- Mensuration
- Volumes of Frustums of Cone, Rectangular-based Pyramid and other Pyramids
- Geometrical Construction
- Triangle: Drawing and Bisection of Line Segment; Construction and Bisection of Angles
- Construction: Construction of Quadrilateral Polygon; Construction of Equilateral Triangle; Locus of Moving Points
- Deductive Proof: Sum of Angles of a Triangle; Revision of Angles on Parallel Line cut by a Transversal Line
- Collection, Tabulation and Presentation of Grouped Data
- Calculation of Range, Median and Mode of Ungrouped Data
- Mean deviation, Variance and Standard Deviation
Mensuration
What is Mensuration?
Mensuration is a branch of mathematics that deals with the measurement of areas and volumes of various geometrical figures. Figures such as cubes, cuboids, cylinders, cones and spheres have volume and area. Mensuration deals with the development of formulas to measure their areas and volumes.
CUBES
A cube is a solid of uniform cross-section. It is formed by squares and has 8 vertices. An example is processed cubed sugar.
The length of a side of a cube is ‘e’ which is the length of all sides since a cube is formed with squares.
TOTAL SURFACE AREA OF A CUBE
A cube has 6 faces. The surface area of each side = e2 as each side is a square.
Therefore, Total surface area of a cube (all 6 sides) = 6e2
The surface area of a cube is gotten by the formula
Surface area= 6e2 sq. units
VOLUME OF A CUBE
In a cube all sides are equal. Length=e, height=e and width=e
Therefore, Volume of a cube= length× width× height
Volume of a cube= e3 cubic units
CYLINDERS
A Cylinder is a uniform circular cross-section. Examples of cylinders are unsharpened pencils like HB or 2B pencils, garden rollers, tins of milk or tomato et cetera
TOTAL SURFACE AREA OF A CYLINDER
There are two types of cylinders;
(1) A closed cylinder and
(2) An open cylinder
TOTAL SURFACE AREA OF A CLOSED CYLINDER
The total surface area of a closed cylinder consists of a sum of the areas of (i) the curve surface and (ii) The two circular end faces.
The curved surface when opened out is a rectangle. This rectangle has length equal to the length of the
Cylinder and the width are equal to the circumference of the circular end face.
Area of curved surface of a cylinder = area of rectangle of dimensions length (L) and width
(Circumference of base)
= 2πrl
Area of the two circular end faces= twice the area of one circular face
= πr2
Hence the total surface area of the closed cylinder
= 2πrl + 2πr2 sq. units
TOTAL SURFACE AREA OF AN OPEN CYLINDER
The total surface area of an open cylinder is the area of the curved surface which is the area of the rectangle the cylinder forms when spread.
Sometimes we are given a thick hollow cylinder. The total surface area is the sum of;
(i) the area of the external curved surface
(ii) the area of the internal curved surface and
(iii) the area of the end annular faces which will be shaded.
VOLUME OF A CYLINDER
A right circular cylinder is a solid of uniform cross-section. If a paper is wrapped round a cylinder, on opening it, a rectangle will be found.
Thus if the height of a cylinder
= h units
And the base radius
= r units
Then the volume of a cylinder
= Area of base by height
= πr2h cubic units
TRIANGULAR PRISMS
A prism is a solid with uniform cross-section of a shape of a triangle or a trapezium or any other polygon.
TOTAL SURFACE AREA OF A TRIANGULAR PRISM
In the case of a triangular shaped prism, the total surface area is the sum of the surface areas of the five faces that make up the prism.
VOLUME OF A TRIANGULAR PRISM
The volume of a prism is the area of its cross-section multiplied by the distance between the end faces. Examples are funnel, Chinese hat, cut periwinkle shell et cetera.
CONES
A cone is a figure with circular base and sides slanting to a common point or vertex. There are two types of cones (i) a right circular cone, where the line joining the vertex is symmetrical and perpendicular to the base of the cone and (ii) the non-right circular cone but this isn’t in the syllabus.
SURFACE AREA OF A CONE
Since a cone is formed from a sector of circle, then the surface area of a cone is equal to the area of the sector that formed it. Let L be the radius of the sector, then L becomes the slant height of the cone. If r is the radius of the base of the cone, then the length of arc of the sector is equal to 2pi.r which equals the circumference of the base of the cone.
If the sector subtends an angle- which it always does- then the area of the sector will be equal to
= θ /360 ×πl2 = curved surface of a cone
But 2πr = length of arc of a sector.
Therefore, 2πr = θ /360 x 2πl
Finally, surface area of a cone = πrl
TOTAL SURFACE AREA OF A CONE
The total surface area of cone is the sum of (i) the curved surface area πrl sq. units and (ii) the area of the base of the cone πr2 sq. units.
Therefore, total surface area of a cone = πr2 + πrl
ASSESSMENT
- What is mensuration?
- What are the two types of cylinder?
- What is the formula for the total surface area of a cone?
- What is the formula for the volume of a cylinder?
- Volumes of Frustums of Cone, Rectangular-based Pyramid and other Pyramids
Mathematics SS1 First Term
Week
CONES
To create a cone we take a circle and a point, called the vertex, which lies above or below the circle. We then join the vertex to each point on the circle to form a solid.
If the vertex is directly above or below the centre of the circular base, we call the cone a right cone. In this section only right cones are considered.
If we drop a perpendicular from the vertex of the cone to the circular base, then the length of this perpendicular is called the height h of the cone.
The length of any of the straight lines joining the vertex to the circle is called the slant height of the cone. Clearly l2 = r2 + h2 , where r is the radius of the base.
Surface area of a cone
Suppose the cone has radius r, and slant height l, then the circumference of the base of the cone is 2πr.
To find the area of the curved surface of a cone, we cut and open up the curved surface to form a sector with radius l, as shown below.
In the figure to the right below the ratio of the area of the shaded sector to the area of the circle is the same as the ratio of the length of the arc of the sector to the circumference of the circle.
Thus the fraction of the area of the whole circle taken up by the sector is
=
.
Hence, the area of the sector is
× πl2 = πrl.
In conclusion, the area of the curved surface of the cone is πrl
Adding this to the base, we have
Surface area of a cone = πrl + πr2.
EXERCISE
Find the surface area of the solid with dimensions shown.
Volume of a cone
When developing the formula for the volume of a cylinder in the module Area Volume and Surface Area, we approximated the cylinder using inscribed polygonal prisms. By taking more and more sides in the polygon, we obtained closer and closer approximations to the volume of the cylinder. From this, we deduced that the volume of the cylinder was equal to the area of the base multiplied by the height.
We can use a similar approach to develop the formula for the volume of a cone.
Given a cone with base radius r and height h, we construct a polygon inside the circular base of the cone and join the vertex of the cone to each of the vertices of the polygon, producing a polygonal pyramid. By increasing the number of sides of the polygon, we obtain closer and closer approximations to the cone. Hence,
Volume of a cone =
× area of the base × height
=
πr2h
OBLIQUE PRISMS, CYLINDERS AND CONES
We have seen that the volume of a right rectangular prism is area of the base multiplied by the height. What happens if the base of the prism is not directly below the top?
Cavalieri’s first principle states that if the cross-sections of two solids, taken at the same distance above the base, have the same area, then the solids have the same volume.
We will not give a proof of Cavalieri’s principle here. To present a rigorous proof requires integration and slicing ideas.
It allows us to say that the volume of any rectangular prism, right or oblique, is given by the area of the base multiplied by the height.
The same applies to oblique cylinders and cones.
EXAMPLE
Find the volume of the cylinder shown in the diagram.
SOLUTION
The volume of the cylinder is V = π × 52 × 15 = 375π cm2.
Cavalieri’s second principle states that if the cross-sections of two solids, taken at the same distance above the base, have areas in the ratio a:b, then the solids have the volumes in the ratio a:b.
EXERCISE 6
We showed earlier that the volume of a square based pyramid with base length 2x and height x has volume
× 2x × 2x × x. Use Cavalieri’s second principle to show that the volume of a pyramid whose base is a rectangle with side lengths c and d and height h is
× cd × x.
PYRAMIDS
A pyramid is a polyhedron with a polygonal base and triangular faces that meet at a point called the vertex. The pyramid is named according to the shape of the base.
square-based pyramid triangular-based pyramid hexagonal-based pyramid
If we drop a perpendicular from the vertex of the pyramid to the base, then the length of the perpendicular is called the height of the pyramid.
Surface area of a right pyramid
The faces bounding a right pyramid consist of a number of triangles together with the base. To find the surface area, we find the area of each face and add them together. Depending on the information given, it may be necessary to use Pythagoras’ Theorem to calculate the height of each triangular face. If the base of the pyramid is a regular polygon, then the triangular faces will be congruent to each other.
EXAMPLE
VABCD is a square-based pyramid with vertex V and
base ABCD, with V vertically above the centre of the
square base. The height of the pyramid is 4 cm and
the side length of the base is 6 cm, find the surface
area of the pyramid.
SOLUTION
We need to find the height VE of triangle VBC,
using Pythagoras’ Theorem.
| VE2 | = VO2 + OE2 |
| = 42 + 32 | |
| = 25 |
Hence VE = 5 cm.
| Area of VCB | = |
| = | |
| = 15 cm2 | |
| Area of base | = 6 × 6 |
| = 36 cm2 | |
| Surface area | = 4 × 15 + 36 |
| = 96 cm2 |
The surface area of the pyramid is 96 cm2.
EXERCISE 1
When it was built, the Great Pyramid of Cheops in Egypt had a height of 145.
m and its base was a square of side length 229 m. Find its surface area in square metres, correct to three significant figures.
Volume of a pyramid
Here is a method for determining the formula for the volume of a square-based pyramid.
Consider a cube of side length 2x. If we draw the four long diagonals as shown, then we obtain six square-based pyramids, one of which is shaded in the diagram.
Each of these pyramids has base area 2x × 2x and height x. Now the volume of the cube is 8×3. So the volume of each pyramid is
× 8×3 =
x3. Since the base area of each pyramid is 4×2 it makes sense to write the volume as
Volume =
× 4×2 × x =
× area of the base × height.
We can extend this result to any pyramid by using a geometric argument, giving the following important result.
Volume of a pyramid =
× area of the base × height.
See the appendix on the pyramid for details.
EXERCISE 2
Find the volume of the Great Pyramid of Cheops whose height is 145.75 m and whose base is a square of side length 229 m. Give answer in cubic metres correct to two significant figures.
EXERCISE 3
Find the volume of the ‘diamond’, with height 24 cm and side length 10 cm as shown.
Lesson Two
How to Prove that the sum of the angles in a triangle is 180 degrees
Theorem
If ABC is a triangle then <)ABC + <)BCA + <)CAB = 180 degrees.
Proof
Draw line a through points A and B. Draw line b through point C and parallel to line a.
Since lines a and b are parallel, <)BAC = <)B’CA and <)ABC = <)BCA’.
It is obvious that <)B’CA + <)ACB + <)BCA’ = 180 degrees.
Thus <)ABC + <)BCA + <)CAB = 180 degrees.
Lemma
If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.
Proof
Assume to the contrary that AB and DC are not parallel.
Draw a line trough A and B and draw a line trough D and C.
These lines are not parallel so they cross at one point. Call this point E.
Notice that <)AEC is greater than 0.
Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.
Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.
Contradiction. This completes the proof.
Definition
Two Triangles ABC and A’B’C’ are congruent if and only if
|AB| = |A’B’|, |AC| = |A’C’|, |BC| = |B’C’| and,
<)ABC = <)A’B’C’, <)BCA = <)B’C’A’, <)CAB = <)C’A’B’.
Geometrical Construction
GEOMETRICAL CONSTRUCTION
When making a geometrical constructions, it is important to remember to do the following-
1. Use a hard pencil with a sharp point. This gives thin lines which are more accurate.
2. Check that your ruler has good straight edge. A damaged ruler is useless for construction work.
3. Check that your compasses are not too loose. Tighten loose compasses with a small screw driver.
4. All construction lines must be seen. Do not rub out anything which leads to the final result.
5. Always take great care, especially when drawing a line through a point.
6. Where possible, arrange that the angles of intersection between lines and arcs are about 900.
Perpendicular bisector of a line segment
The locus of a point which moves so that it is an equal distance from two points, A and B, is the perpendicular bisector of the line joining A and B.
Perpendicular means at right angles to.
Bisector means cuts in half.
To construct this locus, you do the following (try this yourself on a piece of paper):
Draw the line segment XY.
Put your compass on X and set it to be over half way along the line. Draw an arc.
Without adjusting your compass put it on Y and draw another arc.
Label these points A and B.
Draw a straight line through A and B.
The point M where the lines cross is the midpoint of XY. And AB is perpendicular to XY.
Bisecting an angle
V is the vertex of the angle we want to bisect.
Place your compass on V and draw an arc that crosses both sides of the angle.
Label the crossing points A and B.
Place your compass on A and draw an arc between the two sides of the angle.
Without adjusting your compass place it on B and draw another arc that cuts the one you just drew. Label the point where they cross C.
Draw a straight line through V and C.
The line VC bisects the angle. Angles AVC and BVC are equal.
Constructing a 900 Angle
We can construct a 90º angle either by bisecting a straight angle or using the following steps.
Step 1: Draw the arm PA.
Step 2: Place the point of the compass at P and draw an arc that cuts the arm at Q.
Step 3: Place the point of the compass at Q and draw an arc of radius PQ that cuts the arc drawn in Step 2 at R.
Step 4: With the point of the compass at R, draw an arc of radius PQ to cut the arc drawn in Step 2 at S.
Step 5: With the point of the compass still at R, draw another arc of radius PQ near T as shown.
Step 6: With the point of the compass at S, draw an arc of radius PQ to cut the arc drawn in step 5 at T.
Step 7: Join T to P. The angle APT is 90º.
Constructing a 300 Angle
We know that: ½ of 600 = 300
So, to construct an angle of 30º, first construct a 60º angle and then bisect it. Often, we apply the following steps.
Step 1: Draw the arm PQ.
Step 2: Place the point of the compass at P and draw an arc that passes through Q.
Step 3: Place the point of the compass at Q and draw an arc that cuts the arc drawn in Step 2 at R.
Step 4: With the point of the compass still at Q, draw an arc near T as shown.
Step 5: With the point of the compass at R, draw an arc to cut the arc drawn in Step 4 at T.
Step 6: Join T to P. The angle QPT is 30º.
Constructing a 600 Angle
We know that the angles in an equilateral triangle are all 60º in size. This suggests that to construct a 60º angle we need to construct an equilateral triangle as described below.
Step 1: Draw the arm PQ.
Step 2: Place the point of the compass at P and draw an arc that passes through Q.
Step 3: Place the point of the compass at Q and draw an arc that passes through P. Let this arc cut the arc drawn in Step 2 at R.
Step 4: Join P to R. The angle QPR is 600, as the ∆PQR is an equilateral triangle.
Try your understanding regarding the explanations above over and over again.
ASSESSMENT
- What are the important steps to take when doing geometric construction?
- Construct a 900 angle.
Triangle: Drawing and Bisection of Line Segment; Construction and Bisection of Angles
Drawing and Bisection of Line Segment
Bisection means to divide the line segment in two equal parts. In the real world, the majority of lines we see are line segments since they all have an end and a beginning. We can define a line segment as a line with a beginning and an end point. Below are some of instructions and procedures to follow when constructing a line segment-
Draw a line segment measuring 5 cm and its perpendicular bisector. Write the steps of construction.
Step 1 : Draw a line segment AB of length 5 cm.
Step 2 : Mark two arcs with radius more than half of AB with centers A and B respectively.
Step 3 : These arcs intersect at P and Q respectively.
Step 4 : Join P and Q.
Step 5 : PQ is required perpendicular bisector of line segment AB.
________________________________________________________________
Construction of a line perpendicular to a given line at a given point using Ruler and Compass
Draw any line segment AB . Mark any point P on it. Through P, draw a perpendicular to segment AB with the help of ruler and compasses.
Step 1: Given a point P on a line l.
Step 2: With P as center and a convenient radius, construct an arc intersecting the line l at two points A and B.
Step 3: With A and B as centers and a radius greater than AP construct two arcs, which cut each other at Q.
Step 4: Join PQ.
Thus PQ is perpendicular to l.
_________________________________________________________________
Construction of a line perpendicular to a given line and passing through a given point not lying on it by using ruler and compasses.
Draw any line segment AB . Take any point P outside it. Through P, draw a perpendicular to segment AB.
Step 1: Given a line l and a point P outside it.
Step 2: With P as center, draw an arc, which intersects line l at two points A and B.
Step 3: Using the same radius and with A and B as centers, construct two arcs that intersect at a point, say Q, on the other side.
Step 4: Join PQ.
Thus, segment PQ is perpendicular to line l.
_________________________________________________________________
Geometrical Constructions
• Basic Geometric Constructions
• Construction of Line Segment
• Bisecting a Line Segment
• Constructing Angles
• Bisecting Angles
• Constructing Parallel Lines
• Construction of Triangle (SSS)
• SAS Triangle Construction
• ASA Triangle Construction
• HL Triangle Construction (Rhs -construction)
• Constructing Quadrilaterals
• Constructing Triangles(when sum of sides or perimeter is given)
Construction and Bisect of Angles
| Definition The bisector of an angle is a ray whose end point is the vertex of the angle and which divides the angle into two equal angles.In the diagram to the right, the ray CD is the bisector of the angle ACB if and only if the angles ACD and BCD have equal measures. |
| Bisect Angle. To construct the Angle Bisector of an angle follow the following steps. | |
| Given. An angle to bisect. For this example, angle ABC. | |
| Step 1. Draw an arc that is centered at the vertex of the angle. This arc can have a radius of any length. However, it must intersect both sides of the angle. We will call these intersection points P and Q This provides a point on each line that is an equal distance from the vertex of the angle. | |
| Step 2. Draw two more arcs. The first arc must be centered on one of the two points P or Q. It can have any length radius. The second arc must be centered on whichever point (P or Q) you did NOT choose for the first arc. The radius for the second arc MUST be the same as the first arc. Make sure you make the arcs long enough so that these two arcs intersect in at least one point. We will call this intersection point X. Every intersection point between these arcs (there can be at most 2) will lie on the angle bisector. | |
| Step 3. Draw a line that contains both the vertex and X. Since the intersection points and the vertex all lie on the angle bisector, we know that the line which passes through these points must be the angle bisector. | |
Now, try to do this construction yourself.
Applet Instructions
- Drawing lines. Start by depressing the ruler button. Then, click on the point where the line should begin. You can then move the mouse to the other point and click again.
- Drawing arcs. Start by depressing the compass button. Click on the center of the arc. Use the up and down arrow keys to increase or decrease the angle of the arc (or use the method listed below). Click again to place the arc.
- Drawing arcs with same radius. If you hold down the “Shift” key when you select the first point of the arc the radius of your new arc will be same as that of last arc drawn.
- Selecting Items. Make sure both the ruler and compass are not depressed. Then, you can select items by clicking on them. The color of the marks will change from red to green. To deselect something click on it again.
- Adjusting lines and arcs. After placing a line or arc you can make adjustments to them. First, selecting the object you want to change. Then, by clicking (not holding down mouse button) on different points you can make different adjustments.
- Lines. Clicking on either endpoint of a line will release that point and thus allow you to move.
- Arcs. When you select an arc four points will be drawn in addition to the arc. By clicking on each of these points you can modify a different aspect of the arc.
- Center of the circle containing the arc. You can move the center while leaving the center of the curve in the same place. If you hold down the shift key while moving this point, the entire arc will move and keep the same relative position to the center. Be careful not to move the curve off of the screen.
- Center of the curve. You can move the position of the curve while leaving the center of the circle in the same place. If you hold down the shift key while you are moving, the radius is remain constant.
- Ends of the curve. You can adjust the length of the curve.
The best way to see what can be adjusted is to just try things.
- Starting Over. To start over click the reset button.
- Checking construction. To check your construction select the “Check” button. A message indicating the first error that is found or that the construction is correct will be shown in the window to the right.
ASSESSMENT
- Define bisection
- What is the bisector of an angle?
Construction: Construction of Quadrilateral Polygon; Construction of Equilateral Triangle; Locus of Moving Points
Topic: Construction of Quadrilateral Polygon
We can identify different quadrilaterals based on the properties sides, diagonals and angles.
Quadrilaterals are made up of ten parts. However, to construct them, you do not need to know the measurements of all of them.
In case of special quadrilaterals, like the rectangle, just two measurements, the lengths of its adjacent sides are enough to construct it.
A kite can be constructed if the lengths of its distinct adjacent sides and one diagonal are known.
Similarly, a square can be constructed with just the length of its side, while a rhombus can be constructed when the lengths of its diagonals are known.
Steps to Construct a Rectangle:
Step 1: Draw a side of given length (say) CL
Step 2: Draw side LU (say) of given length perpendicular to CL at L.
Step 3: Draw side CE (say) of length equal to LU and perpendicular to CL at C.
Step 4: Draw side UE.
Steps to Construct a Kite:
Step 1: Draw diagonal (say) AY and its perpendicular bisector.
Step 2: Draw sides say AL and AZ of given length.
Step 3: Draw sides LY and YZ.
Constructing a Quadrilateral when four sides and one of its diagonals are given.
Step 1: Construct a triangle ABD (say).
Step 2: Find point C opposite to the vertex A as follows. With B as the centre and given radius, draw an arc on the other side of BD. Similarly with D as the centre and given radius, draw another arc intersecting the previous arc. The point of intersection of these arcs is marked as C.
Step 3: join points B and C, and D and C.
Constructing a quadrilateral when lengths of its three sides and two diagonals are given.
Step 1: Construct a triangle PQR (say).
Constructing a quadrilateral when lengths of its adjacent sides and three angles are given.
Step 1: Draw a line segment of EF (say) of given length.
Step 2: Construct a given angle at E.
Step 3: Construct a given angle at F.
Step 4: Locate point G.
Step 5: Locate point H.
Constructing a quadrilateral when lengths of its three sides and two included angles.
Step 1: Draw a line segment MN (say) of given length.
Step 2: Construct a given angle at M.
Step 3: Construct an angle 90° at N.
Step 4: Locate vertices L and O.
Step 5: Join L and O.
Constructing an equilateral triangle
Constructing an equilateral triangle also known as drawing an equilateral triangle using only a straightedge and a compass is what I will show you here
Step #1:
Take your ruler and a pencil and construct a segment of any length on a piece of paper as shown below
Then, you will try to set your compass opening to match the length of segment AB
Take your compass. Make your sure that the pencil is included in it.
Put the needle of the compass at endpoint A and adjust your compass so that the tip of your pencil touches endpoint B
Step #2:
Put the needle of your compass at A and draw an arc
Put the needle of your compass at B and draw an arc
The two arcs should meet as shown below:
Step #3:
Draw the segments from the two endpoints to the point where the two arcs intersect
Locus of Moving Points Including Equidistance from Two Lines of Two Points and Constant Distant From the Point
When a point moves in a plane according to some given conditions the path along which it moves is called a locus. (Plural of locus is loci.).
CONDITION 1 :
A point P moves such that it is always m units from the point Q
Locus formed: A circle with centre Q and radius m.
Example :
Construct the locus of a point P at a constant distance of 2 cm from a fixed point Q.
Solution:
Construct a circle with centre Q and radius 2 cm.
CONDITION 2:
A point P moves such that it is equidistant from two fixed points X and Y
Locus formed: A perpendicular bisector of the line XY.
Example:
Construct the locus of point P moving equidistant from fixed points X and Y and XY= 6 cm.
Solution:
Construct a perpendicular bisector of the line XY.
CONDITION 3:
A point P moves so that it is always m units from a straight line AB
Locus formed: A pair of parallel lines m units from AB.
Example:
Construct the locus of a point P that moves a constant distant of 2 cm from a straight line AB.
Solution:
Construct a pair of parallel lines 2 cm from AB.
CONDITION 4:
A point P moves so that it is always equidistant from two intersecting lines AB and CD
Locus formed: Angle bisectors of angles between lines AB and CD.
Example:
The following figure shows two straight lines AB and CD intersecting at point O. Construct the locus of point P such that it is always equidistant from AB and CD.
Solution:
Construct angles bisectors of angles between lines AB and CD.
ASSESSMENT
- Define a quadrilateral
- Construct a quadrilateral
Deductive Proof: Sum of Angles of a Triangle; Revision of Angles on Parallel Line cut by a Transversal Line
Angles can be Added
Just like regular numbers, angles can be added to obtain a sum, perhaps for the purpose of determining the measure of an unknown angle. Sometimes we can determine a missing angle because we know that the sum must be a certain value. Remember — the sum of the degree measures of angles in any triangle equals 180 degrees. Below is a picture of triangle ABC, where angle A = 60 degrees, angle B = 50 degrees and angle C = 70 degrees.
If we add all three angles in any triangle we get 180 degrees. So, the measure of angle A + angle B + angle C = 180 degrees. This is true for any triangle in the world of geometry. We can use this idea to find the measure of angle(s) where the degree measure is missing or not given.
Draw a triangle ABC and cut out the three angles.
Step 2:
Rearrange the three angles to form a straight angle on a straight line.
So, the angle sum of a triangle is 180°.
Task 2
To investigate if this works for all triangles, repeat the above process for four different triangles.
Task 3
Copy and complete the following statements:
a. The three angles of a triangle can be arranged to form a ……….. angle.
b. The sum of three angles in a triangle is ………..
Finding the Third Angle of a Triangle
If the measurements of two angles of a triangle are known, then the third angle can be calculated.
Example 3
Calculate the size of the missing angle in the following triangle.
Solution:
Let the missing angle be x
.
So, the missing angle is 80°.
Example 4
Find the values of the pronumerals x and y in the following diagram:
Solution:
Example 5
Find the value of the pronumeral x in the following diagram:
Solution:
Example 6
Find the value of the pronumeral x in the following diagram:
Solution:
Example 7
Find the values of the pronumerals x and y in the following diagram:
Solution:
Example 8
Find the value of the pronumeral x in the following diagram:
Solution:
Revision of Angles on Parallel Line cut by a Transversal Line
| A Line Crosses A Pair of Parallel Lines If a set of 2 parallel lines, line l and line m, are crossed or cut by another line, line n, we say “a set of parallel lines are cut by a transversal.” Each of the parallel lines cut by the transversal has 4 angles surrounding the intersection. These are matched in measure and position with a counterpart at the other parallel line. At each of the parallel lines, there are two pairs of vertical angle. Each angle in the pair is congruent to the other angle in the pair. 1 4, angle 1 is congruent to angle 4. 3, angle 2 is congruent to angle 3. 5 8, angle 5 is congruent to angle 8. 6 , angle 6 is congruent to angle 7. |
| Adjacent Supplementary Angles At each of the parallel lines adjacent angles are supplementary. |
| Names for the Matched Angles The angles have special names identifying their positions with respect to the parallel lines and transversal. They are corresponding angles, alternate interior angles, or alternate exterior angles. An angles is congruent to its matched angle. |
ASSESSMENT
- What is the total angles in a triangle?
Collection, Tabulation and Presentation of Grouped Data
Collection, Tabulation and Presentation of Grouped Data
In some investigations you may collect an awful lot of information. How can you use this raw data and make it meaningful? This section will help you to collect, organise and interpret the data efficiently.
Explaining your results
Imagine that you are asked to carry out a survey to find the number of pets owned by pupils in your school. You decide to ask 50 people, and record your results as follows:
0 2 1 2 0 4 1 0 2 2 1 6 1 1 2 8 0 12
2 1 2 0 3 2 0 1 3 0 1 4 0 3 0 2 3 6
3 3 0 1 2 0 1 1 3 0 2 0 3 2
You now have the information you need, but is this the most efficient way to collect and display the data?
Tallying
Tallying is a method of counting using groups of five.
|= 1
||= 2
||| = 3
|||| = 4
Because we have used groups of five, it is easy to find the total.
Question
Which numbers do the following tally marks represent?
Answer
Using the tally system to record results is much faster than writing out words or figures all the time. For example, if you had to investigate the most popular type of vehicle that passed the school gates, it would be easier to draw tally marks in one of three columns than write: car, car, lorry, bike, car, car, and so on.
By using a tally chart, the data is already collected into groups, and will not require further grouping at a later date.
Collecting data
The easiest way to collect data is to use a tally chart.
When collecting data for the number of pets survey, it would have been useful to draw a table similar to this one.
As each person answers the question, we put a tally next to the appropriate number of pets. The frequency column is completed once all of the data has been collected. The table below shows the results of a new pets survey.
These frequencies can be displayed in a bar chart, as shown.
Frequency means the ‘number of times it occurs’.
In this example, three people had no pets, so the frequency of 0 pets was three.
Remember that the total frequency should be the same as the number of people in your survey. Always check that this is correct.
In this example, we know that 26 people were questioned in the survey. Check this by adding up the frequency totals: 3 + 8 + 12 + 1 + 2 = 26
Here is the same information but this time we have two tables, one for the number of pets owned by boys and one for the number of pets owned by girls.
Number of pets owned by boys
Number of pets owned by girls
These frequencies can be displayed in a dual bar chart.
We can find more information from looking at this graph.
Question
How many pets were owned by the same number of boys as for girls?
Answer
Four pets.
We can see that the height of the bar is the same for both boys and girls for four pets.
Question
How many more girls than boys were there in the survey?
Answer
Two more girls.
Adding the bars for girls and for boys we find:
the total for girls is: 1 + 5 + 7 + 0 + 1 = 14
the total boys is: 2 + 3 + 5 + 1 + 1= 12
Grouping data
When a large amount of data has to be collected, use a grouped frequency distribution.
The following tally chart represents the ages of 200 people entering a park on a Saturday afternoon.
The ages have been grouped into the classes 0-9, 10-19, 20-29, and so on.
| Age | Tally | Frequency |
| 0-9 | III | 8 |
| 10-19 | II | 12 |
| 20-29 | IIII | 24 |
| 30-39 | III | 43 |
| 40-49 | I | 41 |
| 50-59 | II | 27 |
| 60-69 | III | 23 |
| 70-79 | III | 18 |
| 80-89 | III | 3 |
| 90-99 | I | 1 |
These frequencies can also be shown in a histogram.
ASSESSMENT
- What is tallying?
- What is frequency?
Calculation of Range, Median and Mode of Ungrouped Data
Range of Ungrouped Data
The Range is the difference between the lowest and highest values.
Example: In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9.
So the range is 9 − 3 = 6.
It is that simple!
But perhaps too simple …
The Range Can Be Misleading
The range can sometimes be misleading when there are extremely high or low values.
Example: In {8, 11, 5, 9, 7, 6, 3616}:
- the lowest value is 5,
- and the highest is 3616,
So the range is 3616-5 = 3611.
The single value of 3616 makes the range large, but most values are around 10.
So we may be better off using Interquartile Range or Standard Deviation.
| Range of a Function Range can also mean all the output values of a function, see Domain, Range and Codomain. |
| Problem: | Cheryl took 7 math tests in one marking period. What is the range of her test scores? | |
| 89, 73, 84, 91, 87, 77, 94 | ||
| Solution: | Ordering the test scores from least to greatest, we get: | |
| 73, 77, 84, 87, 89, 91, 94 | ||
| highest – lowest = 94 – 73 = 21 | ||
| Answer: | The range of these test scores is 21 points. | |
| Definition: | The range of a set of data is the difference between the highest and lowest values in the set. | |
| In the problem above, the set of data consists of 7 test scores. We ordered the data from least to greatest before finding the range. We recommend that you do this, too. This is especially important with large sets of data. Let’s look at some more examples. |
| Example 1: | The Jaeger family drove through 6 midwestern states on their summer vacation. Gasoline prices varied from state to state. What is the range of gasoline prices? | |
| N1.79, N1.61, N1.96, N2.09, N1.84, N1.75 | ||
| Solution: | Ordering the data from least to greatest, we get: | |
| N1.61, N1.75, N1.79, N1.84, N1.96, N2.09 | ||
| highest – lowest = N2.09 – N1.61 = N0.48 | ||
| Answer: | The range of gasoline prices is $0.48. | |
| Example 2: | Ms. Kaiser listed 9 integers on the blackboard. What is the range of these integers? 14, –12, 7, 0, –5, –8, 17, –11, 19 | |
| Solution: | Ordering the data from least to greatest, we get: –12, –11, –8, –5, 0, 7, 14, 17, 19 | |
| highest – lowest = 19 – –12 = 19 + +12 = +31 | ||
| Answer: | The range of these integers is +31. |
| Example 3: | A marathon race was completed by 5 participants. What is the range of times given in hours below? | |
| 2.7 hr, 8.3 hr, 3.5 hr, 5.1 hr, 4.9 hr | ||
| Solution: | Ordering the data from least to greatest, we get: | |
| 2.7, 3.5, 4.9, 5.1, 8.3 | ||
| highest – lowest = 8.3 hr – 2.7 hr = 5.6 hr | ||
| Answer: | The range of swim times is 5.6 hr. | |
| Summary: | The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set. |
Exercises
| Directions: Find the range of each set of data. Click once in an ANSWER BOX and type in your answer; then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR. |
| 1. | Find the range of these distances run by 6 marathon runners: 10 km, 15 km, 12 km, 14 km, 8 km, 16 km |
ANSWER BOX: km RESULTS BOX:
| 2. | Find the range of these quiz scores: 93, 79, 83, 89, 90, 71, 85 |
ANSWER BOX: RESULTS BOX:
| 3. | Find the range of these race times given in seconds: 7.3 s, 8.4 s, 8.0 s, 7.5 s, 9.4 s, 8.7 s, 9.1 s |
ANSWER BOX: s RESULTS BOX:
| 4. | Employees at a retail store are paid the hourly wages listed below. What is the range of these hourly wages? N7.50, N9.25, N8.75, N9.50, N7.25, N8.50 |
ANSWER BOX: N RESULTS BOX:
| 5. | In a game, points were won and lost, represented by integers. What is the range of points given below? –14, +21, –17, +25, 0, –19, +11, –20, +18 |
ANSWER BOX:
Median of Ungrouped Data
Median is defined as the mid value of the data set. It is a value that falls in the middle-most position of the whole data. Median of an ungrouped data is determined by arranging the given numbers in ascending order and then selecting exactly middle value. In other words, the median is the value that divides the observations (in ascending order) into two equal divisions. The median is a kind of measures of central tendency.
The formulas for calculating the median of an ungrouped data which has total “n” number of observations arranged in increasing order, are:
Case I: When n is odd
Median = [Math Processing Error](n+12)th observation
Case II: When n is Even
Median = [Math Processing Error](n2)th [Math Processing Error]observation + ([Math Processing Error]n2 + 1)[Math Processing Error]th observation
Mode of Ungrouped Data
The mode is defined the value that most frequently occurs in the given data; i.e. the number whose frequency is more than others, is called the mode. It is usually denoted by “Z”.
In order to find the mode of an ungrouped data, we have to find the frequency of each number in the given data set. Then, we have to choose the number having the highest frequency as the mode. The mode is also one of the three measures of central tendency. We can write as:
Mode = Value with highest frequenc
ASSESSMENT
- Define Range
- Define Mode
- Define Median
Mean deviation, Variance and Standard Deviation
Mean Deviation
This is the average sum of the deviations from the arithmetic mean (i.e. the sum of the differences between the scores and the mean) divided by the total frequency.
Example 1
Find the mean deviation in the following test scores,
| Score (x) | Frequency (f) |
| 95 | 1 |
| 85 | 1 |
| 80 | 1 |
| 75 | 4 |
| 70 | 1 |
| 65 | 3 |
| 55 | 1 |
| 40 | 3 |
Solution
The mean of the score X = 66.7
| 1 |
Scores (X)2
Frequency (f)3
X – X҄ = x = d4
Fx = fd95195-66.7 = 28.31 X 28.3 = 28.385185-66.7 = 18.31 X 18.3 = 18.380180-66.7 = 13.31 X 13.3 = 13.375475-66.7 = 8.34 X 8.3 = 33.270170-66.7 = 3.31 X 3.3 = 3.365365-66.7 = -1.73 X (-1.7) = -5.155155-66.7 = -11.71(-11.7) = -11.740340-66.7 = -26.73(-26.7) = -80.1 ∑F = 15 ∑FD = -0.5
So we subtract each score (X) from the mean (66.7) as shown in column above (3). Hence column (3) shows the deviation of each score form the mean. Column (4) shows the product of the deviations and their frequencies which we then sum as ∑fd. Hence the mean deviation is given by the formula ∑fd/∑f
where ∑fd = the sum of the products of deviations and the frequencies
∑f = total frequency
Mean deviation = ∑fd/∑f = -05/15 = -0.0333
We always take the positive sign regardless of the sign. The value of the mean deviation calculated above is so small that it is ignored usually in computing measures of dispersion which is not the true situation. So we talk about the Mean Absolute Deviation to give the true picture.
Mean Absolute Deviation
From the above mean deviation, the negative sign in column 3 were not ignored so we had an insignificant value for the mean deviation. This leads to the finding of the mean absolute deviation where the modulus (or absolute) of the values are considered in the computation.
Mean absolute deviation is defined as the arithmetic mean of the absolute (modulus) value of the difference of each score form the mean (that is adding the values of the deviations from the mean as if they were all positive) and dividing by their total frequency. That is the sum of the modulus of the deviations from the arithmetic mean divided by the total frequency. Hence the mean absolute deviation is given by the formula
MD = ∑f|x|/∑f or ∑f|d|/∑f
where |x| or |d| = modulus of the deviations
∑f|x| or ∑f|d| = sum of the product of the modulus of the deviation and their frequencies
∑f = total frequency
Standard Deviation
The standard deviation is the most important of all the measures of dispersion. Equally important is the variance which is the square of standard deviation. The standard deviation tells us how far or near a score is to the mean score. The standard deviation is denoted with s when we consider only a sample of a group to study or with σ (sigma) when we study the entire group. At times it is shortened as S.D. meaning standard deviation.
The formula for its computation is as follows:
S.D = √∑fd2/∑f or √∑fx2/∑f or √∑x2/N
and variance usually denoted by
s2 or σ2 = ∑fd2/∑f or ∑fx2/∑f
In other words S.D. = √v
where ∑f = sum of frequencies
N = sum of the scores, if each score occurs once only
d2 or x2 = square of the deviations from the area
i.e. d2 or x2 = (X – X҄)2, where X = score and X҄ = arithmetic mean.
Examples
Compute (i) the variance and (ii) the standard deviation of the following test scores:
| Score | 95 | 85 | 80 | 75 | 70 | 65 | 55 | 40 |
| Frequency | 1 | 1 | 1 | 4 | 1 | 3 | 1 | 3 |
Solution
The scores can be arranged in a frequency table as follows:
| (1) |
Scores (X)(2)
Derivation from the mean d = X – X҄ = x(3)
Frequency (f)(4)
fd2 = f(X – X҄)29595-66.7 = 28.311 x (28.3)2 = 806.898585-66.7 = 18.311 x (18.3)2 = 334.848080-66.7 = 13.311 x (13.3)2 = 176.897575-66.7 = 8.344 x (8.3)2 = 275.567070-66.7 = 3.311 x (3.3)2 = 8.696565-66.7 = -1.733 x (-1.7)2 = 8.675555-66.7 = -11.711(-11.7)2 = 136.894040-66.7 = -26.733(-26.7)2 = 2138.67
In the above column (2) we use the value of the already computed mean X҄ = 66.6, hence so applying the formula for computing the standard deviation, √∑fd2/∑f = √∑(X – X҄)2/∑f
where √∑fd2 = 3883.35 and √∑f = 15
we have variance = 3883.25/15 = 258.89
SD = -√258.89 = 16.09
S the standard deviation for the rest scores is 16.09 while the variance, which is the square of the standard deviation = 258.89 i.e. (16.09)2= 258.89 i.e. (S.D)2 = variance.
Question
| Score | 50-54 | 55-59 | 60-64 | 65-69 | 70-74 | 75-79 | 80-84 | 85-89 | 90-94 |
| Frequency | 3 | 5 | 8 | 10 | 7 | 6 | 3 | 2 | 1 |
Calculate for questions 1 – 5 using the table above
1. Determine ∑f
A. 45 B. 50 C. 55 D. 60
2. ∑fd, if d = X – A
A. -160 B. -167 C. -155 D. -201
3. Mean X҄, as X҄ = A + ∑fd/∑f
A. 70.05 B. 68.56 C. 23.55 D. 65.75
4. What is the Variance?
A. 97.03 B. 87.75 C. 92.03 D. 50.08
5. What is the Standard Variation?
A. 12.54 B. 9.59 C. 25.75 D. 32.23
Answers
1. A.
2. C
3. B
4. C
5. B