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

Volume
w = width
A rectangular box is specified by three sides:
the length, the width, and the height.
We say that the dimension of the box
is “l by w by h”. l = length
h = height

Volume
w = width
h = height
Here is a “4 by 3 by 2” box.
4 3
2

Volume
w = width
h = height
Assuming the unit is inch, then the box
may be cut into 2 x 3 x 4 = 24
1–inch cubes so its volume is 24 in3. 4 3
2

Volume
w = width
h = height
Assuming the unit is inch, then the box
may be cut into 2 x 3 x 4 = 24
1–inch cubes so its volume is 24 in3.
We define the volume V of a box whose sides are l, w, and h
to be V = l x w x h unit3.
In particular the volume of a cube whose sides equal to s is
V = s x s x s = s3 unit3.
4 3
2

Volume
Example F. a.
How many cubic inches are there in a cubic foot?
(There are 12 inches in 1 foot.)

Volume
Example F. a.
One cubic foot is
1 ft x 1 ft x 1ft

Volume
Example F. a.
One cubic foot is
1 ft x 1 ft x 1ft or
12 in x 12 in x 12 in = 1728 in3.

Volume
Example F. a.
Example b. How many cubic feet are there in the following
solid?
One cubic foot is
12 in x 12 in x 12 in = 1728 in3.

Volume
Example F. a.
solid?
One cubic foot is
12 in x 12 in x 12 in = 1728 in3.
We may view the solid is consisted of
two solids I and II as shown.
II
I
Method 1.

Volume
Example F. a.
solid?
One cubic foot is
12 in x 12 in x 12 in = 1728 in3.
II
I
The volume of I is 3 x 3 x 3 = 27,
the volume of II is 10 x 3 x 6 = 180.
Method 1.

Volume
Example F. a.
solid?
One cubic foot is
12 in x 12 in x 12 in = 1728 in3.
II
I
The volume of I is 3 x 3 x 3 = 27,
the volume of II is 10 x 3 x 6 = 180.
Hence the volume of the entire solid is
180 + 27 = 207 ft3.
Method 1.

Volume
The solid may be viewed as a box
with volume 3 x 10 x 9 = 270
with a top portion removed.
Method 2.
9 ft
10 ft
3 ft

Volume
The dimension of the removed portion
is also a box with volume 3 x 3 x 7 = 63.
Method 2.
9 ft
10 ft
3 ft
3 ft
3 ft
7 ft

Volume
The dimension of the removed portion
is also a box with volume 3 x 3 x 7 = 63.
Hence the volume of the given solid is
270 – 63 = 207 ft3.
Method 2.
9 ft
10 ft
3 ft
3 ft
3 ft
7 ft

51 basic shapes and formulas

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51 basic shapes and formulas