Module 11 Lesson 1 Notes and Practice
Perimeter and Area of Quadrilaterals and Triangles
The perimeter (P) of a polygon is the sum of the lengths of its sides. That’s easy to find if you
know or can solve for all of the sides. We’ll encounter some perimeter problems as we work
through our area problems.
The area (A) of a polygon is the number of square units it encloses. Each type of special
quadrilateral you studied earlier has a formula to find the area. You should already be familiar
with many of these!
Area of a Parallelogram:
A = bh
(b = base, h = height)
h
b
**Any side can be the considered the base of a parallelogram. The altitude is the
perpendicular segment to the line containing the base. The length of the altitude is our
height.
Area of a Rectangle:
A = bh
(b = base, h = height)
h
b
A = s2
(s = side)
Area of a Square:
s
s
Area of a Rhombus or a Kite:
A=
1
d1d 2
2
d1
d2
( d1 = diagonal 1, d 2 = diagonal 2)
b1
Area of a Trapezoid:
A=
1
h(b1  b2 )
2
h
(Remember: The bases of a trapezoid
are the parallel sides)
Area of a Triangle:
A=
1
bh
2
b2
h
b
Example Set 1: Parallelograms, Rectangles and Squares Look at these examples –Study them carefully as some require you to find some
missing parts before you can plug into the formula.
Find the area.
a.
15 m
32 m
Step 1– Identify the figure →This is a parallelogram!
Step 2 – Write down the formula → A = bh
Step 3 – Do you have all of the necessary information needed to solve? Yes!
Step 4 – Substitute into the formula and solve:
A  bh
 (32)(15)
 480
2
Thus the answer is 480 m (NOTE: Area is always measured in units SQUARED!)
b.
8.5 ft
6 ft
60°
Step 1– Identify the figure →This is a parallelogram!
Step 2 – Write down the formula → A = bh
Step 3 – Do you have all of the necessary information needed to solve? No!
We know the base – since opposite sides of a parallelogram are congruent,
we can move 8.5 to the bottom. However, we are missing the height! We
must find that before we can plug our information into the formula.
The altitude (height) forms a 30-60-90 right triangle inside our parallelogram. If we
solve this triangle, we can determine that the height must be 3 3 .
Step 4 – Substitute into the formula and solve:
A  bh
 (8.5)(3 3)
6 ft
 25.5 3
2
2
Thus the answer is 25.5 3 ft or 44.2 ft
c.
60°
30
3 3
60°
3
Step 1– Identify the figure →This is a rectangle!
Step 2 – Write down the formula → A = bh
Step 3 – Do you have all of the necessary information needed to solve? No!
Once again we are missing the height! We must find that before we can plug
our information into the formula.
The diagonal in the rectangle forms a 30-60-90 right triangle. If we solve this triangle,
You must divide
we can determine that the height must be10 3 .
Step 4 – Substitute into the formula and solve:
30 by the 3 to
A  bh
solve for the
shorter leg of the
 (30)(10 3)
30-60-90 triangle
 300 3
Thus the answer is 300 3 or 519.6
So we have
30
3
-- rationalize it to
get 10 3
Find the area and the perimeter.
d. Square with
side length 2.5 cm.
Step 1– Identify the figure →This is a square!
Step 2 – Write down the area formula → A  s
Step 3 – Do you have all of the necessary information needed to solve? Yes!
Step 4 – Substitute into the formula and solve:
2
A=s 2
 2.52
 6.25
Thus the answer for the area is 6.25 cm
2
To find perimeter – Simply take the
side length 2.5 and multiply it by 4
since a square has 4 equal sides.
The perimeter is 10 cm
Now, “You Try” these: (Answers to the “You Try” problems can be found at the end of
this practice session.)
(Figures for problems 1 - 4 from http://glencoe.mcgraw-hill.com/sites/dl/free/0078738181/518676/geosg.pdf)
Find the area of the given figures.
1.
2.
Find the perimeter and area of the given figures.
3.
4.
Example Set 2: Rhombuses, Kites, Trapezoids and Triangles a. Find the area of the given rhombus.
5
7
Step 1– Identify the figure →This is a rhombus!
Step 2 – Write down the formula → A=
1
d1d 2
2
Step 3 – Do you have all of the necessary information needed to solve? Not quite.
We are given half of our diagonals. We can double each number in the given figure to
get the entire length of the diagonals. This gives one diagonal = 10 and one diagonal
equal to 14
Step 4 – Substitute into the formula and solve:
1
A= d1d 2
2
1
 (10)(14)
2
 70
Thus the answer is 70.
16
Step 1– Identify the figure →This is a trapezoid!
14
b.
Step 2 – Write down the formula → A=
1
h(b1 +b 2 )
2
Step 3 – Do you have all of the necessary information needed to solve? Yes!
Step 4 – Substitute into the formula and solve:
27
1
A= h(b1 +b 2 )
2
1
= (14)(16  27)
2
1
 (14)(43)
2
 301
Thus the answer is 301.
Step 1– Identify the figure →This is a triangle!
c.
Step 2 – Write down the formula → A=
14
1
bh
2
Step 3 – Do you have all of the necessary information needed to solve? Yes!
Step 4 – Substitute into the formula and solve:
1
A= bh
2
1
= (17)(14)
2
 119
17
Thus the answer is 119.
Step 1– Identify the figure →This is a triangle!
d.
Step 2 – Write down the formula → A=
17
8
1
bh
2
Step 3 – Do you have all of the necessary information needed to solve? No! We
are missing our height. When you know 2 sides of a triangle you can use Pythagorean
Theorem to solve for the missing side.
Step 4 – Substitute into the formula and solve:
2
2
2
1
A= bh
2
1
= (8)(15)
2
 60
Thus the answer is 60.
c  a b
17 2  82  x 2
289  64  x 2
225  x 2
15  x
Step 1– Identify the figure →This is a triangle!
Step 2 – Write down the formula → A=
1
bh
2
Step 3 – Do you have all of the necessary information needed to solve? No! We
need the base and the height. This is a 30-60-90 triangle and the rules are
Hypotenuse = 2 * short leg
e.
Long leg = 3 * short leg
28
30
The hypotenuse = 28 so we will use the top formula first.
Hypotenuse = 2*short leg
28 = 2 * short leg
14 = short leg (which is the side opposite of the 30 degree angle)
Now we can use the second formula:
Long leg = 3 * short leg
Long leg = 3 * 14
So the long leg (which is the bottom side) = 14 3
Step 4 – Substitute into the formula and solve:
1
*b*h
2
1
A  * 14 3 * 14
2
A  98 3
A
Step 1– Identify the figure →This is a triangle!
Step 2 – Write down the formula → A=
f.
10
Step 3 – Do you have all of the necessary information needed to solve? No! We
are missing our height and our base. This is a 45-45-90 triangle so we will use those
rules:
2 *leg
Hypotenuse =
45
1
bh
2
Legs are equal to each other.
Since 10 is the hypotenuse, we will use that formula:
2 *leg
Hypotenuse =
10 =
Leg =
2 *leg
10
2

(Divide each side by
10
2
*
2
2

2 )
10 2
5 2
2
Both legs are the same so the base and height are both 5 2 .
Step 4 – Substitute into the formula and solve:
1
*b*h
2
1
A  *5 2 *5 2
2
25 2
A
2
A
Now, “You Try” these: (Answers to the “You Try” problems can be found at the end of
this practice session.)
(Figures for problems 5 - 8 from http://glencoe.mcgraw-hill.com/sites/dl/free/0078738181/518676/geosg.pdf)
Find the area of the given figures.
5.
6.
7.
8.
Example Set 3: Sometimes you are given the perimeter or the area and must solve for
missing parts or other information. Here are a few examples of those types of problems.
a. The perimeter of a rectangle is 36 cm and the length is 8 cm. Find the area.
Step 1: Draw a picture and fill in what you know.
information you need to solve the problem.
P=36
Step 2: Determine and find what
Perimeter of rectangle: P  2l  2w
Substitute given information and solve for the width (height) of the
rectangle.
P  2l  2w
36  2(8)  2 w
20  2 w
10  w
8
Step 3: Solve – Be sure you answer the question that is asked! This question asked us to
find the area of the rectangle.
A  (8)(10)
A  80
The answer is 80.
b. The area of a square is 256 in 2 . Find the perimeter of the square.
Step 1: Write the formula for the area of a square and solve for the side length.
A  s2
Step 2: Find the perimeter.
256  s 2
The perimeter of a square = 4s (since all
the sides are the same).
Perimeter of a square = 4s = 4(16) = 64
256  s 2
16  s
The answer is 64 in.
c. The area of a trapezoid is 152 m 2 . If the height is 8 m and one base is 17 m. Find the
length of the missing base.
Step 1: Write the formula for the area of a trapezoid and fill in the information you know.
1
h(b1  b2 )
2
1
152  (8)(17  b2 )
2
A
Step 2: Solve for the missing base.
1
152  (8)(17  b2 )
2
152  4(17  b2 )
152  68  4b2
84  4b2
21  b2
The answer is 21 m.
Now, “You Try” these: (Answers to the “You Try” problems can be found at the end of
this practice session.)
9. The area of a triangle is 66 square inches. If the height is 6 inches, find the length of the
base.
10. The area of a rhombus is 375 square meters. If one diagonal is 25 meters, find the length
of the missing diagonal.
11. The perimeter of a square is 56 cm. Find the area of the square.
12. A trapezoid has an area of 1870 square inches. The bases are 60 in and 110 in. Find the
height.
Answers to “You Try” problems:
2
1. 288 ft
2
2. 49 2 or 69.3 yd (must find height first by
7. 400
8. 25 (triangle is a 45-45-90 triangle since legs
are congruent. Therefore, each leg is 5 2 )
solving 45-45-90 triangle, height = 3.5 2
9. 22 in
2
3. A = 166.5 km , P = 55 km
2
10. 30 m
4. A = 11.6 m , P = 13.6 m
11. 196 cm
5. 1640
12. 22 in.
6. 72 3 or 124.7 (Must solve 30-60-90 triangle
first)
2