# Chapter 10

### Stephanie Blair

## Section 1- Area of parallelograms and trapezoids

__Vocab__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

__Vocab__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