Chapter 2



Chapter 6 Geometry

This lecture note is based on the text book Mathematical Literacy in a numerate Society – by Matthew A. Isom and Jay Abramson

We will discuss following concepts in this chapter.

Two Dimensional Geometry: Straight lines (parallel and perpendicular), Rays, Angles (acute, obtuse, straight, vertical, complementary, supplementary), Similar and congruent shapes, Polygons (Triangles (scalene, equilateral, isosceles, right, obtuse angled, acute angled), Quadrilateral (parallelogram, rectangles, squares, rhombus, trapezoids), Pentagons (5 sided polygons), Hexagons (6 sided polygons), Heptagons (7 sided polygons), Octagons (8 sided polygons), Nonagons (9 sided polygons), Decagons (10 sided polygons)), Regular polygons, Circle (radius, diameter, circumference), Area, Perimeter.

Geometry of Three Dimensions: Volume and surface area of box, cube, Polyhedron (prism, pyramid, cone), cylinder, sphere.

Trigonometry: soh, cah, toa from a right triangle.

Triangle: Let a, b, c be the measures of three sides and h be the height from a vertex to the side b of a triangle. Then

Perimeter = [pic]units, and semi-perimeter [pic] units

Area = [pic]square units. Area (in terms of sides) =[pic]square units, this formula is known as Herron’s formula.

Quadrilateral: Let a, b, c, d be the sides of a quadrilateral, then perimeter =[pic]

Area of parallelogram = base[pic] height

Area of rectangle = length[pic] width

Circle: Circumference = [pic], we have also [pic]

Oval (Ellipse): Area = [pic], where a and b are the lengths of the semi-major and semi-minor axes.

Volume and surface area:

Box (rectangular solid): Let a, b, c be the length, width, and height of a box. The surface area of the box is equal to [pic] ands the volume is [pic]

Volume of a prism or cylinder[pic], where A is the area of the base and h is the height.

Surface area of a closed cylinder [pic]

Volume of a pyramid or a cone[pic], where A is the area of the base and h is the height.

Volume of a sphere[pic], where A is the area of the base and h is the height. Surface area of sphere [pic]

Trigonometry: soh, cah toa stands for the following results

The right triangle has hypotenuse h, opposite o and adjacent a

Then [pic] h o

[pic] and [pic] x a

Important relations in right triangles: Angles are in degrees.

2 60 [pic] 45

1 1

30 45

[pic] 1

Look at all examples and practice from the text book exercise problems: Page # 216-217: #s 11-27. Page # 230: #s 11-20

Conjecture: The CSR for a regular hexagon is a smaller regular hexagon that has been rotated 30 degrees with respect to the original hexagon. The area of the CSR is one third of the area of the original hexagon, and the length of a side is approximately .577 the length of the side of the original. (L. Miller)

[pic]

Figure 3: The CSR for a Regular Hexagon

The ease with which the conjecture was made is due in large part to the construction and measurement tools available in GSP. The fact that the conjecture contains an approximation is to be expected, as the conjecture was based on GSP measurements. In proving this conjecture, the students discovered that 0.577 is an approximation for[pic].

Note: The students whose work is cited in this article have all given permission for their names to be used.

At this point some students decided to investigate other regular polygons. In addition, proofs of parts of this conjecture were made by several different students using a variety of arguments. They all assumed the CSR of a regular polygon is a hexagon, a point that was only informally argued. Given the time limits of the class, it seemed reasonable to assume this result. The results at this point in the project are summarized in the following conjectures and theorems. In proving parts of his original conjecture, L. Miller's understanding of the relationships deepened, as exemplified by his recognition of the actual ratio of the side of a regular hexagon to the side of its CSR.

Theorem: Assuming the CSR region for a regular polygon is a hexagon, it is an equiangular hexagon. (F. Rocha)

Theorem: The ratio of the area of a regular hexagon and its CSR is 3:1 and the ratio of the side of the regular hexagon to the side of the CSR is [pic]. (L. Miller, L. Cooke, R. Rocha)

Conjecture: The CSR for a regular 2n-gon is a regular 2n-gon, while the CSR for a regular (2n  + 1)-gon is a regular (4n  + 2)-gon. (M. Castro, L. Cooke)

In the course of the next few weeks, several students actively worked on the question of the CSR for equilateral polygons. F. Rocha proved that the CSR for a regular hexagon is not only equiangular, but is in fact regular. L. Cooke extended his analysis to regular 2n-gons and (2n  + 1)-gons. This work can be summarized by the following conjectures. L. Cooke proved the last three conjectures only for a number of specific values of n , but his methods would lend themselves to a proof of the general case stated in each conjecture.

Conjecture: The sides of the CSR for a 4n-gon are parallel to the sides of the 4n -gon.

Conjecture: Let [pic] be the central angle (the angle formed by joining two adjacent vertices to the center of the regular n-gon) of a regular 2n-gon, n  > 2. Then the ratio of the length of the sides of the 2n -gon to its CSR is [pic]. The ratio of the areas is [pic].

Conjecture: The length of the apothem (the segment measuring the distance from the center to a side) of the CSR for any regular n-gon is half the length of the side of the n -gon.

Conjecture: For any (2n+1)-gon, n  > 1, and its CSR, the ratio of the lengths of the sides is [pic]and the ratio of the areas is [pic].

1. Find the supplementary angle of [pic].

2. For a square whose diagonal is [pic], find perimeter and the area.

3. ladder that is 9.5 feet long casts a shadow that is 19 feet long at the same instant that a tree casts a shadow that is 100 feet long. How tall is the tree?

(Solution: use similar triangle notion. [pic] then x = 50, the tree is 50 feet tall.)

4. A smaller right circular cylinder is cut out of a larger solid that is also a right circular cylinder, but 2 inches about its base, as shown below. Find the volume of the remaining solid.

[pic]

[pic]

[pic] [pic]

[pic]

[pic]

5. The oval is inside of a square that is not drawn to scale.

Find the area of the shaded region.

6. Draw the figure of a propane gas tank that consists of a cylinder with a hemisphere at each end. The overall length of the tank is 10 feet and the diameter of the cylinder is 4 feet. Find also the volume of the tank.

7. A water storage tank is an upside down cone. If the diameter of the circular top is 12 feet and the length of the sloping side is 10 feet, how much water will the tank hold? Graph the tank showing length of the diameter and the sloping side.

8. Find the area of the triangle. The angles are in degrees.

60

30

5[pic]

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