Transcript 10-3 & 10-4 : Areas of special quadrilaterals

3.3b: Area and Perimeter
- Quadrilaterals and composite shapes
CCSS:
G-GPE.7 Use coordinates to compute perimeters of polygons and areas of triangles and
rectangles, e.g., using the distance formula.?
G-GMD.1 Give an informal argument for the formulas for the circumference of a circle, area of a
circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri’s
principle, and informal limit arguments.
GSE’s
M(G&M)–10–6 Solves problems involving perimeter,
circumference, or area of two dimensional figures (including
composite figures) or surface area or volume of three
Area of Parallelograms
Base
height
Base
Base (b) = One side of the parallelogram
Height (h) = distance between the bases
(must be perpendicular)
Area of a Parallelogram = (b)(h)
Why ?
Base
height
Base
What shape will it make when we cut off the triangle on the side
and put in on the other side?
A rectangle with the area= (base)*(height)
How many square yards of carpeting are needed to
cover the family room, hallway, and bedroom?
Area if Triangles
1
A  bh
2
*b= base of the triangle
But
Why ?
*h = height of the triangle
* Both are touching the 90 degree angle in
the triangle
h
h
b
It is half of a parallelogram with the same exact base and height
Area if a Rhombus
Area =
1
d1 d 2
2
d1
d1  one diagonal
d2
d 2  the second diagonal
Example
A rhombus has an area of 50 square mm.
If one diagonal has a length of 10 mm,
How long is the other diagonal.
Area of a Trapezoid
1
A  h(b1  b2 )
2
Base
Height
(has to be perpendicular
to bases)
Base (parallel side)
h  height (the distance between th e bases)
b1  one of the bases (a parallel side)
b2)  the other base (opposite parallel side
Example
#1- Find the area
#2 - Find the area
M(G&M)–10–9 Solves problems on and off the coordinate plane involving
distance, midpoint, perpendicular and parallel lines, or slope.
Find the area
3
Use a geometric (area) approach
Instead of algebraic (slope, distance)
2
6 units
1
Now you have a rectangle
With dimensions of 4 by 6.
Arearectangle  (4)(6)  24 units 2
4 units
1
A1  (2)( 4)  4 u 2
2
1
A 2  (2)(6)  6 u 2
2
1
A 3  (2)( 4)  4 u 2
2
To get the area of the original
triangle, subtract the new
triangles from the overall
rectangle.
This will leave you with the area
of the original triangle.
AOriginal Triangle  Arectangle  Asmaller triangles
 24 - 14
 10 u 2
Now you try an example
Does it work with other shapes?
The end
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