Geometrical construction
IWILL, in this week’s lesson, present a review of aspects of geometry, specifically, geometrical construction and application of theorems.
The following is extracted from the syllabus:
Draw and measure angles and line segments accurately, using appropriate instruments.
Construct lines, angles and polygons, using appropriate instruments. a. Parallel and perpendicular lines. b. Bisecting line segments and angles. c. Constructing a line perpendicular to another line, L, from a point that is not on the line, L. Triangles, quadrilaterals, regular and irregular polygons.
d. Angles include 300, 450, 600, 900, 1200 and their combinations.
You are invited to use the above to guide your review and study. Please note the following:
1. You should ensure that you are comfortable with the use of the protractor to measure angles, the pair of compasses to draw arcs representing specific distances, and the ruler to determine distances.
2. It is vital that you learn the methods of constructing the five angles listed above using the pair of compasses, pencil and ruler only.
3. All methods are linked to the fact that the three internal angles of an equilateral triangle, each 600. Bisecting this angle will result in the construction of a 300 angle.
4. Being competent in bisecting line segments and angles, also in constructing parallel and perpendicular lines, is critical to completing the exam-type questions.
5. Repeated practice of each activity must be done to build competence.
Other points to note:
1. Construction arcs and lines must be shown to indicate the methods used.
2. The figure representing the answer should be bold.
3. The measurement of lines and angles are accepted within a specific degree of accuracy and so every effort must be made to be as accurate as possible. The use of the divider will help.
4. The construction of parallel and perpendicular lines is usually required to construct a rectangle or parallelogram.
5. If the question restricts you to only use a ruler, a pencil and a pair of compasses, then a protractor must NOT be used.
In attempting an example, you are encouraged to sketch the figure and then plan the order of the steps which are required.
EXAMPLE 1
(i) Using a ruler, a pencil and a pair of compasses, construct:
a) A triangle ABC in which AB = 8cm, AC = 6cm and angle A = 60o.
b) The line segment CM which is perpendicular to AB and meets AB at M.
(ii) Measure and state the size of angle BCM.
SOLUTION
The recommended order is: draw AB = 8cm, angle A = 600 and AC = 6cm. Finally, join BC and then construct the perpendicular from C to AB.
NB: The recommended order is: draw PQ = 6cm, angles P and Q = 900 and PR & QS = 6cm. Join RS and, finally, construct PQY = 1200, then extend RS to meet QY at Y.
Angle BCM = 440.
EXAMPLE 2
Using a ruler, a pencil and a pair of compasses:
(i) Construct accurately, the square PQRS, with side 6cm.
(ii) Construct an extension of your drawing in (i), the trapezium RPQY so that angle PQY = 120o.
(iii) Hence, measure and state the length of QY. QY = 6.8cm
Please attempt the following on your own.
PRACTICE EXAMPLE
(i) Using a ruler, a pencil and a pair of compasses:
Construct the perpendicular bisector of a line segment PQ, of length 5cm. Construct angle DEF = 450. Construct triangle ABC with AB = 5cm, angle ABC = 600 and angle BAC = 900.
(ii) Measure and state:
The length of AC.
The size of angle ACB.
While the construction lines above are not clearly visible, be sure that yours are as you complete the examples.
We will now proceed to review application of theorems.
A theorem is a geometrical statement which is accepted as fact in solving a problem. For example, ‘the sum of the interior angles of a triangle is 1800’ or ‘ vertically opposite angles are equal’. They are all based on the definition that the angle at a point is 3600 with the proof of each based on logical deductions. The fallacy is widely expressed that you only need to work math problems. You are, therefore, encouraged to pursue the following:
Carefully review the aspects of geometry relating to finding unknown angles in order to identify and list the various theorems.
Emphasis should be placed on intersecting lines, polygons (especially triangles and quadrilaterals) and circles, including the tangents, chords and radius.
List and study all the relevant theorems and ensure you understand their applications. For example, ‘corresponding angles are equal’. You must be familiar with what are corresponding angles.
You should then practise various examples in order to build competencies in their applications.
I will now proceed to provide examples as follows:
EXAMPLE 1
(a) The diagram below shows two straight lines, AB and AC, intersecting a pair of parallel lines, MN and XY.
Determine, giving reasons, the value of: (i) x
(ii) y
(iii) w.
SOLUTION
(i) x0 + 1280 = 1800 (Adjacent angles on a straight line are supplementary.) x0 = 1800 - 1280 x0 = 520.
(ii) x0 + y0 = 1800 (Co-interior angles formed from parallel lines are supplementary.)
520 + y0 = 1800 y0 = 1800 - 520 = 1280 y0 = 1280.
ALTERNATIVE
Y0 = 1280. (Corresponding angles formed from parallel lines MN and XY are equal.)
(iii) w0 + (180 - y0) + ( 180 - 100) = 1800 (Interior angles of a triangle are supplementary.)
w0 + (180 - 128) + ( 180 - 100) = 180 (Interior angles of triangle are supplementary.) w0 + 52 + 80 = 180 w0 + 132 = 180 w0 = 180 - 132 = 48 w0 = 480
EXAMPLE 2
ABCDE is a pentagon inscribed in a circle with centre O. The diameter AD is produced to F. Angle CDF = 135o and angle BAD = 72o.
Determine, giving reasons for your answers, the magnitude of angles,
i) CDA ii) BCD iii) AED.
SOLUTION
i) CDA + CDF = 180 (Adjacent angles on a
straight line), substituting CDA + 135 = 180 CDA = 180 - 135 = 45 CDA = 450.
ii) ACD = 900 (Angle in the semicircle at the circumference is a right angle.)
CDA + CAD + ACD = 180 (Interior angles of a triangle are supplementary.) Substituting, CAD + 45 + 90 = 180
CAD + 135 = 180
CAD = 180 - 135 = 45
CAD = 450
BAC + CAD = 72 substituting,
BAC + 45 = 72
BAC = 72 - 45 = 270
ABC + CDA = 180 (Opposite angles of a cyclic quadrilateral are supplementary.)
ABC + 45 = 180
ABC = 180 - 45 = 135
In triangle ABC,
BAC + ABC + BCA = 180 (Interior angles of a triangle are supplementary.) Substituting, BCA + 135 + 27 = 180
BCA = 180 - 162.
BCA = 180
Since BCD = BCA + ACD and ACD = 90 BCD = 18 + 90 = 108
BCD = 1080
iii) AED = 900 (Angle in the semicircle at the circumference is a right angle.)
EXAMPLE 6
O is the centre of the circle ABCDE and TEF is a tangent to the circle at E.
Given that DEF = 30o, calculate, giving reasons to support your answer, the sizes of the angle:
i) ACD ii) EAD iii) EOD.
SOLUTION
i) ACD = 900 (Angle in the semicircle at the circumference is a right angle.)
ii) Since DEF = 300
EDA = DEF = 30 (Angle between chord and tangent is equal to the angle in the alternate segment.)
EDA = 300 iii) EOD = 60 (Angle at the centre is twice the angle at the circumference subtended by the same chord.) EOD = 600
It is recommended that you establish a group to develop a comprehensive list of theorems. After ensuring that you are comfortable with their use, you should commit them to memory. Please continue to practise other examples.