51 basic geometrical 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.

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.

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.

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.

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.

The loop forms a

perimeter or border

that encloses a flat area, or a plane-shape.

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.

The loop forms a

perimeter or border

that encloses a flat area, or a plane-shape.

The length of the border, i.e. the length of the rope,

is also referred to as the perimeter of the area.

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.

The loop forms a

perimeter or border

that encloses a flat area, or a plane-shape.

The length of the border, i.e. the length of the rope,

is also referred to as the perimeter of the area.

All the areas above are enclosed by the same rope,

so they have equal perimeters.

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.

The loop forms a

perimeter or border

that encloses a flat area, or a plane-shape.

The length of the border, i.e. the length of the rope,

is also referred to as the perimeter of the area.

All the areas above are enclosed by the same rope,

so they have equal perimeters.

Following shapes are polygons:

A plane-shape is a polygon if it is formed by straight lines.

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.

The loop forms a

perimeter or border

that encloses a flat area, or a plane-shape.

The length of the border, i.e. the length of the rope,

is also referred to as the perimeter of the area.

All the areas above are enclosed by the same rope,

so they have equal perimeters.

Following shapes are polygons: These are not polygons:

A plane-shape is a polygon if it is formed by straight lines.

Basic Geometrical Shapes and Formulas

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

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.

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

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.

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

If the sides of a triangle are labeled as

a, b, and c, then the perimeter is

P = a + b + c.

a

b

c

The perimeter of an equilateral triangle is P = 3s.

Triangles with three equal sides are

equilateral triangles.

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

If the sides of a triangle are labeled as

a, b, and c, then the perimeter is

P = a + b + c.

a

b

c

The perimeter of an equilateral triangle is P = 3s.

Rectangles are 4-sided

polygons where the sides

are joint at a right angle as

shown.

Triangles with three equal sides are

equilateral triangles.

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

If the sides of a triangle are labeled as

a, b, and c, then the perimeter is

P = a + b + c.

a

b

c

The perimeter of an equilateral triangle is P = 3s.

Rectangles are 4-sided

polygons where the sides

are joint at a right angle as

shown. s

s

ssA square

Rectangle with four equal sides are squares.

Triangles with three equal sides are

equilateral triangles.

Three sided polygons

are triangles.

Basic Geometrical Shapes and Formulas

If the sides of a triangle are labeled as

a, b, and c, then the perimeter is

P = a + b + c.

a

b

c

The perimeter of an equilateral triangle is P = 3s.

Rectangles are 4-sided

polygons where the sides

are joint at a right angle as

shown. s

s

ssA square

Rectangle with four equal sides are squares.

The perimeter of a squares is

P = s + s + s + s = 4s

Basic Geometrical Shapes and FormulasExample 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

Basic Geometrical Shapes and FormulasExample 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

The area consists of two squares and an equilateral triangle

so all the sides, measured from corner to corner, are equal.

Basic Geometrical Shapes and FormulasExample 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

The area consists of two squares and an equilateral triangle

so all the sides, measured from corner to corner, are equal.

There are 9 sections where each is 20 ft hence we need

9 x 20 = 180 ft of fence.

Basic Geometrical Shapes and FormulasExample 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

The area consists of two squares and an equilateral triangle

so all the sides, measured from corner to corner, are equal.

There are 9 sections where each is 20 ft hence we need

9 x 20 = 180 ft of fence.

If we know two adjacent sides of a rectangle, then we know

all four sides because their opposites sides are identical.

Basic Geometrical Shapes and FormulasExample 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

The area consists of two squares and an equilateral triangle

so all the sides, measured from corner to corner, are equal.

There are 9 sections where each is 20 ft hence we need

9 x 20 = 180 ft of fence.

If we know two adjacent sides of a rectangle, then we know

all four sides because their opposites sides are identical.

We will use the word “height” for

the vertical side and “width” for the

horizontal side.

width (w)

height (h)

Basic Geometrical Shapes and FormulasExample 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

The area consists of two squares and an equilateral triangle

so all the sides, measured from corner to corner, are equal.

There are 9 sections where each is 20 ft hence we need

9 x 20 = 180 ft of fence.

If we know two adjacent sides of a rectangle, then we know

all four sides because their opposites sides are identical.

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?

Basic Geometrical Shapes and Formulas

50 m

70 m

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?

Basic Geometrical Shapes and Formulas

50 m

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

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?

Basic Geometrical Shapes and Formulas

and three widths where each

requires 70 meters of rope.50 m

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

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?

Basic Geometrical Shapes and Formulas

and three widths where each

requires 70 meters of rope.

Hence it requires

3(50) + 3(70) = 150 + 210 = 360 meters of rope.

50 m

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

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?

Basic Geometrical Shapes and Formulas

and three widths where each

requires 70 meters of rope.

Hence it requires

3(50) + 3(70) = 150 + 210 = 360 meters of rope.

50 m

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

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.

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?

Basic Geometrical Shapes and Formulas

and three widths where each

requires 70 meters of rope.

Hence it requires

3(50) + 3(70) = 150 + 210 = 360 meters of rope.

50 m

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

b. What is the perimeter of the following step-shape

if all the short segments are 2 feet? 2 ft

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?

Basic Geometrical Shapes and Formulas

and three widths where each

requires 70 meters of rope.

Hence it requires

3(50) + 3(70) = 150 + 210 = 360 meters of rope.

50 m

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

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.

2 ft

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?

Basic Geometrical Shapes and Formulas

and three widths where each

requires 70 meters of rope.

Hence it requires

3(50) + 3(70) = 150 + 210 = 360 meters of rope.

50 m

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

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.

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.

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

The word “area”

also denotes the amount of surface enclosed.

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

The word “area”

also denotes the amount of surface enclosed.

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.

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

The word “area”

also denotes the amount of surface enclosed.

1 in

1 in

1 in2

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.

Hence the areas of the following squares are:

1 square-inch

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

The word “area”

also denotes the amount of surface enclosed.

1 in

1 in

1 in2

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.

Hence the areas of the following squares are:

1 m

1 m

1 m2

1 square-inch 1 square-meter

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

The word “area”

also denotes the amount of surface enclosed.

1 in

1 in

1 in2

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.

Hence the areas of the following squares are:

1 m

1 m

1 mi

1 mi

1 m2 1 mi2

1 square-inch 1 square-meter 1 square-mile

AreaIf we connect the two

ends of a rope that’s

resting flat in a plane,

the rope form a loop

that encloses an area.

The word “area”

also denotes the amount of surface enclosed.

1 in

1 in

1 in2

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.

Hence the areas of the following squares are:

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 miAreaA 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 mi22 x 3

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

By cutting and pasting, we may find areas of other shapes.

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

By cutting and pasting, we may find areas of other shapes.

Example C. a. Find the area of R as shown.

Assume the unit is meter. 4

4

R12 12

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

By cutting and pasting, we may find areas of other shapes.

Example C. a. Find the area of R as shown.

Assume the unit is meter. 4

4

There are two basic approaches. R

12 12

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

By cutting and pasting, we may find areas of other shapes.

Example C. a. Find the area of R as shown.

Assume the unit is meter. 4

4

There are two basic approaches. R

I. We may view R as a 12 x 12 square

with a 4 x 8 corner removed.

12

8

12

44

R12 12

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

By cutting and pasting, we may find areas of other shapes.

Example C. a. Find the area of R as shown.

Assume the unit is meter. 4

4

There are two basic approaches. R

I. We may view R as a 12 x 12 square

with a 4 x 8 corner removed.

12

8

12

44

R12 12

Hence the area of R is

12 x 12 – 4 x 8

2 mi

3 miAreaA 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).

In general, given the rectangle with

height = h (units)

width* = w (units), h

w

= 6 mi22 x 3

A = h x w (unit2) then its area A = h x w (unit2).

The area of a square is s*s = s2.

By cutting and pasting, we may find areas of other shapes.

Example C. a. Find the area of R as shown.

Assume the unit is meter. 4

4

There are two basic approaches. R

I. We may view R as a 12 x 12 square

with a 4 x 8 corner removed.

12

8

12

44

R12 12

Hence the area of R is

12 x 12 – 4 x 8 = 144 – 32 = 112 m2.

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

I II

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the 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

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown.

2 ft

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown. I

II

III 2 ft

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown.

Area of I is 2 x 2 = 4,

I

II

III 2 ft

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown.

Area of I is 2 x 2 = 4,

area of II is 2 x 6 = 12,

I

II

III 2 ft

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown.

Area of I is 2 x 2 = 4,

area of II is 2 x 6 = 12,

and area of III is 2 x 5 = 10.

I

II

III 2 ft

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown.

Area of I is 2 x 2 = 4,

area of II is 2 x 6 = 12,

and area of III is 2 x 5 = 10.

Hence the total area is 4 + 12 + 10 = 16 ft2.

I

II

III 2 ft

AreaIl. We may dissect R into two

rectangles labeled I and II.

1212

4 4

8

I IIArea of I is 12 x 8 = 96,

area of II is 4 x 4 = 16.

The area of R is the sum of the two or 96 + 16 = 112 m2.

b. Find the area of the following shape R where all the short

segments are 2 ft.

Let’s cut R into three rectangles

as shown.

Area of I is 2 x 2 = 4,

area of II is 2 x 6 = 12,

and area of III is 2 x 5 = 10.

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.

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

hb

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

hb

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

hb

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

hb

h

b

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

Hence the area of the parallelogram is A = h x b where

h = height and b = base.

hb

h

b

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

Hence the area of the parallelogram is A = h x b where

h = height and b = base.

hb

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

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

Hence the area of the parallelogram is A = h x b where

h = height and b = base.

hb

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 12 ft8 ft

AreaA parallelogram is a shape enclosed by two sets of parallel lines.

By cutting and pasting, we may arrange a parallelogram into a

rectangle.

Hence the area of the parallelogram is A = h x b where

h = height and b = base.

hb

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

8 ft8 ft12 ft12 ft

12 ft8 ft

AreaA triangle is half of a parallelogram.

AreaA triangle is half of a parallelogram.

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it

and paste to itself to make a

parallelogram,

h

b

h

b

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it

and paste to itself to make a

parallelogram, and the area of the

triangle is half of the area of the

parallelogram formed.h

b

h

b

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it

and paste to itself to make a

parallelogram, and the area of the

triangle is half of the area of the

parallelogram formed.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.

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it

and paste to itself to make a

parallelogram, and the area of the

triangle is half of the area of the

parallelogram formed.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.For example, the area of all

the triangles shown here is

(8 x 12) ÷ 2 = 48 ft2,

i.e. they are the same size. 12 ft

8 ft 8 ft

AreaA triangle is half of a parallelogram.Given a triangle, we may copy it

and paste to itself to make a

parallelogram, and the area of the

triangle is half of the area of the

parallelogram formed.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.For example, the area of all

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

AreaA trapezoid is a 4-sided figure with

one set of opposite sides parallel.

Area

Example D. Find the area of the

following trapezoid R.

Assume the unit is meter.

A trapezoid is a 4-sided figure with

one set of opposite sides parallel.

125

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

5Example D. Find the area of the

following trapezoid R.

Assume the unit is meter.

A trapezoid is a 4-sided figure with

one set of opposite sides parallel.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

The parallelogram has base = 8 and height = 5,

hence its area is 8 x 5 = 40 m2.

5Example D. Find the area of the

following trapezoid R.

Assume the unit is meter.

A trapezoid is a 4-sided figure with

one set of opposite sides parallel.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

The parallelogram has base = 8 and height = 5,

hence its area is 8 x 5 = 40 m2.

The triangle has base = 4 and height = 5,

hence its area is (4 x 5) ÷ 2 = 10 m2.

5Example D. Find the area of the

following trapezoid R.

Assume the unit is meter.

A trapezoid is a 4-sided figure with

one set of opposite sides parallel.R

Area

By cutting R parallel to one side as shown, we split R into two

areas, one parallelogram and one triangle.

12

Therefore the area of the trapezoid is 40 + 10 = 50 m2.

8

84

The parallelogram has base = 8 and height = 5,

hence its area is 8 x 5 = 40 m2.

The triangle has base = 4 and height = 5,

hence its area is (4 x 5) ÷ 2 = 10 m2.

5Example D. Find the area of the

following trapezoid R.

Assume the unit is meter.

A trapezoid is a 4-sided figure with

one set of opposite sides parallel.R

Area

By cutting R parallel to one side as shown, we split R into two

areas, one parallelogram and one triangle.

12

Therefore the area of the trapezoid is 40 + 10 = 50 m2.

8

84

The parallelogram has base = 8 and height = 5,

hence its area is 8 x 5 = 40 m2.

The triangle has base = 4 and height = 5,

hence its area is (4 x 5) ÷ 2 = 10 m2.

We may find the area of any trapezoid by cutting it into one

parallelogram and one triangle.

5Example D. Find the area of the

following trapezoid R.

Assume the unit is meter.

A trapezoid is a 4-sided figure with

one set of opposite sides parallel.R

Circumference and Area of CirclesA circle has a center x,

x

Circumference and Area of CirclesA 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.

rr

x

Circumference and Area of CirclesA 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.

rr

xThe diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Circumference and Area of CirclesA 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.

rr

x

rr x

d (diameter)

The diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Circumference and Area of CirclesA 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.

rr

x

rr x

d (diameter)

The diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Hence d = 2r.

Circumference and Area of CirclesA 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.

rr

x

rr x

d (diameter)

The diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Hence d = 2r.

The perimeter C of a circle is called

the circumference and

C = πd or C = 2πr where π ≈ 3.14…

Circumference and Area of CirclesA 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.

rr

x

rr x

d (diameter)

The diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Hence d = 2r.

The perimeter C of a circle is called

the circumference and

C = πd or C = 2πr where π ≈ 3.14…

We may use 3 as an under–estimation for π.

Circumference and Area of CirclesA 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.

rr

x

rr x

d (diameter)

The diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Hence d = 2r.

The perimeter C of a circle is called

the circumference and

C = πd or C = 2πr where π ≈ 3.14…

We may use 3 as an under–estimation for π.

Example D. Is 25 feet of rope enough

to mark off a circle of radius r = 9 ft on the ground?

Circumference and Area of CirclesA 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.

rr

x

rr x

d (diameter)

The diameter d of the circle is the length

any straight line that goes through the

center x connecting two opposite points.

Hence d = 2r.

The perimeter C of a circle is called

the circumference and

C = πd or C = 2πr where π ≈ 3.14…

We may use 3 as an under–estimation for π.

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.

Circumference and Area of Circles

r

x

The area A (enclosed by) of a circle is

A = πr2 where π ≈ 3.14…A

Circumference and Area of Circles

r

x

The area A (enclosed by) of a circle is

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

Circumference and Area of Circles

r

x

The area A (enclosed by) of a circle is

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.

Estimating using 3 in stead of 3.14 …

we have the area A to be at least

3 x 52

A

Circumference and Area of Circles

r

x

The area A (enclosed by) of a circle is

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.

Estimating using 3 in stead of 3.14 …

we have the area A to be at least

3 x 52 = 3 x 25 = 75 m2.

A

Circumference and Area of Circles

r

x

The area A (enclosed by) of a circle is

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.

Estimating using 3 in stead of 3.14 …

we have the area A to be at least

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

Circumference and Area of Circles

r

x

The area A (enclosed by) of a circle is

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.

The better approximate answer using π = 3.14

is 3.14 x 52 = 3.14 x 75 = 78.5 m2

Estimating using 3 in stead of 3.14 …

we have the area A to be at least

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

VolumeThe volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

Volume

s

The volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

A cube is a square–box,

i.e. a box whose edges are the same. s

sA cube

Volume

s

The volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

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.

A cube is a square–box,

i.e. a box whose edges are the same. s

sA cube

Volume

s

The volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

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.

A cube is a square–box,

i.e. a box whose edges are the same. s

sA cube

1 in 1 in

1 in3

1 cubic inch

1 in

Volume

s

The volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

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.

A cube is a square–box,

i.e. a box whose edges are the same. s

sA 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

The volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

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.

A cube is a square–box,

i.e. a box whose edges are the same. s

sA 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

The volume of a solid is the measurement of the amount of

“room” or “space” the solid occupies.

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.

A cube is a square–box,

i.e. a box whose edges are the same. s

sA 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.

VolumeA 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

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

A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,

Here is a “4 by 3 by 2” box.

4 3

2

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

A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,

Here is a “4 by 3 by 2” box.

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

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

A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 23 = 8,

Here is a “4 by 3 by 2” box.

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

a 3 x 3 x 3 cube has volume 33 = 27,

a 4 x 4 x 4 cube has volume 43 = 64 (unit3).

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

One cubic foot is

1 ft x 1 ft x 1ft

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

One cubic foot is

1 ft x 1 ft x 1ft or

12 in x 12 in x 12 in = 1728 in3.

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

Example b. How many cubic feet are there in the following

solid?

One cubic foot is

1 ft x 1 ft x 1ft or

12 in x 12 in x 12 in = 1728 in3.

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

Example b. How many cubic feet are there in the following

solid?

One cubic foot is

1 ft x 1 ft x 1ft or

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.

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

Example b. How many cubic feet are there in the following

solid?

One cubic foot is

1 ft x 1 ft x 1ft or

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

The volume of I is 3 x 3 x 3 = 27,

the volume of II is 10 x 3 x 6 = 180.

Method 1.

VolumeExample F. a.

How many cubic inches are there in a cubic foot?

(There are 12 inches in 1 foot.)

Example b. How many cubic feet are there in the following

solid?

One cubic foot is

1 ft x 1 ft x 1ft or

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

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 ft3 ft

Volume

The solid may be viewed as a box

with volume 3 x 10 x 9 = 270

with a top portion removed.

The dimension of the removed portion

is also a box with volume 3 x 3 x 7 = 63.

Method 2.

9 ft

10 ft3 ft

3 ft

3 ft

7 ft

Volume

The solid may be viewed as a box

with volume 3 x 10 x 9 = 270

with a top portion removed.

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 ft3 ft

3 ft

3 ft

7 ft