Chapter 5 Perimeter and Area

1
Chapter 5Perimeter and Area

• 5-1 Perimeter and Area
• 5-2 Areas of Triangles, Parallelograms, and
  Trapezoids
• 5-3 Circumferences and Areas of Circles
• 5-4The Pythagorean Theorem
• 5-5 Special Triangles and Areas of Regular
  Polygons
• 5-6 The Distance Formula and the Method of
  Quadrature
• 5-7 Proofs Using Coordinate Geometry
• 5-8 Geometric Probability

2
5.1 Perimeter and Area

• Perimeterthe distance around an object
• To find the perimeter of a polygon, add the
  lengths of all of its sides.

3
Areathe surface encompassed by a polygon

• Area of Triangle1/2bh, where height, h, is
  perpendicular to the base

4
Area of Squares2 where s is the length of the
side
5
Area of Rectangle l x w, where l is the length
and w is the width
6
5.2 Areas of Triangles, Parallelograms, and
Trapezoids

• Area of Triangle1/2bh, where h is perpendicular
  to the base, b h does not have to touch the base

7
Area of Parallelogramb x h, where height, h, is
perpendicular to base, b
8
Area of Trapezoid1/2(h)(b1b2) or
Ah(midsegment) h is height perpendicular to and
intersecting both bases (b1and b2)
9
5.3 Circumferences and Areas of Circles

• Circumferenceperimeter of a circle
• C2pr or C2d, where r is the radius and d is the
  diameter

10
Area of Circle pr2 If they give you the area
or circumference and ask for the radius, you must
use algebra to solve for r.
11
5.4 The Pythagorean Theorem

• Pythagorean Theorem Given a right triangle, the
  sum of the squares of the legs is equal to the
  square of the hypotenuse
• hyp2leg2leg2 or c2a2b2
• Your longest leg is always c!

12
Pythagorean Triples Sets of lengths that always
make a right triangle3,4,55,12, 137,24,259,
40, 4111, 60, 61
13
Using Side Lengths to Determine if a Triangle is
Right, Acute, or Obtuse Converse of the
Pythagorean Theorem If c2a2b2, then you have
a right triangle.If c2lta2b2, then you have an
acute triangle.If c2gta2b2, then you have an
obtuse triangle.
14
5.5 Special Triangles and Areas of Regular
Polygons

• Special Triangle 1 30-60-90

• Take an equilateral triangle with sides of length
  2a and split it in half. This leaves you with a
  30-60-90 triangle.
• Since everything was split in half, the base of
  this 30-60-90 triangle is half of the original,
  or a.
• Use the Pythagorean Theorem to get the
  height

15
Special Triangle 1 30-60-90

• So, in a 30-60-90 triangle, the side opposite
  30 is a, the side opposite 60 is av3, and the
  side opposite 90 is 2a

• When solving problems with a 30-60-90 triangle,
  set the known side length equal to its ratio.
• Solve for a.
• Then substitute into the other ratios to find the
  missing side lengths.

16
Special Triangle 2 45-45-90

• Take a 45-45-90 triangle with sides of length a.
  (Since the angles are congruent, this triangle is
  isosceles.
• Use the Pythagorean Theorem to find the length of
  the hypotenuse.

17
Special Triangle 2 45-45-90

• So, in a 45-45-90 triangle, the sides
  opposite 45 are a, and the side opposite 90 is
  av2

• When solving problems with a 45-45-90 triangle,
  set the known side length equal to its ratio.
• Solve for a.
• Then substitute into the other ratios to find the
  missing side lengths.

18
Finding the Area of a Regular Polygon

• Lets say we have a hexagon with side length 10.

1. Put the hexagon into a circle. How many
   triangles do we get? What are the angles inside?
2. Draw the height of the triangles (from the vertex
   of the circle to the middle of the edge). This
   is called your apothem in a regular polygon.

19
Finding the Area of a Regular Polygon
3. Use special triangles to find the length of
the apothem and calculate the area of the
triangle. 4. Multiply this area by the number of
triangles in the hexagon to get the total area.
Our side length10, so a5, and the height must
be 5v3.
20
Finding the Area of a Regular Polygon

• Formula for the area of a regular polygon
• A1/2(apothem)(perimeter)
• p(number of sides)(length of side)

21
5.6 The Distance Formula and the Method of
Quadrature

• Distancethe length between two points
• Given points (x1, y1) and (x2, y2)

Choose point one and point two and substitute in
this formula to find distance.
22
5.7 Proofs Using Coordinate Geometry

• Midpointthe point which divides a segment into
  two congruent parts
• Given endpoints (x1, y1) and (x2, y2)
• Slope
• Use properties of polygons to find determine
  lengths of sides and points of vertices.

23
5.8 Geometric Probability

• Probability is the likelihood that an event will
  happen. It is always between 0 and 1. (0 means
  that it is impossible, and 1 means that it always
  has to happen.)
• P (favorable outcomes)
• (possible outcomes)
• In geometry, probability is usually the percent
  of area an event representslike the area
  represented by blue on a spin dial.