Question 1
In a circle, the angle subtended by an arc at the centre is . Find the angle subtended by the same arc at the circumference.
Strand 3 · Geometry Around Us
Mathematics Year 3 Learner Material, Section 3: Circle Theorems & Geometric Construction
In Year 1, you studied angles, their types and constructions, parallel and perpendicular lines, polygons and how to find the perimeter, area and volume of shapes. Skills acquired from these concepts are useful for tasks like designing floor plans, fencing land, and building structures.
In Year 2, you learnt trigonometric ratios and their inverses, applying them to solving real-life situations such as finding building heights, estimating distances, and calculating angles of elevation and depression, which are applicable in areas such as surveying, navigation and construction.
In Year 3, you will explore exciting topics like circles, circle theorems, and the construction of circles, triangles, and quadrilaterals. Circle theorems reveal important relationships between angles, lines and points in and around a circle. This will build on your earlier knowledge of constructions, polygons and trigonometry. These topics are valuable in engineering, architecture and technical drawing. These concepts help greatly in sharpening your mathematical reasoning and equipping you to apply mathematics confidently in schoolwork and everyday problem-solving.
KEY IDEAS
• • In geometry, a point, line or object is said to be equidistant from two or more objects if it is the same distance from each of them.
• • In classical constructions, use only a compass and a straight edge/ruler. This emphasises logical reasoning and geometric relationships.
• • Each step in a construction must be accurate and based on geometric principles (e.g., perpendicular bisector, angle bisector, etc.)
• • Basic constructions include constructing perpendicular bisectors, angle bisectors, perpendicular lines from a point and copying or constructing given angles and triangles.
• • Many constructions are justified using theorems (e.g., the angle in a semicircle is a right angle, or isosceles triangle properties).
• • Constructions are also based on the idea of a locus; a set of points satisfying certain conditions (e. g. all points equidistant) When you look around, what circular objects do you see? Probably objects like bicycle tyres, the rim of a tank or a round bowl, a wall clock, etc. With a classmate, carefully observe the images in Figure 3.1 below and identify the part(s) of each image that is circular.
Bowl Wall clock Roundabout Iris in a human eye Orange Bicycle
Figure 3.1: Real-life objects that have circular shapes.
From your observation and discussion, describe a circle in your own words.
At Junior High School, you learnt about circles and the parts of a circle. In this section, we will explore additional interesting parts of the circle and the relationships among them.
Before we delve into this lesson, write down the “parts of a circle” that you were taught in junior high school. Parts like: the circumference of a circle, radius, diameter, chord and many others. Discuss what you have remembered about these parts of a circle with a classmate. This will help you to better understand the new lesson.
Circles and Circle Theorems
A circle: A circle is a plane (two-dimensional) figure bounded by a circumference such that all the points on the circumference are equidistant from a fixed point within it.
In other words, a circle is a two-dimensional figure (plane figure) formed by a set of points that are at a constant or at a fixed distance (radius) from a fixed point (centre) on the plane.
Figure 3.2. is an example of a circle in which A, B, D and E are points on the circumference. Point O is at the centre of the circle.
Figure 3.2: A diagram illustrating parts of a circle.
Parts of a circle
1. Centre: The centre of the circle is the fixed point from which all points on the boundary of the circle (circumference) are equidistant. In Figure 3.2 and
Figure 3.3, the centre is the point labelled O in the circle.
With this description, you can identify the centre of the wall clock as well.
Figure 3.3: Shows the centre of a circle
2. Circumference: The distance around the circle is called the circumference.
It is the area around the circle or the perimeter of the circle as shown in
Figure 3.4.
Figure 3.4: Circumference of a circle With a classmate, identify the part of each object in Figure 3.4. that represents the circumference.
3. Arc: A part of the circumference is called an arc, as shown in Figure 3.5.
Major arc: A major arc is greater than half the circumference.
Minor arc: A minor arc is less than half the circumference.
Figure 3.5: Illustration of an arc of a circle
4. Radius: A radius is the shortest distance between the centre of a circle and any point on the circumference of the circle.
Any line drawn from the centre of a circle to any part of the circumference is a radius. The plural form of radius is radii.
In Figure 3.5, |OD|,|OA| and |OE| are radii. The radius (r) of a circle can be used to find the area, circumference and volumes of circular shapes.
|OD|=|OA|=|OE|= Radius (r)
Figure 3.6: illustration of a radius
5. Diameter: The diameter of a circle is any straight-line segment that passes through the centre of the circle and whose endpoints lie on the circumference of the circle.
The diameter is special and the longest chord of the circle. The diameter of a circle is twice the length of the radius of that circle.
In Figure 3.7, it can be observed that |AE| is the diameter and it is formed by two radii Since |OA|= |OE| = radius (r), then, diameter =2 × radius = 2r.
Figure 3.7: illustrations of the diameter of a circle
6. Sector: The region bounded by two radii and an arc. Point O is the centre of the circle from which the sector was obtained, and lines OA and OB are radii.
Figure 3.8: Sector of a circle.
Major sector: A major sector has a central angle which is more than 180°.
Minor sector: A minor sector has a central angle which is less than 180°.
Therefore, considering the pizza in the figure above, the portion that has been taken out represents the minor sector and the remaining portion represents the major sector.
7. Semi-circle: Semi circle is exactly half of a circle. The central angle of a semicircle is 180°. The diameter divides the circle into two equal parts (halves/ semi-circles).
Figure 3.9: An illustration of a semi- circle
8. Quadrant: A quarter of a circle created by two radii that are perpendicular to each other is called quadrant. From Figure 3.10, |AD| and |AB| and are radii and are perpendicular to each other.
Figure 3.10: Quadrant of a circle
9. Chord: A straight line connecting two points on the circumference of a circle is called a chord. In the figure below A, B, P and Q are points on the circumference. Lines |AB| and |PQ| are chords.
Figure 3.11: Illustration of a chord Have you observed from the Figure 3.11 that |PQ| is also a diameter? The diameter is also a chord and it is the longest chord in a circle.
10. Segment: A segment of a circle is a region bounded by a chord and the corresponding arc. The space shaded is called a minor segment. This is the region bounded by a chord |AB| and an arc AB.
Figure 3.12: A segment of a circle
• Major segment: a segment where the arc is greater than half the circumference.
• Minor segment: a segment where the arc is less than half the circumference.
From your observation, discuss with a classmate which part of the cake in
Figure 3.13 represents the major segment and minor segment.
11. A Secant: A secant to a circle is a line that crosses or passes through the circle at exactly two distinct points. The line QP is called the secant.
Figure 3.13: Illustrations of a secant line
12. Tangent: A tangent to a circle is a straight line that intersects the circle at exactly one point, i.e. a tangent touches the circumference of the circle at just one/ a single point.
The point of contact of the circle and the tangent line is called the point of tangency.
Figure 3.14: Tangent to a circle Hopefully this has revised the parts of a circle for you and you have also seen the relationship between them.
In Year 1, you learnt how to construct and measure angles, so revise this quickly by discussing the following questions with a classmate.
1. What are the instruments used in constructing angles?
2. How are angles constructed? Try at least two examples (60° and 90°)
3. Which instrument(s) are used for measuring angles?
4. How are angles measured?
Construction of Circles
Circles can be constructed when the centre and the radius are known..
Materials needed: A ruler and a pair of compasses.
Construction of a circle of a given radius To construct a circle with a radius of 6cm, follow the steps below:
Steps
1. Use your ruler to measure a distance of 6cm between the compass tip and the pencil.
2. Mark a point and label it O. This point O will be the centre of your circle.
3. Place the compass pin firmly on point O and slowly rotate the compass 360° to draw a full circle.
Figure 3.15: Construction of a circle of radius 6cm
4. Use a ruler to draw at least three (3) straight lines from point O to different points on the circumference of your circle. Label each line drawn from the centre to the circle.
5. Measure the length of each of the lines (radii) with your ruler and record the measurement of each line.
6. Discuss your observations:
• Do all the lines from the centre to the circle have the same length?
• Record your answer and explain briefly.
7. To extend your learning:
• Repeat the above steps using different radii such as 3cm, 6cm, or 8cm with a classmate.
• Compare your findings and discuss what happens to the circle when the radius changes.
Constructing an infinite number of circles from the same centre Steps
1. Mark the centre point and label this point as O
2. Place the pin of the compass firmly on point O and choose a starting radius (e.g., 2cm) and draw the first circle.
3. With the compass pin still at point O, increase the radius slightly (e.g., by 1 cm) and draw the second circle around the first one.
4. Repeat the process by gradually increasing the radius by equal or chosen intervals (e.g., 1cm each time). For each new radius, keep the compass pin fixed at point O and draw a new circle.
5. Observe the pattern:
• After drawing several circles, observe how the circles become larger as the radius increases.
• Notice how all the circles share the same centre but have different sizes.
Figure 3.16: Many circles with the same centre An infinite number of circles can be drawn around a single point, as illustrated in Figure 3.16. These circles are called concentric circles.
Constructing an infinite number of circles that pass through a single point Steps
1. Mark a fixed-point P on your paper (this is the point through which all the circles will pass).
2. Choose a centre point for a circle - not at P. The centre can be anywhere else on the paper.
3. With the compass pin at the chosen point, set the radius to the point P and draw the circle through P.
4. Repeat Steps 2 and 3 by selecting a new point as the centre (not coinciding with P).
5. Construct new circles with the radius equal to the distance from the new centres to point P.
6. Draw the circles as many as you can.
As there are infinitely many points you can choose as centres (not at P) on your drawing sheet. You can construct infinitely many different circles, each passing through point P.
7. Discuss your observations with a classmate.
Note: Each circle will have a different radius and centre, but will share the same point P on their circumferences.
Figure 3.17: Some circles all passing through point P
Example 3.1
Draw three circles with centres A, B and C respectively passing through a single point D. Expected result
Figure 3.18: Three circles passing through a single point Likewise, infinitely many circles can pass through two points. How can these different circles be constructed such that they all pass through the two fixed points?
Circles passing through two fixed points Click on the link below to watch a video on how to construct infinitely many circles passing through two fixed points.
https://youtu.be/NqSjPJZ3O1E From the video you have watched, try this example in your workbook and show it to your teacher or compare your work with that of a classmate.
Example 3.2
Construct circles passing through two fixed points A and B
Figure 3.19: Circles passing through two points (A and B) Have you ever tried drawing a circle through three points that are not on the same line?
Can a circle be drawn through three points that are not on a straight line (non-collinear)?
Discuss this with your classmate and see if you can construct one.
Click on the link below to watch a video on how to draw a circle through three non- collinear points https://www.youtube.com/shorts/heHjglruyIM?feature=share From the video, you have observed that only ONE circle can pass through three points which are NOT on a straight line (non-collinear). When such a circle is created through the vertices of a triangle, it is called a circumscribed circle and its centre is called a circumcentre. The circumcentre is equidistant from all three vertices.
Construction of a circle through three points
Example 3.3
Construct a circle to intersect three points X, Y, Z which are not on the same line in a plane. Label the centre of the circle, O.
Solution
Steps
1. Draw three points (the points should not be on the same line) and label them X, Y and Z respectively.
2. With a ruler, join the points X to Y with a straight line
3. Draw a perpendicular bisector of |XY| and label the bisector d.
This is the same as constructing the locus of all points equidistant from points X and Y.
Any point on d will be equidistant from X and Y.
4. Join points X to Z with a straight line and draw a perpendicular bisector of |XZ| and label it f.
This is the same as constructing the locus of all points equidistant from points X and Z. Any point on f will be equidistant from X and Z.
5. Indicate the point of intersection of the lines as O.
Point O is equidistant from X, Y, and Z. This means that |OX|=|OY|=|OZ|
6. With your compass pin at O, draw a circle with centre at O and radius |OX| or |OY| or |OZ|.
Figure 3.20: A circle passing through three points What are your observations from the diagram drawn?
Note: All three points X, Y and Z must be equidistant from the circle’s centre, O. This means that we seek to find a point, O, such that |OX|=|OY|=|OZ| Can you recall how to copy a given angle? Try one in your sketch book and then click on the link to watch this video to help revise on how to copy a triangle https://youtu.
be/2DVI5xmIhkI
A circle theorem is a geometric rule that explains special relationships between angles, lines and points in and around a circle.
These theorems describe how angles and lengths behave when they are connected to the centre of a circle, circumference, chords, tangents and arcs. The theorems are also used in geometry to answer questions relating to circles.
Circle Theorem Statements
Statement One
The angle subtended by a chord or arc at the centre of a circle is twice the angle it subtends at the circumference of the circle.
In Figure 3.21, A, B and C are points on the circle. O is the centre of the circle. The angle subtended by the chord |AB|, at the centre is and at the circumference is y.
The circle theorem statement one means that:
x = 2y or
Figure 3.21: Cyclic triangle ABC with centre O
Example 3.4
In the figure below is a circle with centre O, ∠MON=96°, ∠MRN= 2x.Calculate the value of x
Figure 3.22: Cyclic triangle MNR with centre O
Solution
You know that the angle subtended at the centre is twice the angle subtended at the circumference: ⇒ 2∠MRN = ∠MON or ∠MRN = (∠MON) 2(2x) = 96° = ∴ x = 24° Statement Two The angle subtended by the diameter at the circumference is a right angle.
In Figure 3.23, O, A and B are points on the circumference of the circle with centre V.
|AB| represents the diameter of the circle.
Figure 3.23: Cyclic triangle OAB with centre V ∠AOB = 90°
Example 3.5
In the figure below, O is the centre of the circle and |AE| is a diameter.
Given that ∠AED = 56°. Determine the value of ∠DAE
Figure 3.24
Solution
In Figure 3.24, it can be observed that ∠DAE is an angle subtended by the diameter (Theorem 2) 88
Figure 3.24
Solution
In Figure 3.24, it can be observed that ∠AAAAAA is an angle subtended by the diameter |AAAA| (Theorem 2) Therefore, ∠AAAAAA = 90° ⇒ ∠AAAAAA + ∠AAAAAA + ∠AAAAAA = 180° (sum of interior angles of a triangle) ∠AAAAAA + 90° + 56° = 180° ∠AAAAAA + 146° = 180° ∠AAAAAA = 180° − 146° ∠AAAAAA = 34° Statement Three The angles subtended at the circumference by the same chord or arc in the same segment are equal.
In Figure 3.25 M, N, O and P are points on the circumference of the circle.
If you consider |MMMM| as the chord , then ∠MMMMMM = ∠MMMMMM since they are subtended at the circumference by the same chord |MN|.
This implies aa = bb Again, if |MMMM| is the chord , then ∠MMMMMM = ∠MMMMMM.
Statement Three
The angles subtended at the circumference by the same chord or arc in the same segment are equal.
In Figure 3.25 M, N, O and P are points on the circumference of the circle.
If you consider |MN| as the chord ,then ∠MPN = ∠MON since they are subtended at the circumference by the same chord |MN|.
This implies a = b Again, if |OP| is the chord ,then ∠PMO = ∠PNO.
Figure 3.25: A circle with points M, N, O, P on its circumference
Example 3.6
In the figure below ∠MON = 40°, ∠MPN = y.
Determine the value of y
Figure 3.26:
Solution
From the diagram, chord |MN| subtends ∠MON and ∠MPN at the circumference ⇒ ∠MPN = ∠MON = 40° ∴ y = 40° Statement Four Two equal chords subtend equal angles at the centre of the circle.
If chords |XY| = |AB|, then ∠XOY = ∠AOB
Figure 3.27: A circle of two equal chords Statement Five If the angles subtended by two chords at the centre are equal, then the two chords are equal.
Statement five is similar to statement four. One implies the other.
In Figure 3.28, point O is the centre of the circle and |XY| and |AB| are chords in the circle.
From statement four, if |XY| = |AB|, then ∠XOY = ∠AOB.
From statement five, if ∠XOY = ∠AOB, then |XY|=|AB| In short, we can say that “equal chords subtend equal angles at the centre of a circle.”
Figure 3.28: Circle with points Y, X, A and B on its circumference with centre O.
Example 3.7
In the figure below, ABCD are points on the circle whose centre is O. |AB| = |BC|.
Given that ∠BOC = 52° Find the value of:
i. ∠AOB
ii. ∠ADC
iii. ∠OBC
Figure 3.29
Solution
From the Figure 3.29, since |AB| = |BC| then ∠BOC=∠AOB (equal chords subtend equal angles at the centre of a circle)
i. ⇒∠BOC = ∠AOB= 52°
ii. From theorem statement 1, it implies 2∠ADC = (∠AOC) Where ∠ADC = (∠AOC) But ∠AOC = ∠AOB + ∠BOC = 52°+ 52° = 104° Substituting, ∠ADC = (104°) = 52° ∴ ∠ADC = 52°.
iii. Triangle OBC is isosceles as |OB|=|OC|, both radii. Therefore the base angles are equal, so ∠OBC = = 64°.
Statement Six
Opposite angles in a cyclic quadrilateral are supplementary (add up to 180°)
Figure 3.30 shows a cyclic quadrilateral with interior angles a, b, c, and d. Angles a and c are opposite to each other. Likewise, and .
This means that: a + c = 180° and b + d = 180°
Figure 3.30: A cyclic quadrilateral with interior angles a, b, c and d.
Example 3.8
Figure 3.31 below is a circle with points Q, R, S and T on the circumference of the circle.
Given that ∠RST = 78°, find the value of ∠TQR
Figure 3.31
Solution
From the diagram, ∠RST + ∠RQT = 180° (Opposite angles in a cyclic quadrilateral are supplementary) ∠RQT + 78° = 180° ∠RQT = 180° − 78° ∴ ∠RQT = 102° Statement Seven The angle between a tangent and a chord at a point of contact is equal to the angle in the alternate segment.
In Figure 3.32, ABC is a circumscribed triangle. The circle is tangent to Line DAE at point A. This means that ∠BAE is equal to ∠ACB. This is the same as saying x = y.
Figure 3.32: Line DAE tangent to a circle at point A
Example 3.9
In Figure 3.33 |TP| is a tangent to the circle at point Q. Given that ∠SQT = 70° and ∠QSR = 30°, work out for the value of the angle marked a
Figure 3.33
Solution
From theorem statement seven, ∠SRQ = ∠SQT = 70°(angle between tangent and a chord) From the diagram, ∠SQR + ∠QSR + ∠SRQ = 180° (sum of interior angles of a polygon) ⇒a + 30° + 70° = 180° a + 100° = 180° a = 180° − 100° = 80° ⇒a = 80° Proof of Circle Theorem Statements It is now time to prove some of these statements.
Proof of Statement One
Statement The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.
Figure 3.34: Cyclic triangle ABC with centre O.
Proof From Figure 3.34, we want to prove that x = 2y.
Draw a straight line from point C, through point O, to intersect the circle at point D.
Figure 3.35: Cyclic triangle ABC with centre O and CD as a diameter.
This will generate two triangles: AOC and BOC.
Both triangles are isosceles since two of the sides of each triangle are radii. For instance, in ∆AOC,<OAC = <OCA = p and for ∆BOC, ∠OBC = ∠OCB = s Remember that the “exterior angle of a triangle theorem” states that, for any triangle, the measure of the exterior angle of a triangle is equal to the sum of the measures of its two non-adjacent interior angles. Therefore, from this theorem, we have:
∠AOD = 2p and ∠BOD = 2s
Figure 3.36: Cyclic triangle ABC with centre O and CD as a diameter.
This means that the angle at the centre x = 2p + 2s x = 2(p + s) …………………….(1) but y = p + s substitute y = p + s into equation (1) This gives x = 2y which proves the theorem.
Proof of Statement Two
Statement: The angle subtended by the diameter at the circumference is a right angle.
Figure 3.37: Cyclic triangle OAB with centre V.
Proof:
To prove that ∠AOB = 90°:
From Statement 1, ‘The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.’
So, you have . ∠AVB = 2 × ∠AOB You remember also that the angle at a point on a straight line is equal to 180°.
This implies that ∠AVB = 180° as is a diameter.
So, we have which implies that ∠AOB = 90°.
Proof of Statement Three
Statement: The angles subtended at the circumference of a circle by the same arc or chord are equal.
Proof Consider the figure 3.38 below, which shows an arc AB subtending and at two arbitrary points, C and D, on the circumference. O is the centre of the circle.
Figure 3.38: Circle ABCD with centre O.
To prove that ∠ACB = ∠ADB.
Using the circle theorem statement one; ‘The angle subtended by a chord at the centre of a circle is twice the angle subtended by it at the circumference of the circle’ ∠ACB = × ∠AOB …………………… (1) Like wise ∠ADB = × ∠AOB ………… (2) From equations (1) and (2), we get ∠ACB = ∠ADB which proves theorem three.
Proof of Statement Four
Two equal chords subtend equal angles at the centre of the circle.
Proof Consider the circle given below with centre O and two chords |AB| and |XY|, such that |AB| = |XY|.
To prove that ∠AOB = ∠XOY, consider the Figure 3.39 below
Figure 3.39: Circle with centre O with points Y, X, A and B on its circumference.
In triangles XOY and AOB, you have |OA|=|OB| (Radii), |OX|=|OY| (Radii) and |AB|=|CD| (Given) Triangles XOY and AOB are congruent by the Side-Side-Side (SSS) congruence rule.
So, you have ∠XOY = ∠AOB since corresponding sides and angles are the same.
Proof of Statement Five
Statement: If the angles subtended by two chords at the centre are equal, then the two chords are equal.
Proof Consider Figure 3.40; It is a circle with centre O and two chords, |AB| and |CD|, such that ∠AOB = ∠COD.
Figure 3.40: Circle ABCD with centre O To prove that |AB| = |CD| Consider triangles AOB and COD, you have the radius |OA| = |OC|=|OB| = |OD|= radii and ∠AOB = ∠COD (Given) This means that triangles AOB and COD are congruent by the Side-Angle-Side congruence rule.
So, you have |AB| = |CD| (Corresponding parts of congruent triangles).
We will explore the other proofs in later lessons.
The tangent of a circle Remember that a tangent is a straight line drawn from an external point that touches a circle at exactly one point at the circumference of the circle. There can be an infinite number of tangents to a circle. These tangents follow certain properties/ theorems that can be used as identities to perform mathematical computations on circles.
Consider the theorems of the tangent to a circle and we will then look at how to prove them.
Theorem 1 The tangent at any point of a circle is perpendicular to the radius at the point of contact (tangency). In other words, a tangent is perpendicular to a radius at the point of contact (tangency) from the centre of a circle.
To prove that the tangent at any point of a circle is perpendicular to the radius at the point of contact, consider the Figure 3.41 below.
O is the Centre of the circle and |XL| is the tangent to the circle at point A.
Figure 3.41: A tangent to a circle at point A In Figure 3.41, |OA|=|OC| as they are radii of the circle.
In addition, |OB| =|OC| + |CB|.
It can be observed clearly that, |OC| < |OB| and |OA| < |OB| This logic holds for every point B on the tangent L, except point A.
So |OA| is shorter than any other line segment joining O to any point on |XL|.
Hence, OA is perpendicular to |XL| With your drawing tools and materials, let us perform the activity on the next page.
Activity 3.1 Tangent to a Circle
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
1. Draw a circle with centre O and radius OA.
2. Draw a tangent line |XL| through point A.
3. Choose a point B on |XL|, other than A.
4. Join O to B, intersecting the circle at C.
5. Measure |OA| and |OB|
6. Write down your observations, discuss them with your classmate(s) and compare your findings with the conclusion below.
Since every point on |XL|, other than A, lies outside the circle and OB > OA. This implies OA is the shortest line segment from O to L. The shortest line segment drawn to a line from a point outside it, is perpendicular to the line. Therefore, |OA| is perpendicular to line L.
Mathematically, |OA| ⊥ |XL| Therefore, it is true that, the tangent at any point of a circle is perpendicular to the radius passing through the point of contact.
With this understanding, let us look at the examples below.
Example 3.10
In the diagram below, O is the centre of the circle, |NP| is tangent to the circle at R and ∠MOR = 110°.
Calculate the value of ∠MRP
Figure 3.42: A tangent to a circle at point R
Solution
In Figure 3.42, the triangle MOR is isosceles, since |OM|=|OR|= radius r.
⇒∠OMR = ∠ORM 100
Figure 3.42: A tangent to a circle at point R
Solution
In figure 3.42, the triangle MMMMMM is isosceles, since |OM|=|OR|= radius r.
⟹ ∠MMMMMM = ∠MMMMMM Since |MMMM| (radius) meets |NNNN| (tangent) at R (point of contact) ⟹ ∠ MMMMNN = 90° (a tangent is perpendicular to a radius at the point of contact) To find the value of ∠ MMMMNN, find the value of ∠ORM first.
So, let ∠MMMMMM = xx ∠MMMMMM + ∠ MMMMMM + ∠ MMMMMM = 180° xx + xx + 110° = 180° 2xx = 180° − 110° 2xx = 70° 2xx 2 = 70° 2 xx = 35° ∴ ∠MMMMMM = ∠MMMMMM = 35° Now, ∠ MMMMNN = ∠ MMMMMM + ∠ ORP But ∠ ORP = 90° ∠ MMMMNN = 90° + 35° = 125° Theorem 2 Alternate Segment Theorem The Alternate Segment Theorem states that “For any circle, the angle formed between the tangent and the chord at the point of contact of the tangent and the circle is equal to the angle formed by the chord in the alternate segment”.
The alternate segment theorem is also known as the tangent-chord theorem.
100
Figure 3.42: A tangent to a circle at point R
Solution
In figure 3.42, the triangle MMMMMM is isosceles, since |OM|=|OR|= radius r.
⟹ ∠MMMMMM = ∠MMMMMM Since |MMMM| (radius) meets |NNNN| (tangent) at R (point of contact) ⟹ ∠ MMMMNN = 90° (a tangent is perpendicular to a radius at the point of contact) To find the value of ∠ MMMMNN, find the value of ∠ORM first.
So, let ∠MMMMMM = xx ∠MMMMMM + ∠ MMMMMM + ∠ MMMMMM = 180° xx + xx + 110° = 180° 2xx = 180° − 110° 2xx = 70° 2xx 2 = 70° 2 xx = 35° ∴ ∠MMMMMM = ∠MMMMMM = 35° Now, ∠ MMMMNN = ∠ MMMMMM + ∠ ORP But ∠ ORP = 90° ∠ MMMMNN = 90° + 35° = 125° Theorem 2 Alternate Segment Theorem The Alternate Segment Theorem states that “For any circle, the angle formed between the tangent and the chord at the point of contact of the tangent and the circle is equal to the angle formed by the chord in the alternate segment”.
The alternate segment theorem is also known as the tangent-chord theorem.
Theorem 2 Alternate Segment Theorem
The Alternate Segment Theorem states that “For any circle, the angle formed between the tangent and the chord at the point of contact of the tangent and the circle is equal to the angle formed by the chord in the alternate segment”.
The alternate segment theorem is also known as the tangent-chord theorem.
Let us use Figure 3.43; to break down the theorem.
In the Figure, ABC is a cyclic triangle with centre O. The circle is tangent to Line DAE at point A. The shaded part is a minor segment. Its alternative segment is the unshaded part of the circle. According to the alternative segment theorem, ∠BAE is equal to ∠ACB. i.e. x = y Figure 3.43: Tangent to a circle at point A Proof:
From our previous proof, we established that the angle the radius or the diameter makes with the tangent at the point of tangency is a right angle (90°) Therefore, ∠OAE = 90° This implies that ∠OAB + x = 90° solving for ∠OAB, we have ∠OAB = 90°− x Since triangle OAB is isosceles, Figure 3.44: Tangent to a circle at point A ∠OAB = ∠OBA = 90° − x Also, ∠AOB + (90°− x) + (90°− x) = 180° solving for AOB, we have ∠AOB = 180° − 90° − 90° + 2x This implies ∠AOB = 2x ……………......(1) Remember that the angle at the centre of the circle is twice the angle at the circumference of the circle.
This means that:
∠AOB = 2y……………………. (2) Comparing (1) to (2), we have 2x = 2y This means x = y
Example 3.11
In the figure below,|AB |= |AC| and the tangent |BT| touches the circumcircle of triangle ABC at B. Given that ∠ABC = 70°, find:
i. ∠BCT
ii. ∠ATB
Figure 3.45: Tangent to a circle at point A
Solution
i. ∠ABC = ∠BCA = 70° (base ∠s of an Isosceles ∆) ∠ABC + ∠BCA + ∠CAB = 180° 70° + 70°+ ∠CAB =180° 140° + ∠CAB = 180° ∠CAB = 180° − 140° ∠CAB = 40° ∠CAB =∠CBT = 40° ∠BCT = ∠TAB + ∠ABC ∠BCT = 40° + 70° = 110°
ii. ∠ABT + ∠TAB °+ ∠ATB = 180° (70° + 40°) + 40°+ ∠ATB = 180° 150° + ∠ATB = 180° ∠ATB = 180° − 150°= 30° Alternate Segment Theorem for Quadrilaterals In Figure 3.46, line FAE is a tangent to the circle at point A. B, C and D are points on the circumference of the circle and ABCD is a cyclic quadrilateral.
Figure 3.46: Line DAE tangent to cyclic quadrilateral ABCD at point A From the alternate segment theorem, ∠DAF = ∠CBA ⇒w = u………..…(1) We need to prove that ∠CAE = ∠CDA ⇒ x = y Although the alternate angle theorem holds for this, we can try another theorem to prove that this is true.
From the previous theorems we know that opposite angles in a cyclic quadrilateral add up to 180°.
⇒∠y + ∠u = 180° ……………(2) Since ∠w and ∠x are adjacent at point A on a straight line, we can conclude that ∠w+ ∠x = 180° …………….(3) From equations (2) and (3), we have ∠y + ∠u = ∠w+ ∠x but from (1), we have w = u Therefore, ∠y + ∠w = ∠w+ ∠x Eliminating ∠w, ⇒∠y + ∠w − ∠w = ∠x we have ∠y = ∠x or ∠x = ∠y Theorem 3 If two segments from the same external/exterior point of a circle are tangents to the circle, then they are congruent.
This means that two tangents drawn from the same external point/exterior point to a circle are equal when measured from their point of contact.
Figure 3.47: |PA| and |PB| are tangent to a circle with centre O at points A and B.
In Figure 3.47, and are tangents to circle, S, at points A and B. Point O is the centre of the circle. According to the equal tangent theorem, |PA| = |PB|.
Also, ∠POA = ∠POB and ∠APO = ∠BPO
Note: APBO can also be called a tangent–kite and has axis of symmetry OP.
Two tangents to a circle from a point outside the circle are equal in length. |AP|=|BP|.
Thus, ∠APO = ∠BPO and ∠POA = ∠POB Proof All three parts will be proved if we show that ∆PAO is congruent to ∆PBO.
Comparing the two triangles, we see that:
• OA = OB (radii of the same circle)
• OP = hypotenuse of both triangles.
• From our first theorem, ∠OAP = ∠OBP = 90° By the Side-Angle-Side (SAS) property of triangles, triangle OAP is congruent or the same as triangle OBP.
This follows that the lengths of the tangents |AP| and |BP| are equal. Also, |OP| bisects BPA and AOB.
Hence, ∠BOP = ∠AOP and ∠BPO = ∠APO.
Example 3.12
A belt ABCD moves round a shaft D (whose radius is negligible) and a pulley of radius 1.2m. O is the centre of the pulley and |OD| = 3.0m.
The straight portions AD and CD of the belt are tangents at A and C.
i. To the nearest degree, determine the value of ∠AOC
ii. The total length of the belt (ABCD) to the nearest metre (Take π = 3.142)
Figure 3.48: A belt around a pulley and a shaft To solve the above, the figure can be sketched and since the radius of the shaft is negligible, the figure will look like this:
Figure 3.49: Sketch of the pulley system To find ∠AOC, take any of the right triangles (⊿AOD or ⊿COD) Considering ⊿AOD
i. Let ∠AOD be θ 105
Figure 3.49: Sketch of the pulley system To find ∠AAAAAA, take any of the right triangles (⊿AOD or ⊿COD) Considering ⊿AAAAAA
i. Let ∠AAAAAA be θθ AACCCCθθ = AAAAAAAAAAAAAAAA hyyyyCCAAAAAAyyCCAA = |AAAA| |AAAA| AACCCCθθ = 1.2 3.0 = 2 5 θθ = AACCCC⁻¹(2
5) = 66.4218° ∠AAAAAA = ∠AAAAAA + ∠AAAAAA By the tangent theorem, ∠AAAAAA = ∠AAAAAA = 66.4218° ⟹ ∠AAAAAA = 2(66.4218°) = 132.8436° ∴ ∠AAAAAA = 133°(AACC AAhAA AAAAAAnnAACCAA AAAAddnnAAAA)
ii. To find the total length of the belt, you need to find the length of the belt from the external point D to the circle at the points of tangency (A and B) and around the circle.
Considering ⊿AOD again,
ii. To find the total length of the belt, you need to find the length of the belt from the external point D to the circle at the points of tangency (A and B) and around the circle.
Considering ⊿AAAAAA again, By Pythagoras theorem, |AAAA|²= |AAAA|²+ |AAAA|²|AAAA|²= |AAAA|²−|AAAA|²= (3.0)²− (1.2)²|AAAA|²= 9 − 1.44 = 7.56 |AAAA| = √7.56 = 2.7495mm By the tangent theorem, |AAAA| = |CCAA| = 2.7495mm Finding the length of the belt around the circle Length of major Arc AAAACC = 360°−∠AAAAAA 360° × 2ππππ = 360° − 133° 360° × 2 × 3.142 × (1.2) = 4.7549mm Total length of the belt= |AAAA| + |CCAA| + Length of major Arc AAAACC 2.7495mm + 2.7495mm + 4.7549mm = 10.2539mm ∴ The total length of the belt ABCD = 10mm (to the nearest metre) Perform Activity 3.2 below with your classmate.
Activity 3.2 Proof of Some Circle Theorems
Part 1: Angles at the centre and the circumference
i. Draw a circle and label the centre O.
ii. Mark two different points A and B on the circumference and join them to form a chord AB.
iii. With the ruler, join O to A and O to B. This forms a central angle ∠AOB.
iv. Mark a point C on the same arc AB and join A to C and B to C with a ruler.
v. Measure ∠AOB and ∠ACB with a protractor and record them.
vi. Discuss what have observed with your classmate.
This verifies that ∠AOB ≈ 2∠ACB.
Part 2: Angles in the same segment | | Perform Activity 3.2 below with your classmate.
Activity 3.2 Proof of Some Circle Theorems
Part 1: Angles at the centre and the circumference
i. Draw a circle and label the centre O.
ii. Mark two different points A and B on the circumference and join them to form a chord AB.
iii. With the ruler, join O to A and O to B. This forms a central angle ∠AOB.
iv. Mark a point C on the same arc AB and join A to C and B to C with a ruler.
v. Measure ∠AOB and ∠ACB with a protractor and record them.
vi. Discuss what have observed with your classmate.
This verifies that ∠AOB ≈ 2∠ACB.
Part 2: Angles in the same segment
i. With the same circle and chord , choose another point D on the same arc AB as point D.
ii. Join A to D and B to D using a ruler.
iii. Measure ∠ACB and ∠ADB and record each.
Verify that they are equal.
Part 3: Cyclic quadrilateral theorem
i. Draw another circle
ii. Mark four points A, B, C, D on the circumference so they form a quadrilateral.
iii. Join them in this order: A to B, B to C, C to D, D to A.
iv. Measure opposite angles ∠ABC and ∠ADC, find their sum and record the value.
v. Verify that their sum is about 180°. Do the same for the other pair of opposite angles.
Part 4: Tangent–Radius Theorem
i. Draw a new circle with centre O.
ii. Choose a point P on the circumference.
iii. Join O to P to form a radius.
iv. At P, draw a tangent to the circle (make sure the line touches the circle at one point P only).
v. With the protractor, measure the angle between OP and the tangent.
vi. Verify that the angle is 90°.
• What patterns do you notice for each theorem?
• Do the results change if you choose different points?
Understanding Geometric Constructions
In Year 1, we learnt how to construct various angles such as 75°, 105°, 135° and 150°.
We will revise this with a reminder on the construction of angles .
Remember that to have an accurate outcome in construction:
• Make a rough sketch of the task assigned. This will help anticipate challenges associated with the construction.
• Do not clean anything that contributes to the final result. Leave all construction lines visible.
• For lines and points to be as clear and accurate as possible, use hard pencils with sharp tip/point.
1. Construction of angle 105° Materials needed: pair of compasses, ruler, pencil Steps:
i. With your ruler and pencil, draw a straight horizontal line
ii. Choose a point on the line as your centre and any suitable radius,
iii. Construct angle 90°, see the diagram below to remind you how to do this.
Figure 3.50: Construction of 90°
iv. Construct 120° adjacent to 90°
Figure 3.51: Construction of 90° adjacent to 120°
v. Bisect AOB
Figure 3.52: Construction of 105°
Use your protractor to measure the angle to verify
2. Construction of 135°
You can achieve this if you construct 90° and bisect the angle between the other 90°.
Steps:
i. With your compass and ruler, construct angle 90°, as above.
ii. Bisect the adjacent to 90° obtain angle 45, 90°+ 45° = 135°
Figure 3.53: Construction of 135°
Click on this link www.math-only-math.com to watch more tutorials on how to construct angles for better understanding.
Construction of quadrilaterals: Concept, examples and visualisation of surds We have learnt how to construct lines, angles, circles and some triangles. We will now continue with the construction of other interesting shapes, including revising how to construct triangles.
Construction of Triangles
A triangle can be constructed using a ruler, pencil and compass only if the following information is provided:
1. All three sides of the triangle are provided, or
2. Two sides and the included angle are given, or
3. One side and two angles are given.
Use the following examples to perform some activities on constructing triangles.
Materials needed: pair of compasses, ruler, pencil, eraser and workbook.
Example 3.13
Using a ruler and a pair of compasses only, construct triangle ABC such that:
|AB|=10cm, |AC|=8cm, and |BC|=6cm
Solution
We can choose to construct any of the lines first.
Steps To construct line |AB|, first.
1. Draw a line of 10cm and label the line AB
2. Open your compass to 8cm on the ruler and with the compass pin at A, draw an arc
3. Open the compass to a length of 6cm on your ruler and with your compass point now at B, draw an arc to intersect the other arc you have drawn earlier.
Indicate the point of intersection of the two arcs as C.
4. Finally, with your ruler, join the points A to C and B to C with straight lines to get |AB| and |BC|
Figure 3.54: Triangle ABC
Example 3.14
Using a ruler and a pair of compasses only, construct triangle ABC such that: |AB|= 7cm, |AC| = 5cm and ∠CAB= 90°. Measure |BC|
Solution
Steps:
1. Construct line |AB|, indicating the points A and B, which are 7cm apart.
2. At point A, construct a 90° angle.
3. Measure 5cm along this line and mark point C.
Figure 3.55: Triangle ABC (Not drawn to scale)
4. With a ruler, measure line |BC| ≈ 8.5cm
Example 3.15
Using a ruler and a pair of compasses only, construct triangle XYZ such that:
|XY| = 5.5cm, ∠YXZ = 30°, XYZ = 105°.
Solution
Steps:
1. First construct |XY| and indicate clearly the points X and Y which are 5.5cm apart.
2. Construct ∠30° at X and ∠105° at Y.
3. Name the point of intersection of the two lines Z. This gives you the triangle XYZ
Figure 3.56: Triangle XYZ
Construction of a circumscribed circle when given a triangle To construct a circumscribed circle when given a triangle (i.e., a circle where the circumference touches the vertices of the triangle), you have to determine the centre of the triangle. The centre cannot be obtained by just observing and guessing. Below are the steps to guide you to locate the exact centre that will help you draw the circle accurately.
Example 3.16
Copy the triangle below and use it to construct a circumscribed circle.
Figure 3.57: Triangle ABC
Steps:
1. Construct a copy of the triangle.
2. Construct a perpendicular bisector of any two sides of the triangle.
3. Label the point of intersection of the bisectors as O
4. With the compass pin at O and using |OA| or |OB| or |OC| as the radius, construct a circle to pass through A, B and C
Figure 3.58: A cyclic triangle ABC with centre O.
Construction of quadrilaterals We know that a quadrilateral is a four-sided polygon bounded by four straight lines and it has four sides, four vertices and four angles. We know that common quadrilaterals include rectangles, squares, rhombuses, trapeziums, kites. These figures can be constructed using a pair of compasses and a ruler and are easy to construct just as the triangles.
Follow the steps in the examples below on how to construct some of them.
Example 3.17
Using ruler and a pair of compasses only, construct a square ABCD of side 5cm.
Solution
To construct a square, all you need to know is the length of a side and then apply the properties of a square to get the expected outcome.
Steps:
i. Using a ruler and pencil, draw line AB of length 5cm.
ii. With a compass, construct perpendicular lines (angles 90°) at points A and B, as the corner angles of a square are right angles.
iii. Measure 5cm with your compass. With the pin at A, draw an arc on the perpendicular line and label the point D.
iv. With the pin at B, draw another arc (same radius) on the perpendicular line and label it C, sides of a square are equal and parallel.
v. Join C to D to form square ABCD.
Figure 3.59: Construction of square ABCD
Example 3.18
Using a ruler and a pair of compasses only, construct a trapezium MNOP, with |MN| = 7.5cm, |NO|=5.5cm, ∠PMO = 60°, ∠MNO = 90°.
Solution
Steps:
1. With your ruler and pencil, construct line |MN| which is 7.5cm
2. With your compass pin at point M, construct angle 60°
3. Again, with your compass pin at point N, construct angle 90°
4. Measure on the ruler with your compass and with your compass pin at N, draw an arc to intersect the perpendicular line at point O
5. Since it is a trapezium, a pair of opposite sides must be parallel. Therefore, |MN| ∥ |PO|.
With the compass pin again at point M and with the same measure, draw an arc to cut the perpendicular line at point Q
6. Join point Q to point O. Mark the point of intersection of the 60° line and |OQ| as point N.
Figure 3.60: Construction of trapezium MNOP
Example 3.19
Using a ruler and a pair of compasses only, construct a parallelogram ABCD such that:
|AB| = 6cm, |BC| = 5cm ∠DAB = 60°.
Measure |CD|
Solution
To construct a parallelogram, remember that it has its pairs of opposite sides are equal and parallel.
Steps:
1. Draw the line AB, 6cm
2. With the compass pin at point A, construct angle 60°
3. At point B construct angle 60° (parallel to point A) but from the other direction (anticlockwise)
4. Open your compass to 5cm on your ruler and with your compass pin at point B, draw an arc to intersect the line at C
5. With the same radius and with your compass pin at point now at A, draw an arc to intersect the line at D
6. Join the point C to D
7. At point C, open your compass to point D and measure the radius on the ruler.
8. |CD| ≈ 6cm
Figure 3.61: Construction of a parallelogram ABCD Have you ever thought of determining a point or a line that is of equal distance from a point, points, line or lines in a plane?
Earlier we constructed a circle through three points that are non-collinear (not on the same line). This circle was achieved using the idea of the construction of a locus (plural is loci). Now, we will look at how to construct a locus of a point or points, a line or lines in a plane.
Locus A locus is a set of all points that satisfy a given condition or set of conditions. Think of it as a path traced by a point moving according to some specified rules.
Here we will explore the construction of locus in two ways:
• Theorems and conditions of locus
• Construction of locus Locus theorem 1 Locus of points equidistant from a single point Method: Using the point as the centre and the given distance as the radius, construct a circle. The locus is the circle.
Example 3.20
Draw a locus of all points equidistant from a point O.
Solution
Steps:
1. in your workbook, draw a point and label it O
2. With your compass pin at O and opened to a suitable radius, for example, 4cm, and draw a circle.
Figure 3.62: A circle equidistant from O The circle is the locus of all points equidistant from O. Any point drawn on any part on the circumference of the circle above, will be the same distance away from the centre O.
Example 3.21
Construct a locus of all points that are 3cm from A on the quadrilateral ABCD in the
figure below.
Figure 3.63: A quadrilateral ABCD
Solution
With the aid of a ruler and a pair of compasses only, make a copy of the quadrilateral ABCD.
Open your compass to a radius of 3cm and with your compass pin at A, construct a complete circle.
The circle is the locus of the point A.
Figure 3.64: A quadrilateral ABCD and a circle with centre A Locus Theorem 2 Locus of points equidistant from two points.
Method: Construct a perpendicular bisector of the line segment determined by the two points.
The locus is the perpendicular bisector of the line.
Example 3.22
Draw a locus of points which are equidistant from A and B, with |AB| = 8cm.
Solution
The solution is drawing a perpendicular bisector of line AB.
Draw a straight line and two points on the line 8cm apart and label them A and B Using a ruler and a pair of compasses, construct a bisector of AB.
The bisector is the locus, of points equidistant from A and B as shown in Figure 3.65
Figure 3.65: Locus of two points A and B You would observe that the locus of points equidistant from the points A and B is the perpendicular bisector of |AB|.
Example 3.23
Construct a locus of points equidistant from A and C in Figure 3.66
Figure 3.66: A triangle ABC
Solution
Figure 3.67: Triangle ABC and line p bisecting line AC Line p is the locus of the point A and C Locus theorem 3 Locus of points equidistant from two intersecting lines.
Method: Bisect the angle of intersection of the two lines. The locus is the line bisecting the angle. It can also be called the angle bisector.
To construct an angle bisector of any two segments |AB| and |BC|.
Steps:
1. Draw any two intersecting lines (|AB| and |BC|)
2. Locate the point of intersection of the two lines which is B
3. With your compass point at the B and with a suitable radius, draw an angle between the two lines.
4. With your compasses again, bisect the angle between the lines (|AB| and |BC|)
Figure 3.68: Angle bisector of ABC The bisector of the angle at B (the line that bisects the angle) is the locus, which is equidistance from lines AB and BC
Example 3.24
Construct the locus of points equidistant from AB and CD as shown in Figure 3.69 below.
Figure 3.69: lines AB and CD
Solution
Using a ruler and a pair of compasses only, make a copy of the figure Construct the locus P, of points equidistant from AB and CD
Figure 3.70: Construction of loci equidistant from AB and CD Locus theorem 4 Locus of points equidistant from two parallel lines Method: The locus equidistant from two parallel lines, |AB| and |CD| is a line parallel to both AB and CD and halfway between them.
Figure 3.71: line m equidistant from AB and CD In the diagram above, the locus is line m. Line m is parallel to and equidistant from lines AB and CD. Line m is also parallel to AB and CD.
Example 3.25
The diagram below represents a construction of two parallel lines m₁ and m₂.
Construct the locus, d equidistant from m₁ and m₂.
Figure 3.72: parallel lines m₁amd m₂
Solution
Steps:
1. Construct a perpendicular at any point O on line m₁ to intersect line m₂ at point A.
Figure 3:73: A perpendicular transversal intersecting parallel lines m₁and m₂at point O and A
2. Construct a perpendicular bisector of line |OA|
Figure 3.74: Line d equidistant from parallel lines m₁and m₂
3. Line d is the locus of points parallel to lines and .
Locus theorem 5 Locus of points equidistant from a line.
Method: This is a pair of parallel lines that are a distance away from the given line.
This theorem is illustrated in the Figure 3.75.
The figure illustrates the locus of points, p, of points m units from line AB.
The locus, p, represents the dotted lines on opposite sides of line AB.
Figure 3.75: Loci p equidistant from line AB Locus theorem 6 Constructing a perpendicular from a point O to a line.
Method: Open the compass to a distance more than the perpendicular distance from the point O, to the line.
Example 3.26
Construct a perpendicular bisector of |AB| from a point C
Solution
Steps:
1. Draw a line.
2. Locate a point C anywhere above the and with your compass pin at point C, draw an arc to intersectline at the two points and label the points A and B.
Figure 3.76: Construction of an arc from point C to |AB|
3. With the compass point at A and using the radius |AC|, draw an arc on the other side of the line.
4. With the compass point at B and using the same radius as used in step 3, draw an arc to intersect the one drawn in step 3.
5. Finally, join point C with a straight line to the intersection of the two arcs.
Figure 3.77: Construction of a perpendicular from point O to line AB Locus Theorem 7 Locus of points equidistant from three fixed points.
Method: First, join the three points to make a triangle.
Construct the perpendicular bisectors of any two sides of the triangle.
The intersection of the two perpendicular bisectors is the locus. This locus is a single point.
NOTE: This locus is a single point called the circumcentre. This point can be used to construct a circumscribed circle.
Example 3.27
Three points X, Y and Z are positioned at different positions in a plane. Given that all these points are non-collinear (not on the same line), determine the precise centre of the three points in the plane.
Solution
Steps:
1. Draw three points that are far apart from one another and are not on the same line (non-collinear).
Label the points X, Y and Z.
2. With your ruler, join the three points to form a triangle.
3. Construct the perpendicular bisectors of any two sides of the triangle or for even the three lines.
4. Locate the point of intersection of the perpendicular bisectors and label it O
Figure 3.78: Perpendicular bisectors of sides XY and ZY of triangle XYZ The point of intersection of the two perpendicular bisectors is the locus. This locus is a single point (O) and it is equidistant from the three points X, Y and Z Now have a go at summarising the locus theorems in your own words for easy recall.
Write them in your jotter and discuss them with a classmate or compare it with the ones below.
In summary, the Locus theorems can be considered under these five (5) rules.
Five Rules of Locus Theorems
Rule 1: Given a point, the locus of points is a circle.
Rule 2: Given two points, the locus of the points is a straight line midway between the two points.
Rule 3: Given a straight line, the locus of points is two parallel lines.
Rule 4: Given two parallel lines, the locus of points is a line midway between the two parallel lines.
Rule 5: Given two intersecting lines, the locus of points is a pair of lines that cut the intersecting lines in half.
Read through the example provided below carefully and perform the activity that follows, either with a classmate or on your own.
Example 3.28
Using a ruler and a pair of compasses only, construct
a. triangle XYZ with |XY| = 6cm, |XZ| = 8cm and ∠YXZ = 75° b.
i. the locus d₁ of the points equidistant from |XY| and |XZ|
ii. the locus of d₂ points equidistant from X and Y
c. Locate
i. the point of intersection O of d₁and d₂inside the triangle
ii. with your compass pin at point O and a radius |OX| construct a circle.
Figure 3.79: Solution to example 3.28
Example 3.29
Using a ruler and a pair of compasses only, construct:
i. A quadrilateral PQRS, |PQ| = 8cm, |PS| = 6cm, |QR| = 10cm, ∠QPS = 60° and ∠PSR = 135°
ii. The locus L1, of points equidistant from |QR| and |RS|
iii. The locus L2, from Q, perpendicular to L1 Locate O, the point of intersection of L1 and L2.
Solution
Given |PQ| = 8cm, |PS| = 6cm,|RS| = 10cm, ∠QPS = 60° and ∠PSR = 135°
Figure 3.80: Solution to example 3.29 Applications of Construction The idea of construction especially the concept of locus is very important and applicable in areas like road construction. It helps to check:
• Safety: Proper median and lane divisions to reduce accidents.
• Efficiency: Parallel drainage systems to prevent erosion and flooding.
• Traffic Flow: Roundabouts and bisector-aligned intersections minimise congestion in the urban areas.
Remember, construction can be applicable in many areas. Look for other areas where these concepts are applicable.
Activity 3.2 Application of Locus
Discuss the following scenarios with a classmate. For each of the scenarios, investigate which theorem(s) of locus construction will be applicable.
1. Mr Abotsi instructs students to move around a point, keeping a distance of 3 metres away at all times.
2. Your constituency’s MP wants to site a market equally accessible to three separate towns by placing it at their central point.
3. Students are to construct a straight horizontal line through the exact centre of an A4 sheet without folding it.
4. A contractor plans to build gutters equally spaced on both sides of a 1500m road to control flooding.
5. At a 60° walkway intersection in a school, the headmistress wants lights positioned to brighten both paths equally.
Expected Responses
1. Locus of points equidistant from a single point.
2. Locus of points equidistant from three fixed points.
3. Locus of points equidistant from two parallel lines
4. Locus of points equidistant from a line.
5. Locus of points equidistant from two intersecting lines.
The activity below will help you practice what you have learnt.
Activity 3.3 Constructing a circle through a triangle using the idea of locus Steps
1. Draw a line segment XY of length 6cm using your ruler.
2. With your compass pin at X, construct an angle 75°.
3. From point X, construct Z of length 8 cm.
4. Join points Y and Z to form triangle XYZ.
5. For locus L1, Construct the angle bisector of ∠YXZ to find the locus of points equidistant from |XY| and |XZ|.
6. For Locus L2, construct the perpendicular bisector of to find the locus of points equidistant from X and Y.
7. Locus L3, with your compass pin at Y as the centre and a radius of 4.5 cm, draw a circle to find the locus of points 4.5cm from Y.
8. Identify the point of intersection of L1 and L2 inside the triangle and label it as d1
9. Identify the point of intersection of L1 and L3 inside the triangle and label it d2
10. Identify the point of intersection of L3 and L2 inside the triangle and label it as d3 With your protractor, measure angle d2 d1 d3 (the angle at d1)
11. Compare your diagram and also the angle value of d1 to that of your classmate(s).
12. Compare your diagram with the one in the figure below.
Figure 3.81: Solution to activity 3.3
1. In the figure below, ABCD is a cyclic quadrilateral. |EF| is tangent to the circle at point C.
∠DCE = 46° and ∠ABC = 98°
a. Calculate:
i. ∠DAC
ii. ∠ADC
iii. ∠ACE iv.
Figure 3.82: A cyclic quadrilateral 2.
a. Using a ruler and a pair of compasses only, construct triangle ABC such that: |AB| = 6cm, |AC| = 6cm and ∠CAB = 60°.
b. Construct a perpendicular from C to meet |AB|.
c. Construct |CD|= 6cm, parallel to |AB| through C.
d. With a ruler, join B to D.
e. What type of quadrilateral is formed?
3.
a. Using a ruler and a pair of compasses only, construct triangle ABC such that: ∠ABC = 30° and ∠CAB = 60° and |AB| = 10cm
b. Construct a perpendicular from the point C to meet the line AB at P.
c. Extend the line CP to meet point D such that BC = BD.
Join A to D and B to D.
d. What type of quadrilateral is ADBC?
4. In the figure below, |AD| and |CD| are tangents to a circle centre O.
If ∠ABC = 70°, calculate:
i. ∠AEC
ii. ∠AOC
iii. ∠ADC
Figure 3.83: Diagram for question 4
5. In the diagram below, O is the centre of the circle, BD is a tangent to the circle, ∠CBD = 26° and AO is parallel to BC
i. Find, ∠AOB
ii. Calculate the value of the reflex angle AOC
Figure 3.84: Diagram for question 5
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A tangent touches a circle at point . The radius is drawn to . What is the angle between and the tangent?
From an external point , two tangents and are drawn to a circle, touching it at and . If , what is ?
In a circle with centre , is a diameter and is a point on the circumference. If , find .
Kofi is constructing a quadrilateral . He wants point to be equidistant from and . Which construction should he make?