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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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Solution. We begin with a guess. Suppose we guess the width of the garden to be 25 m. The length will be 45 m and the area 45 x 25 = 1125 m2. The area is too big, so we reduce the width and try again. A convenient way of handling this problem is to put the numbers in a table so we have an orderly record of our progress.
Width, in m (= width + 20 m), in m Area, in m2 Comment
25 45 1125 Too big, reduce width
15 35 525 Too small, increase width
20 40 800 Too small, increase width
22 42 924 Too big, reduce width by a small amount
21.7 41.7 904.89 Too big, reduce width by a very small amount
21.6 41.6 898.56 Too small, increase width by an extremely
small amount
You may consider the answer of 21.6 m to be accurate enough, or you may wish to try 21.65 m, say, and try further. The actual width, correct to four decimal places, is 21.6228 m.
References: Area, Rectangle.
A triangle is a polygon with three sides. See the entry Polygon for the angle sum of a triangle. The following triangles are studied under the respective entries: isosceles triangle, equilateral triangle, and scalene triangle.
Triangles will always tessellate. Suppose we draw any triangle. The one chosen in figure a is a scalene triangle and is shaded with an arrow drawn on it facing downward so the pattern of the tessellation is clear. To achieve the pattern the shaded tile is rotated through half a turn to obtain the white tile. In this way the shaded tile and the white tile placed together form a parallelogram, which is easy to tessellate. A tiling pattern is drawn in the figure.
The area of a triangle is obtained using the length of its base and its height.
Area = \ base x height
In the tessellation pattern in figure a, the area of the triangle is equal to half the area of the parallelogram. The area of a parallelogram is equal to base x height.
Figure b has different kinds of triangles that may help you to identify the base and the perpendicular height so that the area can be calculated. In fact any side of the triangle can be taken as the base, but choose one that matches a known perpendicular height.
Example. Find the area of the triangle drawn in figure c on the grid of squares of side 1 cm.
Solution. The triangle is scalene. Extensive calculations would have to be done to calculate the lengths of the base and the perpendicular height to find the area of the triangle. An alternative method is to draw a rectangle to enclose the triangle; in this example the rectangle happens to be a square.
Area of square = 9 cm2 Area of A = 1.5 cm2
Area of 15 = 1 cm2 Area of C = 3 cm2
Area = length x width
Area = half the base (which is 3) x height (which is 1)
Similar to finding area of A
Similar to finding area of A
The areas in figure c are connected by the following equation:
Area of square = Shaded triangle + A + B + C
9 = Shaded triangle + 1.5 +1 + 3
9 = Shaded triangle + 5.5
Area of shaded triangle = 3.5 cm2 Solving the equation
The idea used to solve the above example is adapted to prove Pythagoras’ theorem,
as follows.
Figure d shows a larger square with sides of length (a + b) units, a small one with sides of length c units, and four congruent right-angled triangles. First write
Area of larger square = (a + b)(a + b) = a2 + 2 ab + b2 Area of smaller square = c2 Area of each right-angled triangle
Area = length x width
Expanding the brackets
Area = length x width
Area of triangle = half-base x height
Sum of areas of 4 triangles = 2ab The areas in the figure are connected by the following equation:
Area of larger square = Area of smaller square + sum of areas of 4 triangles a2 + 2 ab + b2 = c2 + 2 ab
+ b2
Subtracting 2ab from both sides of the equation
This is the formula for Pythagoras’ theorem in each of the four right-angled triangles.
For other methods of calculating the area of a triangle check see the entries Heron’s Formula and Cosine Rule.
References: Angle Sum of a Triangle, Cosine Rule, Heron’s Formula, Linear Equation, Polygon, Pythagoras’ Theorem, Tessellations.
References: Difference Tables, Pascal’s Triangle.
References: Amplitude, Circular Functions, Cycle, Frequency.
Trigonometry, like Pythagoras’ theorem, has a wide variety of uses and appears in many entries in this book. In this entry a brief introduction will be given; then trigonometry will be applied to solving right-angled triangles. It is essential to have a scientific calculator available, set in the degree mode. For other aspects of trigonometry see the following entries Angle between a Line and a Plane, Angle between Two Planes, Circular Functions, Cosine Rule, Sine Rule, Trigonometric Graphs.
A triangle has three angles and three sides, and the trigonometry we study in this entry uses two sides and one angle at a time. The sides of a right-angled triangle are given names in the following way (see figure a). Suppose the angle in the triangle we are focusing on is called theta, which is a letter of the Greek alphabet and is written as 9. The side of the triangle that is opposite to 6 is called the opposite side, which is written as O. The side that is opposite to the right angle is called the hypotenuse, which is written as H. This is the longest side in the triangle. The remaining side, which is adjacent to angle 6, is called the adjacent side, and is written as A. In the two right-angled triangles in figure a the names of the sides are marked on the triangles. Depending on where angle 6 is, the names of the sides O and A are interchanged.
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