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18.2 Polygons
A polygon is a flat closed figure
described by straight-line segments
and angles.
• Polygons are named by the number of
sides they contain and other relations.
• The base of any polygon is the horizontal
side or a side that would be horizontal if
the polygon’s orientation is modified.
• The adjacent side is the side that has an
endpoint in common with the base.
The kinds of polygons we will work
with will be quadrilaterals and
triangles. We will use formulas to
find both perimeter and area of
these shapes.
The following are quadrilaterals for
which you need to be able to
calculate the perimeter and area:
• Parallelogram- opposite sides are parallel
a
h
b
• A = bh
• P = 2(a + b)
Base and height
always form
right angles
The following are quadrilaterals for
which you need to be able to
calculate the perimeter and area:
• Rectangle- a parallelogram with 4 right
angles.
h
b
• A = bh or lw
• P = 2(b + h) or 2(l + w)
The following are quadrilaterals for
which you need to be able to
calculate the perimeter and area:
• Square- a rectangle with 4 equal sides
b
2
• A = b or s
• P = 4b or 4s
2
The following are quadrilaterals for
which you need to be able to
calculate the perimeter and area:
• Rhombus- a parallelogram with 4 equal
sides
b
• A= bh
• P = 4b
h
b
The following are quadrilaterals for
which you need to be able to
calculate the perimeter and area:
• Trapezoid- a quadrilateral with one pair of
parallel sides.
a
c
• A = ½ (a + b)h
• P=a+b+c+d
d
h
b
The area of a rectangle is 280 sq
cm. Its width is 14 cm. Find its
length.
A 150 sq/ft roll of fiberglass is 30 in
wide. What is its length in feet?
If the cost is $9.75/sq yd, what is
the cost of the roll?
Find the area.
20 ft
14 ft
48 ft
168 ft
12 ft
2
2
6 ft
8 ft
168  48  216 ft
2
The replacement cost for construction
of the building below is $75/sq.ft.
Determine how much insurance should
be carried for full replacement.
75 ft
26  44  1144 ft 2
40 ft
44 ft
16 ft
12 ft
24 ft
26 ft
24  40  960 ft 2
28  25  700 ft
2804  75  $210300
2
A piece of sheet metal has a
rectangular hole in it as shown.
Find (a) the area of the piece
punched out, and (b) the area of
metal left.
40 in
12 in
5.5 in
21 in
Triangles
The formulas for perimeter and
area of triangles are:
• P=a+b+c
• A = ½ bh
c
a=h
b
c
a
h
b
h
c
a
b
If only the sides (not the height) of
a triangle are known, the area can
be found using a formula called
Heron’s formula given by
A  ss  as  bs  c
where a, b and c are the sides of
the triangle and s = ½ (a+b+c).
Find the area of the triangle below.
24 in
1
s  24  25  40
2
 44.5
25 in
40 in
A  44.544.5  2444.5  2544.5  40
 44.520.519.54.5
 80049.9375
 283in 2
A square hole is cut from the
equilateral triangle as shown. Find
the area remaining in the triangle.
A  Atri  Asq
 60202020  1010
 480,000  100
 693  100
 593cm
2
10.0 cm
40 cm