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The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
Download (direct link): theatozofmath2002.pdf
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In all transformations the image shape of F is denoted by F', using the appropriate letter.
Suppose the translation that maps a point A onto A! is defined by the vector (^2) and A! —r A" under a further translation (2). The vector that maps A onto A" is found by adding together the two vectors. Therefore, A A” under the translation (q). The translation that maps the image A! back to the object A is (“4), and is found by changing the signs of the two numbers in the vector that maps A to A1.
Properties of Translations
♦ There are no invariant points.
♦ The object shape and the image shape are congruent shapes. We call this an isometry transformation.
♦ If the lines are parallel in the object, they will also be parallel in the image. We call this an affine transformation, which is one that preserves parallelism.
♦ Translation is a direct transformation.
References: Cartesian Coordinates, Composite Transformations, Congruent Figures, Escher, Indirect Transformation, Invariant Points, Transformation Geometry, Vector.
Reference: Changing the Subject of a Formula.
Reference: Alternate Angle.
A trapezium is a quadrilateral, which means a four-sided polygon, with one pair of opposite sides parallel. In figure a, the angles a and b are called cointerior angles and their sum is 180°, which means that a + b = 180°. The two triangles ACD and BCD are equal in area. The trapezium has rotational symmetry of order one. The trapezium on the right is an isosceles trapezium, which has two sides equal in length, two pairs of congruent angles, and one axis of symmetry.
To find the area of a trapezium we need to know the lengths of each of the two parallel sides and their perpendicular distance apart. The formula for the area of the shaded trapezium in figure b is
(a + b)
In words, the area is expressed as the mean length of the two parallel sides multiplied by the perpendicular height.
Example. The trapezium in figure c is drawn on a square grid, and the side of each square is 1 cm. Find the area and perimeter of the trapezium.
4 4
□ □
Solution. From figure c we can see that a = 3, b = 6, and h = 4, all in centimeters. Write
(3 + 6) (a + b)
Area = —-— x 4 Substituting into the formula A = —-—h
= 4.5 x 4 = 18
The area of the trapezium is 18 cm2.
The trapezium is divided into two right-angled triangles and a rectangle, as shown in figure c. Write
Perimeter = ?> + y + 6 + x Adding together the four sides of the trapezium.
x is found in the first right-angled triangle and y in the second right-angled triangle:
*2 = 12+42> =22+42 Pythagoras’ theorem
JC2 = 17, y1 = 20
JC = Vl7, lea II Taking square roots.
Perimeter = 3 + V20 + 6 + VT7
The perimeter of the trapezium is 17.60 cm, to 2 dp.
See the entry Prism for the volume of a swimming pool with a trapezium as the cross section.
References: Area, Base, Congruent Figures, Hexagon, Mean, Parallel, Perimeter, Perpendicular Lines, Polygon, Pythagoras’ Theorem, Quadrilateral, Rectangle, Regular Polygon, Symmetry, Triangle.
These graphs are also known as distance-time graphs, in which time is the horizontal axis and distance is the vertical axis. Travel graphs usually depict a journey. The first example explains how to set up a distance-time graph to solve a problem.
Example 1. David and his son William decide to have a race. David can run at a constant speed of 5 meters per second (m/s) and William can run at a constant speed of 2 m/s. David gives his son a 10 meters head start. After how many seconds will David catch up with his son?
Solution. We draw a distance-time graph for each runner using a suitable table of values. The distances are measured from the place where David starts the race; therefore we must add 10 meters to the distances run by William so the graph of each runner starts at t = 0.
Time in seconds (f) 0 1 2 3 4
David’s distance in meters 0 5 10 15 20
William’s distance in meters 10 12 14 16 18
The straight-line graphs of the two athletes are both drawn on the same axes, using the following sets of coordinates (see figure a):
♦ David: (0, 0), (1, 5), (2, 10), (3, 15), (4, 20)
♦ William: (0, 10), (1, 12), (2, 14), (3, 16), (4, 18)
David catches up with William at the point on the graph where the two lines intersect. This time is estimated to be about 3.3 seconds.
There are some important points to note from this problem.
♦ A distance-time graph of constant speed is a straight line.
♦ The greater the speed, the steeper is the line.
♦ The slope (gradient) of a distance-time graph gives the speed. For example, suppose we calculate the gradient of William’s line graph:
2 meters
1 second
2 m/s
The formula for gradient
In the right-angled triangle rise = 2 meters,, run = 1 second
This is William’s speed
Example 2. Harry loves cycling and goes for a ride. For the first 3 hours he cycles at a speed of 20 km/h. He stops 1 h for lunch, and then continues at a more leisurely pace of 15 km/h for 2 h. Harry stops for | h to take photographs of a waterfall. Realizing how late it is, he turns for home, and arrives there 3 h later. Draw a distance-time graph of his trip. At what constant speed did he return home? What was his average speed for the whole trip, assuming each of his speeds is constant?
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