Chapter 10
Stephanie Blair
Section 1- Area of parallelograms and trapezoids
Base of a Parallelogram- the length of any one of its sides
Height of a Parallelogram- perpendicular distance between the base and the opposite side
Bases of a Trapezoid- It's two parallel sides
Height of Trapezoid- perpendicular distance between the bases
Formulas
Area= bh AREA OF A PARALLELOGRAM
Area=1/2(b1+b2)h AREA OF A TRAPEZOID
Practice Problems
1. Find the area of the parallelogram
A=bh
=8*10
=80
A= 80 inches
2. The Winslow House in River Forest, Illinois, was designed by Frank Lloyd Wright.The front part of the roof is a trapezoid, as shown. What is its area?
A =1/2(b1+b2)h
=1/2(31+77)25
=1350
A= 1350 feet
Real Life Scenarios
Trapezoid~~~ As stated in the problem above, a trapezoid could be a roof. How much siding do you need for this front portion? Use the area formula, 1/2(b1+b2)h to find out. Save your money by not buying too much siding but just the right amount.
Parallelogram~~~ You could have a parallelogram in a parking spot. You know what I mean, how there are sometimes slanted. Well you might have to ask yourself if your car will fit within the parking space. Well, how much space does your car take up and how big is the parking spot. Use the area formula, A=bh to find out. Measure the parking spot to see if your car will fit. Its that easy!
Section 2 - Area of circles
Area- the number of spuare units covered by a figure
Circle- the set of all points in a plane there are all the same distance from a fixed point
Radius- the distance between the center and any point of the circle
Diameter- the distance across the circle through the center
Circumference- the perimeter of a circle C= pi(r)^2
π (pi)- 3.14, irrational decimal used to find measurements of a circle
Formulas
A= pi(r)^2 Area of a circle
Practice Problems
1. Find the radius of the circle that has the area of 530.66 sq ft
A= π (r)^2
530.66= (3.14)r^2
169= r^2
Square root of 169= r
13 feet= r
2. Find the area of a circle with a diameter of 10 inches.
A=π r^2
= π (5)²
= 78.5 inches squared
Real Life Scenarios
Circle~~ Finding the right size for a cover of a pool. Most people cover their pool during the winter. Well, how do you find the right size for a cover? Use the formula A=π r² to find out
Section 3- Three Dimensional Figures
Solid- 3-D figure that encloses a part of space
Polyhedron- a solid that is enclosed by polygons
Face- the polygons that form a polyhedron
Prism- A polyhedron with 2 bases that lie parallel planes & the other faces are rectangles
Pyramid- a polyhedron with one base and the other faces are triangles
Cylinder- a solid with 2 congruent circular bases that lie in parallel planes
Cone- a solid with one circular bases
Sphere- the set of all points in a plane there are all the same distance from a fixed point
Edge- the segments where faces of a polyhedron meet
Vertex-a point where three or more edges meet
Practice Problems
1. How many faces, edges, and vertices does a pentagonal pyramid have?
6 faces
10 edges
6 vertices
2. How many faces, edges, and vertices does a rectangular prism have?
6 faces
9 edges
8 vertices
Section 4- Surface area of Prisms and Cylinders
Net- 2-D drawing of a solid when it is unfolded
Surface Area- sum of all its faces
Formulas
2pir^2 + 2(pi)(rh) SURFACE AREA OF A CYLINDER
S= 2B + Ph SURFACE AREA OF A PRISM
Practice Problems
1. You are painting a storage chest with the given dimensions. Find the surface area to be painted. Base- 30 in. Width- 15 in. Height- 15 in.
draw a net of the chest
Add up all the areas of each given shape
The surface area is 2250 square inches
2. Find the surface area of the can with a height of 10.7 cm and a width of 8 cm.
S= 2pi r^2 + 2pi rh
=2pi(4)^2 + 2pi(4)(10.7)
=369.45 square centimeters
Section 5- Surface area of prisms and cones
Slant Height- The height of a lateral face, any face that is not the base
Formulas
S= B+1/2Pl SURFACE AREA OF A PYRAMID
s= pir^2+ pi(r)l SURFACE AREA OF A CONE
Practice Problems
1. Find the surface area of a cone with radius 4 m and a slant height 9 m.
S= pi(r)^2 + pi(r)l
= pi(4)^2 + pi(4)(9)
= 163.4 square meters
2.Find the surface area of the regular pyramid. Round to the nearest tenth.
Find the perimeter of the base-- P= 8+8+8= 24
S= B + 1/2Pl
=27.7 + 1/2(24)(6)
= 99.7 square meters
Section 6- Volumes of cylinders and prisms
Volume- measure of the amount of space it occupies
Formulas
V= Bh VOLUME OF A PRISM
V=Bh VOLUME OF A CYLINDER
Practice Problems
1. Find the volume of the prism. Height 2 in Length 12 in Width 8 in
V=Bh
=lwh
=12(8)(2)
=192 cubic inches
2. The radius is one half the diameter, so r-3 cm. Find the volume of the cylinder
V=Bh
=pir^2
=pi(3)^2(9)
=81pi
=254.469 cubic centimeters
Section 7- Volume of Pyramids and Cones
Pyramid-
Cone-
Volume-
Formulas
V=1/3Bh VOLUME OF A PYRAMID
V=1/3Bh VOLUME OF A CONE
Practice Problems
1.Find the volume of the pyramid Height 12 Length 24 Width 10
V= 1/3Bh
=1/3(1/2*24*10)(12)
=480 cubic centimeters
2. The radius is one half the diameter, so r= 6ft. Find the volume of the cone
V=1/3pir^2h
=1/3pi(6)^2(12)
=144pi
=452.389 cubic feet