Download Introductory Algebra

§ 3.3
Problem Solving in Geometry
Geometry
Geometry is about the space you live in and the shapes that surround you. For
thousands of years, people have studied geometry in some form to obtain a better
understanding of the world in which they live.
In this section, we will look at some basic geometric formulas for perimeter, area, and
volume.
Remember that perimeter is a linear measure, that is, a measure of length. It will be
measured in units such as feet, or meters.
Area is a measure of square units and it will be measured in units such as square feet,
square miles, or square meters. How many floor tiles are in your kitchen? That’s an area
measure.
Volume is a measure of cubic units. It measures the number of cubic units in a three
dimensional object or rather the amount of space occupied by the figure. A liter is a
measure of volume. Cups and gallons are also measures of volume. The volume of a solid
is the number of cubic units that can be contained in the solid.
Blitzer, Introductory Algebra, 5e – Slide #2 Section 3.3
Perimeter and Area of Rectangle
Common Formulas for Perimeter and Area
Square
s
l
s
As
P = 4s
Rectangle
2
w
A = lw
P = 2l + 2w
Blitzer, Introductory Algebra, 5e – Slide #3 Section 3.3
Area of Triangle and Trapezoid
Common Formulas for Area
Triangle
h
Trapezoid
b
h
b
1
A  bh
2
Blitzer, Introductory Algebra, 5e – Slide #4 Section 3.3
a
1
A  h( a  b)
2
Perimeter of Rectangle
Example: A rectangle has a width of 10 inches and a
perimeter of 125 inches. Find the length
1.
2.
3.
Let l represent the unknown length.
l = length of the rectangle.
Represent other unknowns in terms of l.
there are no other unknowns
Write an equation that describes l the condition.
2w + 2l = P
2(10) + 2l = 125
20 + 2l =125
Blitzer, Introductory Algebra, 5e – Slide #5 Section 3.3
Perimeter of a Rectangle
Example Continued: A rectangle has a width of 10 inches and a
perimeter of 125 inches. Find the length.
4.
Solve the equation and answer the question.
20 + 2l = 125
2l = 105
l = 52.5
The length of the rectangle is 52.5 inches.
5.
Check the solution in the original problem.
The perimeter of a rectangle that is 10 inches by 52.5
inches is 125 inches. 2(10) + 2(52.5) = 125
Blitzer, Introductory Algebra, 5e – Slide #6 Section 3.3
The Circle
Formulas for Circles
r
radius  r
area  A   r
circumfere nce  2 r
2
The circumference of a circle is a measure of length. The circumference of a
circle is the distance around it, or its perimeter.
Blitzer, Introductory Algebra, 5e – Slide #7 Section 3.3
Area and Circumference of Circle
Example : Find the area and circumference of a
circle with a radius of 5 inches.
Area  A   r , r  5
2
A 5
A  25 square inches  78.54 square inches
Perimeter  P  2 r
P  2 5
P  10 inches  31.42 inches
2
Blitzer, Introductory Algebra, 5e – Slide #8 Section 3.3
Volume of a Cube and a Rectangular Solid
Common Formulas for Volume
Cube
V s
Rectangular Solid
3
V  lwh
Blitzer, Introductory Algebra, 5e – Slide #9 Section 3.3
Volume of a Cylinder and a Sphere
Common Formulas for Volume
Circular Cylinder
V r h
2
Sphere
4
3
V  r
3
Blitzer, Introductory Algebra, 5e – Slide #10 Section 3.3
Volume of a Cone
Cone
1 2
V  r h
3
Blitzer, Introductory Algebra, 5e – Slide #11 Section 3.3
Volume of a Cylinder
Example : Find the volume of a cylinder that has a
height of 4 feet and a radius of 5 feet.
V r h
r 5
h4
2
V 5 4
V   100
V  100 cubic feet  314.16 cubic feet
2
Blitzer, Introductory Algebra, 5e – Slide #12 Section 3.3
Important Fact about Triangles
Angles of a Triangle
The sum of the measures of the three angles
of a triangle is 180º.
Blitzer, Introductory Algebra, 5e – Slide #13 Section 3.3
Angles of a Triangle
Example : One angle of a triangle is three times another. The
measure of the third angle is 10º more than the smallest angle. Find
the angles.
1.
2.
3.
Let x represent one of the quantities.
x = measure of the first angle
Represent other quantities in terms of x.
3x = measure of the second angle
x + 10 = measure of the third angle
Write an equation in x that describes the conditions.
x + 3x + x + 10 = 180 (Sum of the angles of a triangle is 180.)
Blitzer, Introductory Algebra, 5e – Slide #14 Section 3.3
Angles of a Triangle
Example continued : One angle of a triangle is three times
another. The measure of the third angle is 10º more than the smallest
angle. Find the angles.
4.
Solve the equation and answer the question.
x + 3x + x + 10 = 180
5x + 10 = 180
Combine like terms.
5x =170
Subtract 10 from both sides.
x = 34
5.
Divide both sides by 5.
3x = 102 measure of the second angle
x + 10 = 44 measure of the third angle
The angles are 34º, 102º and 44º.
Check the solution.
102º is three times 34º, thus one angle is three times another. The
measure of the third, 44º is 10º more than the smallest angle. The angles
sum to 180º
Blitzer, Introductory Algebra, 5e – Slide #15 Section 3.3
Special Angles
Complementary and Supplementary
• Complementary Angles: Two angles whose sum is
90º.
• Supplementary Angles: Two angles whose sum is
180º.
Blitzer, Introductory Algebra, 5e – Slide #16 Section 3.3
Complements and Supplements
Algebraic Expressions for Angle Measures
Measure of an angle: x
Measure of the angle’s complement: 90 – x
Measure of the angle’s supplement: 180 – x
Blitzer, Introductory Algebra, 5e – Slide #17 Section 3.3
Complement of an Angle
Example : Find the measure of an angle that is 12 º
less than its complement.
1.
2.
3.
Let x represent one of the quantities.
x = measure of the first angle
Represent other quantities in terms of x.
90 - x = measure of the complement angle
Write an equation in x that describes the conditions.
x = (90 – x) – 12
The
angle
is
12 less
than
the complement.
Blitzer, Introductory Algebra, 5e – Slide #18 Section 3.3
Complement of an Angle
Example continued: Find the measure of an angle that is
12 º less than its complement.
4.
Solve the equation and answer the question.
x = (90 – x) – 12
x = 90 – x – 12
Remove parentheses.
x = 78 – x
Combine like terms.
2x = 78
5.
Add x to both sides.
Divide both sides by 2.
x = 39
x = 39º measure of the first angle
90 – x = 51 measure of the complement angle
The angles are 39º and 51º.
Check the solution.
39º is 12º less than 51º and the angles sum to 90º
Blitzer, Introductory Algebra, 5e – Slide #19 Section 3.3
Area of a Circle, An Application
EXAMPLE
Which one of the following is a better buy: a large pizza with
a 16 inch diameter for $12.00 or two small pizzas, each with a
10 inch diameter, for $12.00?
SOLUTION
STEP 1: Since the cost is the same, we must just determine
which has greatest area – the one big pizza or the two
small ones.
STEP 2: Find the area of the big pizza. The diameter is 16,
so the radius is 8. The area of the big pizza is then:
A=
 (8)  64
2
which is approximately 201 square inches.
Blitzer, Introductory Algebra, 5e – Slide #20 Section 3.3
Application, continued
CONTINUED
Now we must find the area of each of the two smaller pizzas.
A=
 (5) 2   (25)
which is approximately 78.5 square inches
for each smaller pizza. We would have two
of these, so the total area would be
around 157 square inches.
But still… the best buy is the one large. And we know something now that the
average person doesn’t consider when purchasing pizza. And that is, when you
are considering how big a pizza is – don’t just consider the radius. Consider
the square of the radius! And that’s why one sixteen inch pizza would be
bigger than two eight inch ones. It’s even bigger than two ten inch ones as
we discovered above. Hungry yet?
Blitzer, Introductory Algebra, 5e – Slide #21 Section 3.3