The word geometry in Greek means "earth measurement" and had its origins in problems associated with surveying. In these notes we are considering plane or flat two-dimensional geometry. (Solid geometry studies objects in three dimensions.) The fundamental notions used to describe planar objects are points and lines. Two different points define one and only one line that passes through both of them and the shortest distance between the two points is along this line. We shall use capital letters such as A to label points and a pair of letters such as AB to label the segment of the line between A and B. Line segments are measured in units of length such as feet or meters.
Two lines in a plane either intersect at a single point or are parallel. Where the lines meet four "openings" or angles are formed as shown below. The symbol Ð is used for the word angle.
In order to label angles several conventions are used. As shown in the above picture, the point where the lines intersect A is called the vertex of the four angles. The separate angles are then named by indicating a point on each of the two lines which act as sides of the angle. Thus ÐBAD is the angle with vertex at A and with sides that include segments BA and AD. This same angle could also be labeled as ÐDAB. A less precise but more convenient notation is to label angles by a single letter or number, provided that the symbol is placed appropriately in the diagram. For example, in the above diagram Ð1 = ÐEAD. The most commonly used measure for angles is the degree system developed by the Babylonians over 3000 years ago. They decided for reasons of easy divisibility by whole numbers to divide a circle into 360 equal units of angle called a degree and signified by the symbol °. For precise surveying or problems in astronomy a smaller unit is required. This lead to the minute which is 1/60th of a degree and the second which is 1/60th of a minute. The symbol for minute is ', while the symbol for second is ".
Thus, we have
and 
.
An angle to the nearest second could be specified as
Ð4 = 37°29'54", this notation is called
DMS for Degree Minute Second. To avoid fractions this same angle could be
given in DD (Decimal Degrees) as
.
We could also go from DD to DMS as follows:
These conversions have been built into many scientific calculators. For
example, to enter 37°29'54" in DMS notation on the Casio fx-300SA, keystroke
37 
29
54 , where
is the third row
second column key. The display will then automatically show the DD result of 37.49833333°.
To convert back to DMS notation keystroke 
and the calculator will
display 37°29°54° which is the Casio's way of indicating 37°29'54".
To enter 37°29'54" using the TI-30Xa, keystroke 37.29.54 
, where the second decimal point (which does
not show in the
display) separates the seconds from the minutes and
is the
DMS --> DD function. To convert from DD to DMS use
. These conversions can be
embedded into calculations involving angular
measure.
Since there are 360° in a circle, the angular measure of a straight line is 180°, while in a square or right angle there are 90°. Two angles whose sum is 180° make a straight line when combined and are called supplementary. Consider again the intersection of two lines. ÐEAD + ÐCAE = ÐEAD + ÐBAD = 180°. Since ÐEAD is on both sides of this equation, we conclude that ÐCAE = ÐBAD and by a similar argument that ÐCAB = ÐEAD. These equal angles formed by two intersecting lines are called vertex angles.
As an application of this result consider the following diagram. Once the measure of Ðd is known, the remaining three can be determined. Since they are vertex angles, Ðb = Ðd = 144°39'35". Since Ða is the supplement to Ðd, Ða = 180° - 144°39'35" = 179°59'60" - 144°39'35" = 35°20'25" = Ðc.
This same calculation is done on the Casio fx-300SA using the following keystrokes:
180 
144 39
35
and on the TI-30Xa using
180
144.39.35
Consider a pair of parallel lines crossed by a third line (called the transversal) as shown below. If we imagine that point B is superimposed over point F by moving segment CD onto the line through EG, the corresponding angles ÐEFB and ÐCBA are equal. Similarly, ÐEFH = ÐCBF, ÐABD = ÐBFG, and ÐFBD = ÐHFG. From the equality of vertex angles, the alternating interior angles (alternate sides of the transversal, inside the two parallel lines) are equal, i.e., ÐEFB= ÐFBD and ÐCBF= ÐBFG. Similarly, the alternating exterior angles (alternate sides of the transversal, outside the two parallel lines) are equal, i.e., ÐEFH = ÐABD and ÐCBA = ÐHFG.
Polygons are closed figures in the plane whose sides are line segments. The simplest polygon is the three-sided triangle. The points at the corners A, B, and C are the vertices of the triangle, and the angles ÐBAC, ÐBCA, and ÐABC are called the interior angles of the triangle. The triangle is often then labeled as triangle ABC.
Consider the triangle ACB shown below. Through the vertex C we construct a line parallel to the base segment AB. Since they are alternating interior angles, ÐDCA = ÐCAB and ÐECB = ÐCBA. However, ÐDCA + ÐACB + ÐECB = 180°, so we have ÐCAB + ÐACB + ÐCBA = 180° or in words, the interior angles of a triangle always sum to 180°.
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Consider the five-sided pentagon on the right. By drawing three triangles from one of the vertices, we see that the sum of the internal angles of the pentagon is the sum of all the internal angles in the three triangles or 3(180°) = 540°. A similar argument for an n-sided polygon shows that the sum of the internal angles is (n-2)(180°). |
As an application of these principals, consider the following problems. In the triangle below the missing angle Ð1 is calculated as follows :
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Ð1 = 180° - 97°13' - 30°59' = 51°48' |
If L1 and L2 are parallel, then Ðd = 101°, Ðf = 180°- Ðd = 79°, Ðb = Ðc = 38°, Ða =180° - Ðb = 142°, Ðe = 180° - Ðd -38° = 41°, and Ðg = 180° - Ðe = 139°.
We now turn our attention to triangles. The first question we need to consider is what determines a triangle. Every triangle has three sides and three angles, so six numbers (three angle measures and three side lengths) are associated with every triangle. Two triangles are called congruent if one can be superimposed exactly on the other. Essentially, triangles that are congruent are the "same". There are three rules for congruency : SAS, ASA, and SSS.
The notation SAS means two sides and the angle between these two sides have
been specified; the remaining side and two angles are determined. The notation
ASA (or equivalently AAS, since if any two angles in a triangle are known, the
third can be determined from180° minus the sum of the other two) means two angles
and the side between them have been specified. Of course, the sum of the two
specified angles must be less than 180°. Finally, SSS means all three sides have
been specified. Since the shortest distance between any two vertices is the side
joining these points, the longest of the three sides of a triangle must be shorter
than the sum of the other two sides for the triangle to exist.
Note: AAA, while it determines a triangle's shape, does not determine its
size and is therefore not a rule of congruence.
A triangle is called isosceles if two sides are equal. Consider the isosceles triangle ACB with AC = CB. From C construct the segment CD to D the mid point (i.e., AD = DB) of AB. Now by SSS triangle ADC is congruent to triangle BDC. Thus, ÐADC = ÐBDC, and since these two angles sum to180°, CD is perpendicular (makes a right angle) to AB. This is indicated in the diagram by the "little" box at D. Also ÐCAB = ÐCBA; in words, the angles opposite to the equal sides are also equal.
Consider the following isosceles triangle with Ð1 unspecified. Since the angles opposite the equal sides must be equal, Ð1 = 180°- 2(39°) = 102°.
A triangle is called equilateral if all three sides are equal. Of course, equilateral triangles are also isosceles, so AB = BC and the opposite angles ÐBAC and ÐBCA are equal. But AC = BC, so ÐABC and ÐBAC are equal. Thus all three angles are equal and since they sum to 180°, we have that for an equilateral triangle all of the internal angles equal 60°.
A right triangle is a triangle with a 90° interior angle. The large side opposite the 90° angle is called the hypotenuse while the remaining two sides are called legs. In the diagram shown, the hypotenuse is c and the legs are a and b. The two acute (less than 90°) angles are Ð1 and Ð2 with Ð1 opposite to the side of length a and Ð2 opposite to the side of length b. The sum of the three interior angles is Ð1 + Ð2 + 90° = 180°, so we have the result that the two acute angles in a right triangle are complementary, i.e., Ð1 + Ð2 = 90°.
Construct a right triangle with legs a and b and hypotenuse c and Ð1 opposite to the side of length a and Ð2 opposite to the side of length b. Consider the two ways of constructing a square of side a + b shown below.
In the figure on the left we have Ð1 + ÐTQR + Ð2 = 180°, but since Ð1 + Ð2 = 90°, we know that ÐTQR = 90°. A similar argument shows that ÐQTS = ÐQRS = ÐRST = ÐTQR= 90°. So the figure in the center on the left is the square of side c, i.e., the square of the hypotenuse. Thus, the area of the square of side a + b is the square of c plus the area of the four right triangles.
From the figure on the right we see that the area of the square of side a + b is the square of a plus the square of b plus the area of the same four right triangles. Since the area of the two figures must be the same we conclude that the square of the hypotenuse is the sum of the squares of the legs, or in symbols c2 = a2 + b2.
This result is called the Pythagorean Theorem and is probably the most famous and useful result in geometry! The result can be stated a number of different ways. To calculate the hypotenuse knowing the lengths of the legs, we take the square root and get
To calculate a leg, say a, knowing the hypotenuse and the other leg, b, the formula is rearranged to
or 
.
Remember the square root symbol is also a grouping symbol. Parentheses need to be used!
To find the missing hypotenuse of the following right triangle, we compute as follows:
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Note: The implied parenthesis inside the square root has been made explicit as required to get the correct answer on the calculator. |
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The units work out to be linear units as required for a length since
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As a second example consider finding the unknown quantities in the following right triangle. The missing dimension x is a leg and can be calculated from
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| The missing angle is complementary to 41°24'35", so Ð1 = 90° - 41°24'35" = 48°35'25". |
The last topic we consider is the perimeters and areas of figures in a plane. The perimeter is the total linear distance around the boundary of a polygon. For example, the perimeter of the pentagon shown below is computed as P = 3 in + 6 in + 5 in + 3 in + 5 in = 22 in.
Consider a rectangle of width W and length L. The perimeter is computed as
from |
P = W + W + L + L = 2W + 2L |
| The area of a rectangle, which measures the amount of "two dimensional space" inside the rectangle is given by A = W·L. This in fact is the definition of area. The area of all figures is based on this formula. The trick is to be clever enough to rearrange the figure into a rectangle with the same area. |
Note: Perimeter always has units of length, while area always has units of length2. For the following rectangle, we compute P as P = 2(1.85 cm) + 2(2.20 cm) = 8.10 cm. The area is given by A = 1.85 cm × 2.20 cm = 4.07 cm2.
A parallelogram is a four sided polygon (or quadrilateral) with opposite sides parallel. All rectangles are parallelograms, but in a generic parallelogram the angle between adjacent sides is not necessarily 90°. Consider the parallelogram shown below with base b and perpendicular distance h between the top and bottom sides. Imagine cutting off a right triangle from the left end and moving it to the right end. Since the left and right sides are parallel, this right triangle fits perfectly to make a rectangle of dimensions b and h. So we have for a parallelogram that A = b×h.
Next consider a generic triangle triangle with bottom side (base) of length b and perpendicular distance (height) h from the base to the top vertex.
Imagine making an exact copy of this triangle and joining it to the original triangle as shown. The result is a parallelogram of base b and height h. Since the area of the original triangle is half of the area of this parallelogram, we arrive at the result that the area of a triangle is given by
A more detailed argument shows that for a triangle with sides a, b, and c the area can be calculated from Heron's formula:
with the semi-perimeter S given by the formula:
As an application consider calculating the perimeter and area of the following triangle. The perimeter is just the sum of the lengths of the three sides.
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P = 15.6 in + 15.6 in + 16.5 in = 47.7 in |
| To calculate the area, we could calculate the height by dropping a perpendicular from the top vertex. Since the triangle is isosceles, this bisects the 16.5 in base. Using the Pythagorean Theorem we compute h as follows: | |
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|
| Then the area is calculated as half the base times the height. | |
| A = 0.5 × 16.5 in × 13.2 in = 109 in2 |
Another way to calculate the area is to use Heron's formula.
A quadrilateral with two opposite sides parallel is called a trapezoid. Suppose that the two parallel faces have lengths a and b and are separated by a perpendicular distance h.
Imagine making an exact copy of this trapezoid and joining it to the original trapezoid as shown. The result is a parallelogram of base a + b and height h. Since the area of the original trapezoid is half of the area of this parallelogram, we arrive at the result that the area of a trapezoid is given by the following formula:
As an application consider calculating the perimeter and area of the following trapezoid.
The first step is to calculate the length x. Using the Pythagorean Theorem the length of the base of the right triangles that form the sides of the trapezoid is computed as follows :
Then x = 8 cm + 2 x 5.29 cm = 18.58 cm. The perimeter is just the sum of the lengths of the four sides. P = 18.6 cm + 8 cm + 8 cm + 8 cm = 42.6 cm.
To calculate the area, we could add the area of the two right triangles that form the sides of the trapezoid to the area of the 8 cm by 6 cm central rectangle.
We get the same result by using the formula for the area of a trapezoid .
A circle is formed by generating all points in a plane which are a fixed distance called the radius from a center point here labeled as C. A line segment with end points on the circle that passes through the center is called a diameter. D and r symbolize the lengths of the diameter and radius, respectively.
Since AC = CB = r and D = AB = AC + CB = 2r, We have the following formulas :
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All circles are similar. By this we mean that all circles have the same shape. If you've seen one circle, you've seen them all! Looked at another way, all circles are "scale models" of each other, and like in any scale model, the lengths of corresponding parts are in the same ratio. The distance around the boundary of a circle is called its circumference which is like the perimeter of a polygon. Imagine that we have two circles. The first labeled as 1 has radius r1, diameter D1, and circumference C1. The second labeled as 2 has radius r2, diameter D2, and circumference C2. Since all circles are scale models of each, the ratio of circumference to diameter is the same for both circles. The value of this ratio is symbolized by the Greek letter lower case pi.
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| Rearranging this formula gives the results | |
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| or more simply the relationship that circumference is pi times the diameter or twice pi times the radius. | |
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The existence and value of pi strike many people as mysterious. However, a simple argument illustrated below shows that pi is slightly larger than 3. In fact the decimal approximation to pi is known to over a billion digits! There is a pi button on your calculator (eighth row, third column on the Casio fx-300SA; third row, first column on the TI-30Xa). Pressing it gives p » 3.141592654.
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Form six equilateral triangles each of side 1 (the units of length don't matter). Since all of the internal angles equal to 60°, the six equilateral triangles can be joined next to each other with a common vertex as shown. The union of the six equilateral triangles is a regular hexagon with a perimeter of 6. Making the common vertex the center of a circle of radius 1, we see that the circumference of this circle is slightly bigger than the perimeter of the hexagon. |
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The determination of the formula for the area of a circle was done by Archimedes over 2200 years ago. To get some understanding of it, observe the following diagram. A circle is sliced up into an even number of thin slices. The slices "almost" look like triangles. The slices are then rearranged, half pointing up, the rest pointing down. Then they are pushed against each other until they just touch. The resulting figure resembles a rectangle, except the top and bottom have little bumps, but by making the number of slices bigger and bigger (this technique is called the method of "exhaustion"), the bumps get smaller and smaller until we do "get" a rectangle. The height of the rectangle is r and the base is half of the circumference (since half of the slices faced up and the rest down). Thus we get the following "famous" formula for the area of a circle.
As an application the circumference and area of a circle of diameter 2.500 in are computed by the following calculations :
| C = p × 2.500 in = 7.854 in |
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Ð1 = 85°19'56"
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Line L1 is parallel to line L2 and
ÐCAB = 76°59' and
ÐCBA = 46°29' and
ÐCEG = 61°29'.
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BE is parallel to AD and
ÐACB = 64°17' and
ÐBAC = 66°35'.
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Find the measure of the missing angles.
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For each figure below calculate the distance around (perimeter) P.
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For each figure below calculate both the area, A, and the distance around
(perimeter or circumference), P or C.
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For each figure below calculate the requested missing information.
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How many square feet of flooring does the following room have?
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The cross section of a shed is shown below. Find the height h above the
ground and the total area of the building's cross-section, A.
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A 2.25 cm diameter hole is drilled in a 4.5 cm diameter circle. What area is left after the hole has been drilled?
area = ________________
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© 2002 by Al Lehnen; HTML-ized by Kevin Mirus (kmirus@madison.tec.wi.us or kjmirus@execpc.com). |
| This document was last modified Wednesday, August 21, 2002, 4:10 PM. |
