2.7.2: Geometrical Shapes

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Learning Objectives

know what a polygon is
know what perimeter is and how to find it
know what the circumference, diameter, and radius of a circle is and how to find each one
know the meaning of the symbol ππ and its approximating value
know what a formula is and four versions of the circumference formula of a circle
know the meaning and notation for area
know the area formulas for some common geometric figures
be able to find the areas of some common geometric figures
know the meaning and notation for volume
know the volume formulas for some common geometric objects
be able to find the volume of some common geometric objects

Polygons

We can make use of conversion skills with denominate numbers to make measurements of geometric figures such as rectangles, triangles, and circles. To make these measurements we need to be familiar with several definitions.

Definition: Polygon

A polygon is a closed plane (flat) figure whose sides are line segments (portions of straight lines).

Polygons

$Four shapes, each completely closed, with various numbers of straight line segments as sides.$

Not polygons

$Four shapes. One three-sided open box. One oval. One oval-shaped object with one flat side, and one nondescript blob.$

Perimeter

Definition: Perimeter

The perimeter of a polygon is the distance around the polygon.

To find the perimeter of a polygon, we simply add up the lengths of all the sides.

Sample Set A

Find the perimeter of each polygon.

$A rectangle with short sides of length 2 cm and long sides of length 5 cm.$

Solution: $\begin{array} {rcl} {\text{Perimeter}} & = & {\text{2 cm + 5 cm + 2 cm + 5 cm}} \\ {} & = & {\text{14 cm}} \end{array}$

Sample Set A

$A polygon with sides of the following lengths: 9.2cm, 31mm, 4.2mm, 4.3mm, 1.52cm, and 5.4mm.$

Solution: $\begin{array} {rcr} {\text{Perimeter}} & = & {\text{3.1 mm}} \\ {} & & {\text{4.2 mm}} \\ {} & & {\text{4.3 mm}} \\ {} & & {\text{1.52 mm}} \\ {} & & {\text{5.4 mm}} \\ {} & & {\underline{\text{+ 9.2 mm}}} \\ {} & & {\text{27.72 mm}} \end{array}$

Sample Set A

$A polygon with eight sides. It is not an octagon, but can be visualized as one large rectangle with two smaller rectangles connected to it.$

Solution

Our first observation is that three of the dimensions are missing. However, we can determine the missing measurements using the following process. Let A, B, and C represent the missing measurements. Visualize

$A polygon with eight sides. It is not an octagon, but can be visualized as one large rectangle with two smaller rectangles connected to it. The height and width are measured and labeled with variables, A, B, and C.$

$\text{A = 12m - 2m = 10m}$
$\text{B = 9m + 1m - 2m = 8m}$
$\text{C = 12m - 1m = 11m}$

$\begin{array} {rcr} {\text{Perimeter}} & = & {\text{8 m}} \\ {} & & {\text{10 m}} \\ {} & & {\text{2 m}} \\ {} & & {\text{2 m}} \\ {} & & {\text{9 m}} \\ {} & & {\text{11 m}} \\ {} & & {\text{1 m}} \\ {} & & {\underline{\text{+ 1 m}}} \\ {} & & {\text{44 m}} \end{array}$

Practice Set A

Find the perimeter of each polygon.

$A three-sided polygon with sides of the following lengths: 3 ft, 8 ft, and 9 ft.$

Answer: 20 ft

Practice Set A

$A four-sided polygon with sides of the following length: 6.1m, 8.6m, 6.3m, and 5.8m.$

Answer: 26.8 m

Practice Set A

$A seven-sided polygon with sides of the following lengths: 10.07mi, 3.88mi, 4.54mi, 4.92mi, 12.61, 10.76mi, and 3.11mi.$

Answer: 49.89 mi

Circumference/Diameter/Radius

Diameter (d)
A diameter of a circle is any line segment that passes through the center of the circle and has its endpoints on the circle.

Radius (r)
A radius of a circle is any line segment having as its endpoints the center of the circle and a point on the circle.
The radius is one half the diameter.

Circumference (C)
The circumference of a circle is the distance around the circle. It is given by $C = \pi d= 2 \pi r$
$A circle with a line directly through the middle, ending at the edges of the shape. The entire length of the line is labeled diameter, and the length of the portion of the line from the center of the circle to the edge of the circle is labeled radius.$

Sample Set B

Find the circumference of the circle.

$A circle with a dashed line from one edge to the other, labeled d = 7 in.$

Solution

Use the formula $C = \pi d$.

$C = \pi \cdot 7\ in.$

By commutativity of multiplication,

$C = 7\ in. \cdot \pi$

$C = 7 \pi in.$, exactly

This result is exact since $\pi$ has not been approximated.

Sample Set B

Find the perimeter of the figure.

$A cane-shaped object of an even thickness, with one straight portion and one portion shaped in a half-circle. The thickness is 2.0cm, the length of the straight portion is 5.1cm, and the radius of the semicircle portion is 6.2cm.$

Solution

We notice that we have two semicircles (half circles).

The larger radius is 6.2 cm.

The smaller radius is $\text{6.2 cm - 2.0 cm = 4.2 cm.}$

The width of the bottom part of the rectangle is 2.0 cm.

$\begin{array} {rcll} {\text{Perimeter}} & = & {\text{2.0 cm}} & {} \\ {} & & {\text{5.1 cm}} & {} \\ {} & & {\text{2.0 cm}} & {} \\ {} & & {\text{5.1 cm}} & {} \\ {} & & {(0.5) \cdot (2) \cdot (3.14) \cdot \text{(6.2 com)}} & {\text{Circumference of outer semicircle.}} \\ {} & \ \ + & {\underline{(0.5) \cdot (2) \cdot (3.14) \cdot \text{(4.2 com)}}} & {\text{Circumference of inner semicircle.}} \\ {} & & {} & {\text{6.2 cm - 2.0 cm = 4.2 cm}} \\ {} & & {} & {\text{The 0.5 appears because we want the}} \\ {} & & {} & {\text{perimeter of only half a circle.}} \end{array}$

$\begin{array} {rcr} {\text{Perimeter}} & \approx & {\text{2.0 cm}} \\ {} & & {\text{5.1 cm}} \\ {} & & {\text{2.0 cm}} \\ {} & & {\text{5.1 cm}} \\ {} & & {\text{19.468 cm}} \\ {} & & {\underline{\text{+13.188 cm}}} \\ {} & & {\text{48.856 cm}} \end{array}$

Practice Set B

Find the outside perimeter of

$A shape best visualized as a hollow half-circle. The thickness is 1.8mm, and the diameter of the widest portion of the half-circle is 16.2mm.$

Answer: 41.634 mm

The Meaning and Notation for Area

The product $\text{(length unit)} \cdot \text{(length unit)} = \text{(length unit)}^2$, or, square length unit (sq length unit), can be interpreted physically as the area of a surface.

Area
The area of a surface is the amount of square length units contained in the surface.

For example, 3 sq in. means that 3 squares, 1 inch on each side, can be placed precisely on some surface. (The squares may have to be cut and rearranged so they match the shape of the surface.)

We will examine the area of the following geometric figures.

$Triangles, a three-sided polygon, have a height, h, measured from bottom to top, and base, b, measured from one end to the other of the bottom side.$ $Rectangles, a four-sided polygon, have a width, w, in this case the vertical side, and a length, l, in this case the horizontal side.$

$Parallelograms, a four-sided polygon with diagonal sides in the same direction have a height, h, measured as the distance from the bottom to top, and a base, b, measured as the width of the horizontal side.$ $Trapezoids, a four-sided polygon with diagonal sides facing leaning into each other, have a height measured as the distance between the two bases. Trapezoids have two bases of differing lengths, base 1, and base 2.$

$Circles. The distance across the circle is the diameter. The distance from the center of the circle to the edge is the radius.$

Finding Areas of Some Common Geometric Figures

Sample Set A

Find the area of the triangle.

$A triangle with height 6 feet and length 20 feet.$

Solution

$\begin{array} {rcl} {A_T} & = & {\dfrac{1}{2} \cdot b \cdot h} \\ {} & = & {\dfrac{1}{2} \cdot 20 \cdot 5 \text{ sq ft}} \\ {} & = & {10 \cdot 6 \text{ sq ft}} \\ {} & = & {60 \text{ sq ft}} \\ {} & = & {60 \text{ ft}^2} \end{array}$

The area of this triangle is 60 sq ft, which is often written as 60 $\text{ft}^2$.

Sample Set A

Find the area of the rectangle.

$A rectangle with width 4 feet 2 inches and height 8 inches.$

Solution

Let's first convert 4 ft 2 in. to inches. Since we wish to convert to inches, we'll use the unit fraction $\dfrac{\text{12 in.}}{\text{1 ft}}$ since it has inches in the numerator. Then,

$\begin{array} {rcl} {\text{4 ft}} & = & {\dfrac{\text{4 ft}}{1} \cdot \dfrac{\text{12 in.}}{\text{1 ft}}} \\ {} & = & {\dfrac{4 \cancel{\text{ ft}}}{1} \cdot \dfrac{\text{12 in.}}{1 \cancel{\text{ ft}}}} \\ {} & = & {\text{48 in.}} \end{array}$

Thus, $\text{4 ft 2 in. = 48 in. + 2 in. = 50 in.}$

$\begin{array} {rcl} {A_R} & = & {l \cdot w} \\ {} & = & {\text{50 in.} \cdot \text{8 in.}} \\ {} & = & {400 \text{ sq in.}} \end{array}$

The area of this rectangle is 400 sq in.

Sample Set A

Find the area of the parallelogram.

$A parallelogram with base 10.3cm and height 6.2cm$

Solution

$\begin{array} {rcl} {A_P} & = & {b \cdot h} \\ {} & = & {\text{10.3 cm} \cdot \text{6.2 cm}} \\ {} & = & {63.86 \text{ sq cm}} \end{array}$

The area of this parallelogram is 63.86 sq cm.

Sample Set A

Find the area of the trapezoid.

$A trapezoid with height 4.1mm, bottom base 20.4mm, and top base 14.5mm.$

Solution

$\begin{array} {rcl} {A_{Trap}} & = & {\dfrac{1}{2} \cdot (b_1 + b_2) \cdot h} \\ {} & = & {\dfrac{1}{2} \cdot (\text{14.5 mm + 20.4 mm}) \cdot (4.1 \text{ mm})} \\ {} & = & {\dfrac{1}{2} \cdot (\text{34.9 mm}) \cdot (4.1 \text{ mm})} \\ {} & = & {\dfrac{1}{2} \cdot \text{(143.09 sq mm)}} \\ {} & = & {71.545 \text{ sq mm}} \end{array}$

The area of this trapezoid is 71.545 sq mm.

Sample Set A

Find the approximate area of the circle.

$A circle with radius 16.8ft.$

Solution

$\begin{array} {rcl} {A_c} & = & {\pi \cdot r^2} \\ {} & \approx & {(3.14) \cdot (16.8 \text{ ft})^2} \\ {} & \approx & {(3.14) \cdot (\text{282.24 sq ft})} \\ {} & \approx & {888.23 \text{ sq ft}} \end{array}$

The area of this circle is approximately 886.23 sq ft.

The Meaning and Notation for Volume

The product $\text{(length unit)}\text{(length unit)}\text{(length unit)} = \text{(length unit)}^3$, or cubic length unit (cu length unit), can be interpreted physically as the volume of a three-dimensional object.

Volume
The volume of an object is the amount of cubic length units contained in the object.

For example, 4 cu mm means that 4 cubes, 1 mm on each side, would precisely fill some three-dimensional object. (The cubes may have to be cut and rearranged so they match the shape of the object.)

$A rectangular solid, with length l, width w, and height h.$ $A sphere with radius r.$ $A cylinder with height h and radius r.$ $A cone with height h and radius r.$

Finding Volumes of Some Common Geometric Objects

Sample Set B

Find the volume of the rectangular solid.

$A rectangular solid with width 9in, length 10in, and height 3in.$

Solution

$\begin{array} {rcl} {V_R} & = & {l \cdot w \cdot h} \\ {} & = & {\text{9 in.} \cdot \text{10 in.} \cdot \text{3 in.}} \\ {} & = & {\text{270 cu in.}} \\ {} & = & {\text{270 in.}^3} \end{array}$

The volume of this rectangular solid is 270 cu in.

Sample Set B

Find the approximate volume of the sphere.

$A circle with radius 6cm.$

Solution

$\begin{array} {rcl} {V_S} & = & {\dfrac{4}{3} \cdot \pi \cdot r^3} \\ {} & \approx & {(\dfrac{4}{3}) \cdot (3.14) \cdot \text{(6 cm)}^3} \\ {} & \approx & {(\dfrac{4}{3}) \cdot (3.14) \cdot \text{(216 cu cm)}} \\ {} & \approx & {\text{904.32 cu cm}} \end{array}$

The approximate volume of this sphere is 904.32 cu cm, which is often written as 904.32 cm$^3$.

Sample Set B

Find the approximate volume of the cylinder.

$A cylinder with radius 4.9ft and height 7.8ft.$

Solution

$\begin{array} {rcl} {V_{Cyl}} & = & {\pi \cdot r^2 \cdot h} \\ {} & \approx & {(3.14) \cdot (\text{4.9 ft})^2 \cdot \text{(7.8 ft)}} \\ {} & \approx & {(3.14) \cdot (\text{24.01 sq ft}) \cdot \text{(7.8 ft)}} \\ {} & \approx & {(3.14) \cdot \text{(187.278 cu ft)}} \\ {} & \approx & {\text{588.05292 cu ft}} \end{array}$

The volume of this cylinder is approximately 588.05292 cu ft. The volume is approximate because we approximated $\pi$ with 3.14.

Sample Set B

Find the approximate volume of the cone. Round to two decimal places.

$A cone with height 5mm and radius 2mm$

Solution

$\begin{array} {rcl} {V_{c}} & = & {\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h} \\ {} & \approx & {(\dfrac{1}{3}) \cdot (3.14) \cdot (\text{2 mm})^2 \cdot \text{(5 mm)}} \\ {} & \approx & {(\dfrac{1}{3}) \cdot (3.14) \cdot (\text{4 sq mm}) \cdot \text{(5 mm)}} \\ {} & \approx & {(\dfrac{1}{3}) \cdot (3.14) \cdot \text{(20 cu mm)}} \\ {} & \approx & {20.9\overline{3} \text{ cu mm}} \\ {} & \approx & {\text{20.93 cu mm}} \end{array}$

The volume of this cone is approximately 20.93 cu mm. The volume is approximate because we approximated $\pi$ with 3.14.

Table of Formulas for Common Shapes

	Figure	Formulas
Square	$Screenshot (282).png$ Figure $\PageIndex{1}$	\[P=4 s \] \[A=s^{2}\]
Rectangle	$Screenshot (283).png$ Figure $\PageIndex{2}$	\[P=2 l+2 w \] \[A=l w\]
Parallelogram	$Screenshot (284).png$ Figure $\PageIndex{3}$	\[P=2 a+2 b\] \[A=b h\]
Trapezoid	$Screenshot (285).png$ Figure $\PageIndex{4}$	\[P=a+b+c+d\] \[A=\frac{1}{2} h(a+b)\]
Triangle	$Screenshot (286).png$ Figure $\PageIndex{5}$	\[P=a+b+c\] \[A=\frac{1}{2} b h\]
Circle	$Screenshot (287).png$ Figure $\PageIndex{6}$	\[C=2 \pi \] \[r=\pi r^{2}\] Volume (V) is measured in cubic units and surface area (SA) is measured in square units.
Cube	$Screenshot (288).png$ Figure $\PageIndex{1}$	\[S A=6 s^{2}\] \[V=s^{3}\]
Rectangular Solid	$Screenshot (289).png$ Figure $\PageIndex{2}$	\[S A=2 l w+2 l h+2 w h\] \[V=l w h\]
Right Circular Cylinder	$Screenshot (290).png$ Figure $\PageIndex{3}$	\[S A=2 \pi r^{2}+2 \pi r h\] \[V=\pi r^{2} h\]
Right Circular Cone	$Screenshot (292).png$ Figure $\PageIndex{4}$	\[S A=\pi r^{2}+\pi r s\] \[V=\frac{1}{3} \pi r^{2} h\]
Sphere	$Screenshot (293).png$ Figure $\PageIndex{5}$	\[S A=4 \pi r^{2}\] \[V=\frac{4}{3} \pi r^{3}\]