Download Lesson 10:Areas

IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
Lesson 10: Areas.
Objetives:





To know and use surface measurements adequately.
To work out the perimeter of a circle, circumference.
To work out the length of an arc.
To know Pythagoras Theorem and how to prove it geometrically.
To apply Pythagoras Theorem to:
−
−
−
−













The identification of acute, obtuse and right triangles
The calculation of the diagonal of a rectangle
The calculation of the height of an isosceles triangle
The calculation of the apothem of a regular polygon
To know how to deduce the formula of the area of the more common polygons and of the circle.
To learn the formula of the area of parallelograms: rectangle, square, rhombus and rhomboid
To learn the formula of the area of triangles and trapeziums.
To learn the formula of the area of a regular polygon.
To learn the formula of the area of a circle, of a circle sector and of an annulus.
To apply adequately the formulae to find out the area of the more common plane figures.
To know how to calculate the area or a composite figure.
To know how to find out the sum of the interior angles of a polygon.
To know how to calculate the interior angle and the central angle of a regular polygon.
To know the definition of some angles in a circle: interior angle, inscribed angle and interior angle.
To know the measurement of the angles mentioned before.
To get into the habit of writing down the measurement unit in the outcome of an exercise.
To solve word problems by using Pythagoras’ Theorem.
INDEX:
1.
2.
3.
4.
5.
6.
7.
8.
Perimeter
1.1. Perimeter of regular polygons
1.2. Perimeter of a circle: circumference
1.3. Length of an arc of a circle
Pythagoras’s Theorem
Applications of Pythagoras’s Theorem
3.1. Identifying right triangles
3.2. Calculating the diagonal of a rectangle
3.3. Calculating the height of an isosceles triangle
3.4. Calculating the apothem of a regular polygon
Areas of polygons
4.1. Area of parallelograms: rectangle, square, rhombus and rhomboid
4.2. Area of triangles
4.3. Area of a trapezium
4.4. Area of a regular polygon
4.5. Area of a composite 2-D shape
Area of circular figures
5.1. Area of a circle
5.2. Area of circular sector
5.3. Area of an annulus
Area of a composite plane figure
Angles in polygons
7.1. Sum of the angles of a polygon
7.2. Central angle of a polygon
Angles in a circle
1
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
1. Perimeter
The perimeter of a polygon is the distance around the outside of the polygon.
A polygon is 2-dimensional; however, perimeter is 1-dimensional and is measured in linear units
The perimeter of a polygon is the sum of its length sides.
Examples
1. Find the perimeter of a triangle with
sides measuring 5 centimetres, 9
centimetres and 11 centimetres.
P = 5 cm + 9 cm + 11 cm = 25 cm
2. A rectangle has a length of 8 centimeters and a width of 3 centimeters. Find the
perimeter.
P = 8 cm + 8cm + 3 cm + 3 cm = 22 cm
P = 2(8 cm) + 2(3 cm) = 16 cm + 6 cm = 22 cm
We name 8 cm + 3 cm semi-perimeter, S,.
Given a rectangle with length
l and width w we name semi perimeter to S  l  w , so:
P  2  l  w  2  S
1.1. Perimeter of a regular polygon
Regular polygons have equal sized sides. Their perimeter is easy to calculate:
The perimeter of a n -sided regular polygon of length side l , is.
P  nl
Examples
We name l to length of the sides of the following regular polygons. Find their perimeter:
Equilateral triangle
Square
All three sides have equal lengths
Parallelogram with 4 equal sides and
P= 3·
l
4 right angles: P = 4·
Regular hexagon
l
All six sides have equal length
P = 6·
l
2
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
1.2. Perimeter of a circle: Circumference
From Latin: circum "around" + ferre "to carry"
Circumference is the distance around the edge of a circle. This is it is the perimeter of the
circle.
If we divide the perimeter of a circle by its diameter we always obtain the same decimal number.
This is a very important number, named pi and designed by  . This number is non exact and
non periodic. It has infinite decimal figures and they don’t repeat any “period”.
Its value has been calculated with a precision of hundreds of thousands of decimal figures but
we can not know the exact value. An approximated value is:
  3.1415926
We will use:
  3.14
The perimeter of a circle or circumference, L , can be calculated through two formulae:
L   ·d
L  2· ·r
where d is the diameter and r the radius
Examples
3. Find the circumference of a circle with a radius 12 inches.
r = 12
C=2 r
C = 2 (12)
C = 24
inches.



4. Find the circumference of a bike wheel whose diameter is 16 inches.


C=
d
C=
(16)
C = 16
inches.
We can also apply the other formula

If the diameter is 16, the radius is 8 inches. (The radius is ½ of the diameter).
Now use the formula



C = 2 r.
C = 2 (8)
C = 16
inches.
3
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
1.3. Length of an arc of circle

If we have a circle of radius r the length of the AB , which measures n degrees is:
2r·nº
LA B 
360 º
How did we obtain this formula?
If you observe the length of an arc of a circle and the angle intercepted by arc are in direct
proportion. Remember that the arc which subtends 360º has the length of the circumference.
Now think of the ”regla de tres directa simple”:
2r subtends
 360 º
LAB  nº
subtends
So: LAB 
2r  nº
360 º
Example
5. Find the length of a 24º arc of a circle with a 5 cm radius.
2r·nº
LA B 
360 º
If we substitute in the formula:
LA B 
2 5·24 º 2 120 2 6.28



 2.08cm
360 º
360
3
3
2. Pythagoras’ theorem
A right-angled triangle or right triangle is a triangle having a right angle. One of the angles of the
triangle measures 90 degrees. The side opposite the right angle is called the hypotenuse. The two
sides that form the right angle are called cathetus (plural catheti), “cateto” in Spanish, but in English
they use to say simply sides or legs.
Examples
red lines in red are the hypotenuses
4
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
A right triangle has the special property that is known as the Pythagorean theorem or Pythagoras’
theorem.
In a right-angled triangle the sum of the squares of the lengths of the sides equals the square
of the length of the hypotenuse.
a2  b2  c2
To prove this theorem, we must first remember how to find the area of a square which is the product
of the length of its side times it.
Let's call the length of our square's side "c" which makes its area c squared.
The same goes for a square of length "a" and a square of length "b".
Proving Pythagoras' theorem.
The following diagram proves Pythagoras' theorem is true.
In the diagram there are two identical squares whose sides
measure b+c. In each square we have positioned four identical
right-angled triangles in different places whose hypotenuse is a
and other two sides are b and c.
As the original squares are exactly the same, the spaces in each
square also have the same area. In the square on the left, a2
and in the square on the right, b2+c2.
In the square on the right there are two spaces that
are squares whose sides are b and c. Therefore, their
areas are b2 and c2 respectively.
Therefore a2 = b2+c2
In the square on the left, once the four triangles have been
positioned we get a space which is a square in the middle whose
sides are a, equal to the hypotenuse of the triangle. Therefore,
the area of this square is a2.
If we know the lengths of a side and hypotenuse, we can find the length of the other side:
c  a2  b2
b  a2  c2
And if we know the lengths of the two sides we can find the length of the hypotenuse:
a  b2  c2
Example
6. Find the length of the missing side:
The hypotenuse (side opposite the right angle) is the missing side. If this side is
Pythagoras’ theorem
a , then by
a 2  7 2  24 2  49  576  625
a  625  25
5
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
7. Given the hypotenuse, 26 cm length & one side 22 cm length, find the other side
Pythagorean triples
When the side lengths of right triangle satisfy the Pythagorean Theorem, these three numbers are
known as “Pythagorean triples or triplets”
How to identify and create Pythagorean triples
The most common examples of Pythagorean triples are:




3,4,5 triangles
o A 3,4,5 triple simply stands for a triangle that has a side of length 3, a side of length
4 and a side of length 5.
o If a triangle has these side length then it must be a right triangle
5,12,13 right triangles
7,24,25 right triangles
8,15,17 right triangles
We can conclude the following:
 Only right triangles verify the Pythagorean theorem
 If a triangle has a triple as side lengths then it must be a right triangle
Example
8. Check if 7, 24, 25 is a Pythagorean triple. Is 14, 48, 50 a Pythagorean triple? What do you
observe?
We have to check if they satisfy Pythagoras’ Theorem a 2  b 2  c 2 .
a must be the greatest number:
25  24  7
625  576  49
2
2
2
Similarly:
50 2  48 2  14 2
2500  2304  196
NOTE:
The triples above such as 3,4,5 represent the ratios of side lengths that satisfy the Pythagorean
theorem.
Therefore, you can create other triples by multiplying any of these triples by a number.


(3,4,5) ×2 = (8,6,10) and 8,6,10 is also a Pythagorean triple
(5,12,13) ×2 = (10,24,26) and 10,24,26 is also a Pythagorean triple
Any multiple of the ratios above represent the sides of a right triangle.
6
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
3. Applications of Pythagoras’ theorem
3.1. Identifying right triangles
Given any triangle ΔABC, if
a is the biggest side:

If a  b  c  The triangle is
right-angled.

If a  b  c  The triangle is
acute-angled

If a  b  c  The triangle is
obtuse-angled
2
2
2
2
2
2
2
2
2
Recall only right-angled triangles verify Pythagoras’ Theorem.
Example
8. Determine if a triangle of side lengths 7, 24, 25 is a right triangle.
We have to check if they satisfy Pythagoras’ Theorem a 2  b 2  c 2 .
a must be the greatest number:
25  24  7
625  576  49
2
2
2
3.2. Calculating the diagonal of a rectangle
The triangle ΔABC is right-angled. Its hypotenuse is the diagonal of the rectangle, d , and its sides
are the sides of the rectangle.
Applying Pythagoras’ theorem we have:
d 2  l 2  w2  d  l 2  w2
Notice the rectangle is divided into two equal right triangles.
Example
9. Find the diagonal of the rectangle whose length = 12 cm d width = 8 cm.
d 2  12 2  8 2  d  12 2  8 2  208  14,4 cm
3.3. Calculating the height of an isosceles triangle
In isosceles triangle, the height from the vertex to the unequal side divides the unequal side into
equal segments.
7
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
The triangles ΔAMC and ΔMBC are right–angled and equal. Then
2
c
l 2  h2    
2
c
h  l2  
2
2
In equilateral triangles this happens with any height.
In scalene triangle heights divide it into two right triangles (except one height in obtuse-angled), but
they are not equal.
Example
10. Find the height in this isosceles triangle
This is done by Pythagoras’s Theorem
2
6
2
h 2     6.5
2
11. Find the height in this equilateral triangle
Let A, B and C be the vertices of the
equilateral triangle and M the midpoint of
segment BC. Since the triangle is equilateral,
AMC is a right triangle. Let us find h the
height of the triangle using Pythagorean
theorem.
h 2 + 5 2 = 10 2
Solve the above equation for h.
2
2
l
 10 
l 2  h 2     h  10 2    
2
2
 100  25  8.66 cm
8
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
3.3. Calculating the apothem of a regular polygon
An apothem of a regular polygon is a line segment from the midpoint of one of the sides of the
polygon to the centre of that polygon.
This segment is always perpendicular to the side. The term "apothem" also refers to the length
of this line segment.
The radius of a regular polygon is the line segment from the centre of the polygon to any vertex.
To calculate the length of the apothem, you must know the length of the side of a regular polygon,
and the length of the radius of the polygon.
In the figure below – a regular pentagon – you can see that the right triangle in green has as sides
the apothem a , l , and hypotenuse the radius of the polygon, r .
2
If we apply the Pythagoras’ theorem we
have:
r 2  a2 
So:
l2
2
l
 a2  r 2   
2
l
a  r2  
2
2
2
We can generalise this process to a n-sided regular polygon.
The apothem of a regular polygon of length side l and radius r is:
l
a  r2  
2
2
A particular case is the regular hexagon. In a regular hexagon, the radius and the side have
the same length.
In this case the formula is, for r  l ,
l
a  l2   
2
2
Examples
11. Find the apothem of a regular pentagon with side 6 cm and radius 5 cm.
6
r 2  a2   
2
So:
2
 a 2  5 2  32
a  5 2  32  16  4 cm
9
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
12. Find the apothem of a regular hexagon of side 4 cm.
l
a  l2   
2
2
a  4 2  2 2  16  4
 12  3.46 cm
4. Area of polygons
4.1. Area of parallelograms
Area of a rectangle
A rectangle is a four-sided polygon with four right angles, whose
opposite sides are parallel and equal. We name the length basis b , and
the width height h .
The area of a rectangle is the product of its width and length.
Area  length  width
this is
A  b·h
Area of a square
A square is a parallelogram with 4 equal sides and 4 right angles.
A square is a particular case of a rectangle and a rhombus simultaneously.
So, it shows both the properties of rhombus and rectangle simultaneously.
The area of a square of length side l is :
A  l·l  l 2
Examples
13. Find the area of a square having length side five.
A  5·5  52  25
The area measures 25 cm2
14. Find the area of a square of diagonal 6.
We have to apply the Pythagoras’ Theorem:
6 2  l 2  l 2  2l 2
10
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
l2 
36
 18
2
l  18  4.24 cm
Area of a rhombus
A rhombus is a four-sided polygon having all four sides of equal length and whose
opposite sides are parallel.
Each rhombus has two diagonals. The biggest one is called long diagonal and the
smaller one is called short diagonal. (Note that the diagonals of a rhombus are
perpendicular).
The area of a rhombus is half the area of the rectangle of base D and altitude or height d .
The area of any rhombus is equal to one-half the product of the lengths D and d of its diagonals.
D·d
Area 
2
Examples
15. Find the area of the rhombus where the long diagonal is 9 units and the short
diagonal is 8 units long.
Area 
D·d 9·8

 36
2
2
The area measures 36 square units
16. Find the area of a rhombus with short diagonal 6 cm long and side 5 cm long.
The diagonals and the side are related.
The shaded triangle is right angled, so we
can apply Pythagoras’ theorem
2
2
D d 
2
     L
 2  2
2
2
D
D
2
2
    5 2  3 2  25  3  16
 3 5
2
2
D
  16  4  D  8 cm.
2
So: 
Then the area will be: A 
D·d 8·6

 24 cm2.
2
2
11
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
Area of a rhomboid
How could we find the area of this parallelogram?
Make it into a rectangle as we do on the
right side:
The rectangle is made of the same parts as the parallelogram, so their areas are the same. The
area of the rectangle is b  h , so the area of the parallelogram is also b  h .
Warning: Notice that the height h of the rhomboid is the perpendicular distance between two
parallel sides of the rhomboid, not a side of the parallelogram (unless the parallelogram is also
a rectangle, of course). See the picture below
.
The area of a rhomboid that has base b units and height h units, A , is b  h square units
A  bh
Examples
17.Find the area of this rhomboid:
Area of a parallelogram = base height
A  b·h
 25 10
 250 in 2
(The 12 is a side. The height and base are always perpendicular to each other.)
4.2. Area of a Triangle
We define as altitude or height of a triangle as the perpendicular distance from one side (its base) to
the opposite vertex.
12
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
IMPORTANT FACTS:


In equilateral triangles all the altitudes divide the opposite side into two halves.
In isosceles triangles the height over the base divides the opposite side into two halves.
How could we find the area of this triangle?
Make it into a parallelogram. This can be done by making a copy of the original triangle and putting
the copy together with the original.
The area of the parallelogram is b·h , so the area of the triangle is
bh
2
Warning: Notice that the height h (also often called the altitude) of the triangle is the perpendicular
distance between a vertex and the opposite side of the triangle.
The area of a triangle that has base b units and corresponding altitude h is.
bh
A
2
Examples
17. Find the area of an obtuse triangle with bas 8 cm and corresponding height 5 cm.
b  h 8·5

2
2
 20 cm 2
A
18. Find the area of an equilateral triangle with side a cm. Apply it to the case a=8 cm
We draw a diagram of an equilateral
triangle.
Draw a height from the vertex to the midpoint of the base. If a is the length of a side of the
equilateral triangle and h the height then the triangle below, which is half of the rectangle, is a
right triangle
13
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
2
a
a
a  h     h 2  a 2    
2
2
2
and hence, by Pythagoras' theorem:
h 2  82  4 2 
2
2
h  48  6.93 cm
h 2  48 
Then the area will be:
A
b·h 8·6.93

 27.72cm2
2
2
19. Find the area of the given right triangle.
In a right triangle, the base and height are
the two perpendicular legs.
So one side is a base and the other the
corresponding height
A
b  h 6·8

 24 in 2
2
2
4.3. Area of a trapezium
A trapezium is a quadrilateral with exactly
one pair of parallel sides. The parallel sides
are called bases. The nonparallel sides are
called legs (or sides).
The longer base (the bottom) is big B and
the smaller base (the top) is little b...
The height is h
How can I find the area of a trapezium?
Take two copies of the trapezoid (one blue trapezium and one green trapezoid)... Tip one upside
down and stick them together... Now, you've got a parallelogram.
14
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
The area of the parallelogram is: A  base·height  B  b·h
But, this is double of what we need... So, multiply by 1/2!
The area of a trapezium, A , that has B and b lengths of the two parallel bases, and h is its height,
is:
B  b·h
A
2
Example
20. Find the area of an isosceles trapezium of bases 19 cm and 9 cm.
A special trapezium is the isosceles trapezium (like an isosceles triangle)
Properties of the sides of an isosceles trapezoid:
The bases (top and bottom) of an isosceles
trapezium are parallel.
An isosceles trapezium has opposite sides of the
same length.
The angles on either side of the bases have the
same size/measure.
The diagonals (not show here) have the same
length.
We know both bases but we have to know the height too. If we draw the heights (which have
the same length) we observe two equal right triangles with leg 5, because 19  9  10 and the two
triangles are equal. They have also the same hypotenuse, 13, and the other leg is the height h.
13
h
19
So the height is: 13 2  52  h 2 
5
h  13 2  5 2  169  25  144  12 cm
15
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
The area will be:
A
B  b·h  9  19 ·12  168 cm2
2
2
4.4. Area of regular polygon
A regular polygon is a figure that lies on a plane with sides of equal length and angles of
equal measure.
The area of any polygon is the sum of the areas of the triangles it can be divided into.
In the figure on the right, the polygon can
be broken up into four triangles by drawing
all the diagonals from one of the vertex.
.
The calculation is easier in a n-sided regular polygon as it can be divided into n identical triangles.
One of the points (vertices) of the triangle is the centre of the polygon and the other two are two
consecutive vertices of the polygon. Let s the side and l its length.
Hexagon
It is broken up into six equal triangles
The height of each of these triangles (apothem) is the perpendicular line from the centre of the
polygon to one of its sides.
Therefore, we can work out the area of each triangle by multiplying the length of the side by half of
the apothem. Remember the area of a triangle is
1
A  ·b·h
2
1
in our case A  ·l·a
2
If we multiple this result by the numbers of sides we can work out the area of the polygon:
16
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
1
n·l·a
A  n· ·l·a 
2
2
But n·l  p , the perimeter of the regular polygon. So we have:
Area of
a regular polygon 
perimeter · apothem
2
The area of a regular polygon is half the perimeter multiplied by the apothem:
A
perimeter · apothem p·a

2
2
Example
21. Find the perimeter and area of the regular pentagon in figure, with apothem
approximately 5.5 in.
4.5. Area of a composite 2-D shape
A figure (or shape) that can be divided into more than one of the basic figures is said to be a
composite figure (or shape).
For example, figure ABCD is a composite figure as it consists of two basic figures. That is, a figure
is formed by a rectangle and triangle as shown below. We can work out its area by adding up the
area of the rectangle and the area of the triangle whose formulas are well known.
We can also find this area by using the formula for a trapezium area.
The area of a composite figure is calculated by dividing the composite figure into basic figures and
then using the relevant area formula for each basic figure.
Example
21. Find the area of the following composite figure:
17
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
We see this figure consists of two basic
figures:



The triangle A B C
The trapezium CDEF

Area of the triangle A B C :
12  9·9  94.5cm2
ATria 
2
Area of the trapezium CDEF :
8  12 ·4  40cm2
ATrap 
2
Area of the polygon: A  ATria  ATrap  94 .5  40  134 .5cm 2
5. Area of circular figures
5.1. Definitions
A circle is a plane figure, bounded by a single curve line verifying that every part of which is
equally distant from a point within it, called the centre. We name to the distance from the centre
to any point of the boundary radius.
We defined circumference as the distance around the edge of a circle.
In everyday use, the term "circle" may be used interchangeably to refer to either the boundary of
the figure (known as the perimeter) or to the whole figure including its interior.
In academic use when we refer to the whole figure including its interior the term disk is used.
In Spanish we name circunferencia to a line forming a closed loop, every point on which is a
fixed distance from a centre point. This distance is named radius.
There are other figures related to circle.
A circle sector is any piece of the circle between two radial
lines (shaded in both dark and clear grey).
A segment of a circle is the region between a chord of a
circle and its associated arc (shaded in dark grey).
An annulus is the region lying between two concentric
circles. It is also named crown.
5.2. Area of a circle
We can think of a circle as a n-sided regular polygon with n a very large number. This polygon
can be inscribed in a circumference. The perimeter of such a polygon approximates to the
circumference when n increases, and its apothem approximates to the radius of the
circumference. You can see it below.
18
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
Area of
the circle 
perimeter  apothem circumference  r

2
2
As the circumference is L  2    r , then: Area
the circle 
of
2   r 2   r 2

  r2
2
2
The area of the circle is:
A   r2
5.3. Area of a circle sector
We want to find the area of a circle sector. First we observe that the area of a circle sector and
the measure of its arc in degrees are in direct proportion. The circle sector of arc 360º is the
whole circle. Now think of the ”regla de tres directa simple”:
ACirc measures
 360 º
ASect subtends
 nº
A  nº
So: ASect  Circ
360 º
If we substitute ACirc by its value, r 2 , we obtain the formula: ASect 
r 2 nº
360 º
The area A of any circle sector with an arc that has degree measure nº and with radius r is equal
to:
A
r 2 nº
360 º
5.4. Area of an annulus
The following formula whish comes from the definition of annulus as the region lying between two
concentric circles. of radii R and r .Then we work out this area a the difference between the area of
the biggest circle and the smallest circle.
19
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
The area of an annulus of radii R and r is:

A  R 2  r 2   R 2  r 2

Example
22. Find the area of the crown with radius 4 cm and 6 cm.
To work out the area of the crown we find the area of the large circle minus the area of the small
circle.
 r 2 =  6 2 =  36 = 36  cm 2
2
2
2
A s =  r =  (4) =  16= 16  cm
2
A Shaded = 36  - 16  = 20  cm
AL =
We can also apply directly the formula.
Example
22. Find the shaded area
We can find this area by subtracting the circular sectors of the circles of radii
r  5 cm.
A1 
R  8 cm and
  52  50
 10,9 cm2
360
  8 2  50
A2 
 27,91 cm2
360
A  A2  A1  27,91  10,9  17,01 cm2
20
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
6. Angles in polygons
6.1. Sum of the interior angles of a polygon
The sum of the angles of a triangle is 180º.
If you look at the n-sided polygons below you see that can be divided into
Four-sided polygon, quadrilateral
4  2 triangles


Five-sided polygon, pentagon
5  2 triangles
n  2 triangles.
Six-sided polygon, hexagon
6  2 triangles
The sum of the measures of all the interior angles in a triangle is 180º. We can obtain the sum
of the interior angles of a n-sided polygon by adding the sum of the interior angles of each
triangle, this is (n - 2) · 180°.
Moreover in a regular polygon all the angles measure the same.
The sum of the interior angles of a convex n sided polygon is
Each interior angle of a regular n-sided polygon is
n  2 180º
n  2  180º
n
Example
23. Here's a regular pentagon. Find the sum of its interior angles and the measurement of
each interior angle.
From vertex A we can draw two diagonals which separate the pentagon into three triangles. We
multiply 3 times 180 degrees to find the sum of all the interior angles of a pentagon, which is 540
degrees.
sum of angles = n  2  180º

=

5  2 180º  3 180º  540
6.2. Central angle of an inscribed polygon:
A central angle of a regular polygon is the angle formed by two consecutive radii.
All the central angles add up to 360º
21
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
If a regular polygon has n sides:
360 º
n

The central angle measures

The measure of an interior angles is
180 º (n  2)
n
Observe the first result is obvious because all the central angles, which are n , add up to 360º.
The second result was shown in point 6.1.
Example
24. Find
1. The sum of the interior angles of a hexagon
2. The measure of the interior angle of a regular hexagon
3. The measure of the central angle of a regular hexagon
We can reason about this questions and get the solution without using the formulae
Sum of the Interior Angles of a Hexagon:
To find the sum of the interior angles of a
hexagon, we divide it up into triangles... There
are four triangles. Because the sum of the
angles of each triangle is 180 degrees, we get
4 4 180  720
Applying the formula: n  2  180º for
we obtain the same result.
n  6,
So, the sum of the interior angles of a hexagon
is 720 degrees.
Regular Hexagons:
All sides are the same length (congruent) and all
interior angles are the same size (congruent).
To find the measure of the interior angles, we know
that the sum of all the angles is 720 degrees (from
above)... And there are six angles...
720
 120º
6
Applying the formula:
n  2  180º n
6  2  180º  720  120º
6
n
6
6
So, the measure of the interior of an hexagon is 120º
The measure of the central angles of a regular hexagon:
To find the measure of the central angle of a regular
hexagon, make a circle in the middle... A circle is 360
degrees around... Divide that by six angles...
360º n 6 360º

 60º
n
6
So, the measure of the central angle of a regular
hexagon is 60 degrees.
22
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
7. Angles in a circle
You can see this table is the same as the one in your text book. All these angles are interesting
but this year the most important are central angles, inscribed angles and interior angles.
Central angle
Inscribed angle
Semi inscribed angle
It has its vertex on the centre
of the circle
It has its vertex on the circle
(Spanish circumferencia) and
its sides are secant lines
It has its vertex on the circle
(Spanish circumferencia) and
one of its sides is tangent to
the circle and the other is
secant
Its measurement is the same
as the arc it intersects
It measures half the arc it
intersects
It measures half the arc it
intersects

AB

AB
2

AB
2
Interior angle
Exterior angle
Circumscribed angle
It has its vertex on an interior
point of the circle.
It has its vertex on an exterior
point and its sides are secants
It has its vertex on an exterior
point and its sides are
tangent to the circle:
It measures half the sum of
the two arcs it intersects
It measures half the difference
of the two arcs it intersects
 
AB  CD
2
 
AB  CD
2
It measures half the sum of
the two arcs it divides the
circle into.
 
AB  CD
2
23
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
24
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
P = 8 cm + 8cm + 3 cm + 3 cm = 22 cm
P = 2(8 cm) + 2(3 cm) = 16 cm + 6 cm = 22 cm
We name 8 cm + 3 cm Semi-perimeter, S,.
Given a rectangle with length a and width b we name semi-perimeter to S  a  b , so:
P  2  a  b  2  S
1.2. Perimeter of a regular polygon
The perimeter of a n -sided regular polygon of length side l , is.
P  nl
EXAMPLE
We name l to length of the sides of the following regular polygons. Find their perimeter:
Equilateral triangle
Square
All three sides have equal lengths
Parallelogram with 4 equal sides and
P= 3·
l
4 right angles: P = 4·
Regular hexagon
All six sides have equal length
l
P = 6·
l
1.3. Perimeter of a circle: Circumference
From Latin: circum "around" + ferre "to carry"
Circumference is the distance around the edge of a circle. Also 'periphery' , 'perimeter'.
If we divide the perimeter of a circle by its diameter we always obtain the same decimal number.
This is a very important number, name Pi and designed by  . This number is non exact anon
periodic. It has infinite decimal figures and they don’t repeat any “period”.
Its value has been calculated with a precision of hundreds of thousands of decimal figures but
we can not know the exact value. An approximated value is:
  3.1415926
We will use:
  3.14
The perimeter of a circle or circumference, L , can be calculated through two formulae:
L   ·d
L  2· ·r
where d is the diameter and r the radius
25
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
EXAMPLES:
1. Find the circumference of a circle with a radius 12 inches.
r = 12
C=2 r
C = 2 (12)
C = 24
inches.



2. Find the circumference of a bike wheel whose diameter is 16 inches.


C=
d
C=
(16)
C = 16
inches.

Or
If the diameter is 16, the radius is 8 inches. (The radius is ½ of the diameter).
Now use the formula



C = 2 r.
C = 2 (8)
C = 16
inches.
1.4. Length of an arc of circle

If we have a circle of radius r the length of the AB , which measures n degree is:
2r·nº
LA B 
360 º
How did we obtain this formula?
If you observe the length of an arc of a circle and the angle intercepted by arc are in direct
proportion. Remember that the arc which subtends 360º has the length of the circumference.
Now think of the ”regla de tres directa simple”:
2r subtends
 360 º
LAB  nº
subtends
So: LAB 
2r  nº
360 º
EXAMPLE:
Find the length of a 24º arc of a circle with a 5 cm radius
26
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
BLOG
Circumference video http://www.mathtutordvd.com/public/124.cfm
http://www.mathopenref.com/arcsector.html circular sector or circle
sector
http://www.mathopenref.com/annulus.html
http://www.mathopenref.com/rectanglediagonals.html
You can see a video about how to work out the area of a regular hexagon
http://es.video.yahoo.com/watch/3009502/8602185
HATy mas videos para el blog que he ido quitando. Buscar en la leccion
11 del 1º
Blog porblema propuoner
Given an equilateral triangle ABC with side length = s.
Find h
By Pythagorean Theorem (PT),
h² =
h =
.
27
IES “LA Laboral”. Bilingual Section
Author: Mónica García Pinillos
28