Chapter 11
MEASURING LENGTH AND AREA
11.1 AREAS OF TRIANGLES AND PARALLELOGRAMS
11.2 AREAS OF TRAPEZOIDS, RHOMBUSES, AND KITES
11.3 PERIMETER AND AREA OF SIMILAR FIGURES
11.4 CIRCUMFERENCE AND ARC LENGTH
11.5 AREAS OF CIRCLES AND SECTORS
11.6 AREAS OF REGULAR POLYGONS
11.7 USE GEOMETRIC PROBABILITY
11.1 Area of Triangles and Parallelogram
s
Postulate 24: Area of Square Postulate
The area of the square is the square of the length of its sides.
A = s2
Postulate 25: Area Congruence Postulate
If two polygons are congruent, then they have the same area.
Postulate 26: Area Addition Postulate
The area of the region is the sum of the areas of its non-overlapping parts.
Theorem 11.1: Area of the Rectangle
The area of the rectangle is the product of its base
and height.
h
b
A = bh
11.1 Area of Triangles and Parallelogram
Theorem 11.2: Area of a Parallelogram
The area of a parallelogram is the product of the base and
Its corresponding height.
h
A = bh
b
Theorem 11.3: Area of a Triangle
The area of a triangle is one half the product of the base
and its corresponding height.
Parallelograms
Either pairs of the parallel sides can be used as the bases
of the parallelogram. The height is the perpendicular
distance between these bases.
If you transform a rectangle to form other parallelograms
with the same base and height, the area stays the same.
h
b
A = (1/2) bh
Chapter 11 Section 1 Practice Problems
Find the area of the shaded region.
8
4
Which postulate can be used to
prove congruency?
A
12
Area=(12)(8)-42
Area=96-16
Area=80 units2
4
B
4
Area Congruence Postulate
Chapter 11 Section 1 Practice Problems continued…
Find the area of the shapes below.
5
3
9
6
A = (1/2) bh
H = √(52 – 32) = 4
A = (1/2) 6 ∙ 4 = 12
The area of the triangle
is 12 units2.
A = bh
A = 9 ∙ 3 = 27
The area of the
parallelogram is 27 units 2.
11.2 Area of Trapezoids, Rhombuses, and Kites
The height of the trapezoid is the perpendicular distance
between its bases.
Theorem 11.4 Area of Trapezoid
The area of Trapezoid is one half the product
Of the height and the sum of the lengths of the bases.
b1
h
b2
A = (0.5)h(b1 + b2)
Theorem 11.5 Area of a Rhombus
The area of a Rhombus is one half the product of the
Length of its diagonals.
d2
A = (0.5)d1d2
Theorem 11.6 Area of a Kite
The area of a kite is the product of the lengths of its
diagonals.
d2
A = (0.5)d1d2
Chapter 11 Section 2 Practice Problems
The area of the trapezoid is 54ft2. Find
the height of the trapezoid.
Find the area of the rhombus.
4ft
16ft
20ft
8ft
A = (0.5)h(b1 + b2)
54=(0.5)h(8+4)
54=(0.5)(12h)
54=(6h)
h=9ft
202=162+x2
X=12
d1=32ft, d2=24ft
A=(0.5)(32)(24)
A=384ft2
11.3 Perimeter and Area of Similar Figures
Recall…
If two polygons are similar, then the ratio of their perimeters, or of any two
corresponding lengths, is equal to the ratio of their corresponding side lengths.
However, the areas of similar polygons have a different ratio.
Theorem 11.7 Areas of Similar Polygons
The area of two polygons are similar with the lengths of
corresponding sides in the ratio of a : b, then the ratio of
their areas is a2 : b2.
a
Side length of Polygon I
= b
Side length of Polygon II
Area of Polygon I
Area of Polygon II
a2
= 2
b
b
a
I
II
Polygon I ~ Polygon II
Any two regular polygons with the same number of sides are similar. Also, any two circles
are similar.
Chapter 11 Section 3 Practice Problems
Find the height of figure B if it is
similar to figure A (not drawn to scale
obviously).
Write the ratios of the corresponding side
lengths of the 2 similar polygons from
figure A to figure B given their areas.
8ft
Area of Figure A=108in2
Area of Figure B=432in2
A
B
Area=40m2
108 1
=
432 4
Area=360m2
Ratio of Areas=1/9, so the ratios of corresponding side
lengths in 1/3.
The ratio of the
1
corresponding sides in , as
the square root of
1
4
is
2
1
.
2
1/3=8/𝑥
X=24m.
11.4 Circumference and Arc Length
The circumference of a circle is the distance around the
circle. For all circles, the ratio of the circumference to the
diameter is the same. This ratio is known as π, or pi.
Theorem 11.8 Circumference of a Circle
The circumference C of a circle is C = πd or C = 2πr, where
d is the diameter of the circle and r is the radius of the circle
r
d
C
C = πd = 2πr
Arc Length Corollary
In a circle, the ratio of the length of a given arc to
the circumference is equal to the ratio of the
measure of the arc to 360
AB
mAB
= mAB
,
or
length
of
AB
=
X 2πr
360
2πr
360
A
B
Chapter 11 Section 4 Practice Problems
Find the circumference of the circle.
Express your answer in terms of π.
Find the length of AB. Express your answer
to the nearest hundredth.
B
5in
A
Radius=5in, so diameter=10in
Circumference=πd
Circumference=10πin.
30˚
12mm
Circumference=24π
30
𝑥
=
360 24π
x=2π
x=6.28mm
11.5 Area of Circles and Sectors
Theorem 11.9 Area of a Circle
The area of a circle is π times the square of the radius.
Sector of a circle: the region bounded by two radii of
the circle and their intercepted arc.
r
A = πr2
Theorem 11.10 Area of a Sector
The ratio of the area of the sector of a circle to the area
of the whole circle (πr2) is equal to the ratio of the
measure of the intercepted arc to 360◦
A
P
mAB x πr2
mAB
Area of sector APB
,
or
Area
of
sector
APB
=
=
360 ◦
πr2
360 ◦
r
B
Chapter 11 Section 5 Practice Problems
Find the area of sector ABC to
the nearest hundredth.
The area of sector ABC is 24πcm2.
Find the measure of angle ABC.
A
C
𝑥
45
=
64π 360
45˚ B
8m
X=8π
Area of sector ABC=25.13m2
A
C
12cm
B
Area of circle B=144πcm2
24
𝑥
=
144 360
m ∠ABC=60˚
11.6 Areas of Regular Polygons
The center of the polygon (point P) and the radius of the
polygon (MP) are the center and radius of the circle that
circumscribes the regular polygon.
M
Q
P
N
The distance from the center to any side of the polygon is
called the apothem of the polygon (QP).
A central angle of a regular polygon is an angle formed by
two radii drawn to consecutive vertices of the regular
polygon (∠𝐌𝐏𝐍).
Theorem 11.11 Area of a Regular Polygon
The area of a regular n-gon with side length s is one half
the product of the apothem a and the perimeter P,
so A = 12 aP or A = 12 a * ns
s
a
Chapter 11 Section 6 Practice Problems
Find the area of the regular polygon.
Express your answer to the nearest
hundredth. (Hint=use trigonometric
ratios to find the apothem.)
Find the area of the shaded region if
the triangle is equilateral. Express your
answer to the nearest hundredth.
4ft
6in
Apothem=
15*
3
tan 36
3
tan 36
Area=61.94in2
Area=29.48ft2
11.7 Use Geometric Probability
The Probability of an event is the measure of the likelihood
that the event will occur.
A geometric probability is a ratio that involves a geometric
measure such as length or area.
Key concept Probability and Length
Let AB be a segment that contains the segment CD. If a
point K on AB is chosen at random, then the probability
that it is on CD is the ratio of the length of CD to the
length of AB.
Key concept Probability and Area
Let J be a region that contains the region M. If a point K
in J is chosen at random, then the probability that it is in
region M is the ratio of the area of M to the area of J.
A
C
D
B
Length of CD
P(K is on CD) =
Length of AB
M
J
P(K is in region M) =
Area of M
Area of J
Chapter 11 Section 7 Practice Problems
What is the probability that a point
randomly selected on AD is also on BC?
Express your answer as a common
fraction.
A
B
C
What is the probability that a point
randomly selected in region A is also
in region B? Express your answer as a
common fraction. (Region A is the
entire figure).
D
13
B
182
13
182
reduces to
A
1
14
The probability is
1
14
The probability is
1
4