Topic 2: Area and Perimeter - Basic Mathematics Form Four

Topic 2: Area and Perimeter – Basic Mathematics Form Four

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Basic Mathematics For Form Four Full Notes, LINEAR PROGRAMMING. Linear programming – is a branch of mathematics which deals with either minimizing the cost or maximizing the profit. It gives the best way of utilizing the scarce resources available. It is so called because it only involves equations and inequalities which are linear. Simultaneous Equation. One of the methods used in solving linear simultaneous equations is a graphical method. Two linear simultaneous equations in two unknowns can be graphically solved by passing through the following procedures. Draw the two lines which represent the two equations on the xy – plane this is done by deter mining at least two points through which each line passes, the intercept are commonly used Determine the point of intersection of the two lines. This point of intersection is the solution to the system of equations. FACT: If two straight lines are not parallel then they meet at only one point: In case the lines do not meet, there is no solution to the corresponding system of simultaneous equations. Example 1 Graphically solve the following system of simultaneous equations. Example 2 Find the solution to the following system of simultaneous equations by graphical method. Solving Simultaneous Equations Graphically Solve simultaneous equations graphically Example 3 Solve the following simultaneous equations graphically and check your solution by a non-graphical method: Example 4 Find the solution to the following system of simultaneous equations by graphical method. Exercise 1 Find the solution to the following systems of simultaneous equations graphically. Try: Ali paid 34 shillings for 10 oranges and 35 mangoes. Moshi went to the same market and paid 24 shillings for 16 oranges and 18 mangoes. What was the price for a mango and for an orange? Inequalities Forming Linear Inequalities in Two Unknowns from Word Problems Form linear inequalities in two unknowns from word problems Linear inequalities Normally any straight line drawn on xy – plane separates it into two disjoint sets. These sets are called half – planes Consider the equation y = 5 drawn on the xy plane as shown below. From the figure above, all points above the line, that is all points in the half plane A which is above the line satisfy the relation y>5 and those lying in the half plane B which is below the given line, satisfy the relation y< 5. Shading of Regions In linear programming usually the region of interest is left clear that is we shade unwanted region(s). NB: When shading the half planes we consider the inequalities as the equations but dotted lines are used for the relations with > or < signs and normal lines are used for those with ≥ or ≤ signs. Consider the inequalities x>0, y>0 and 2x + 3y >12 represented on the xy-plane In this case we draw the line x=0, y= 0 and 2x+3y=12 but the point about the inequality signs for each equation must be considered. From the figure above, the clear region satisfy all the inequalitiesx>0, y>0 and 2x + 3y >12, these three lines are the boundaries of the region. The Solution Set of Simultaneous Linear Inequalities Graphically Find the solution set of simultaneous linear inequalities graphically Example 5 Draw and show the half plane represented by 8x + 2y ≥16 Feasible Region Definition: In the xy plane the region that satisfies all the given inequalities is called the feasible region (F.R) Example 6 Indicate the feasible region for the inequalities 2x+3y ≥ 12 and y-x ≤ 2. Determine the solution set of the simultaneous inequalities y + x ≥3 and x-2y ≤ 9. Example 7 Fatuma was given 30 shillings to buy oranges and mangoes. An orange costs 2 shillings while a mango costs 3 shillings. If the number of oranges bought is at least twice the number of mangoes, show graphically the feasible region representing the number of ranges and mangoes she bought, assuming that no fraction of oranges and mangoes are sold at the market. Solution:- Let x be the number of oranges she bought and y the number of mangoes she bought. Now the cost of x and y together is 2x + 3y shillings which must not exceed 30 shillings. Inequalities: 2x + 3y ≤30 ……… (i) and x≥2y …………….. (ii), Also because there is no negative oranges or mangoes that can be bought, then x≥ and y≥0 ……….. (iii) Now the line 2x + 3y ≤30 is the line passing through (0, 10) and (15,0) and the line x≥2y or x – 2y ≥ 0 is the line which passes through (0,0) and (2,1). Exercise 2 For practice. Draw the graph of the equation 2x – y = 7 and show which half plane is represented by 2x – y >7 and the one represented by 2x – y 3 – x on the same axes and indicate the feasible region. A post office has to transport 870 parcels using a lorry, The Objective Function An Objective Function from Word Problems Form an objective function from word problems Linear programming components Any linear programming problem has the following: Objective Alternative course (s) of action which will achieve the objective. The available resources which are in limited supply. The objective and its limitations should be able to be expressed as either linear mathematical equations or linear inequalities. Therefore linear programming aims at finding the best use of the available resources. Programmingis the use of mathematical techniques in order to get the best possible solution to the problem Steps to be followed in solving linear programming problems; Read carefully the problem, if possible do it several times. Use the variables like x and y to represent the resources of interest. Summarize the problem by putting it in mathematical form using the variables let in step (b) above. In this step you need to formulate the objective function and inequalities or constraints. Plot the constraints on a graph From your graph, identify the corner points. Use the objective function to test each corner point to find out which one gives the optimum solution. Make conclusion after finding or identifying the optimum point among the corner points. Maximum and Minimum Values Corner Points on the Feasible Region Locate corner points on the feasible region Example 8 A student has 1200 shillings to spend on exercise books. At the school shop an exercise book costs 80shillings, and at a stationery store it costs 120 shillings. The school shop has only 6 exercise books left and the student wants to obtain the greatest number of exercise books possible using the money he has. How many exercise books will the student buy from each site? Therefore the student will buy 6 exercise books from each site. Example 9 A nutritionist prescribes a special diet for patients containing the following number of Units of vitamins A and B per kg, of two types of food f1 and f2 If the daily minimum in take required is 120 Units of A and 70 units of B, what is the least total mass of food a patient must have so as to have enough of these vitamins? Solution: Let x be the number of kg(s) of F1 that patient gets daily and y be the number of kg(s) of F2 to be taken by the patient daily. Objective function: F (x, y) = (x + y) minimum f (C) = 10 + 0 = 10 So f (B) = 6.8 is the minimum Therefore the least total mass of food the patient must have is 6.8 kilograms The Minimum and Maximum Values using the Objective Functio Find the minimum and maximum values using the objective function Example 10 A farmer wants to plant coffee and potatoes. Coffee needs 3 men per hectare while potatoes need also 3 men per hectare. He has 48 hired laborers available. To maintain a hectare of coffee he needs 250 shillings while a hectare of potatoes costs him 100 shillings. . Find the greatest possible land he can sow if he is prepared to use 25,000 shillings. Solution: Let x be the number of hectares of coffee to be planted and y be the number of hectares of potatoes to be planted. Objective function: f (x, y) = (x, + y) maximum 3x + 3y ≤ 48 or x + y ≤16 ………….(i) 250x + 100y≤ 25,000 Or 5x + 2y ≤ 500………(ii) x ≥ 0 ……………………(iii) y≥ 0 ……………………(iv) Using the objective function f (x, y) = (x + y) maximum, f (A) = (0 + 250) = 250 f (B) = (0+16) = 16 f (C) = (16+0) = 16 f(D) = (100+0)= 100 (maximum) Therefore the greatest possible area to be planted is 250 hectors of potatoes. NB: In most cases L.P problems must involve non-negativity constraints (inequalities) that are x ≥ 0 and y ≥ 0. This is due to the fact that in daily practice there is no use of negative quantities. Example 11 A technical school is planning to buy two types of machines. A lather machine needs 3m2 of floor space and a drill machine needs 2m2 of floor space. The total space available is 30m2. The cost of one lather machine is 25,000 shillings and that of drill machine is 30,000 shillings. The school can spend not more than 300,000 shillings, what is the greatest number of machines the school can buy? Solution: Let x be the number of Lather machines and y be the number of drill machines to be bought Objective function: f(x, y) = (x + y) max Inequalities: 3x + 2y ≤ 30.. ………………….(i) 25,000x + 30,000y ≤300,000 Or 5x + 6y ≤ 60……………………..(ii) x ≥ 0 ……………………………….(iii) y ≥ 0…………………… ………….(iv) Since the incomplete machine can’t work, then B = (8, 3) or (7, 4).That is approximating values of x and y to the possible integers without affecting the given inequalities or conditions. Now by using the objective function, f (A) = 0 + 10 = 10 f(B) = 7 + 4 0r f (B) = 8 + 3 = 11 f (C) = 10 + 0 = 10 f (D) = 0 + ) = 0 So f (B) gives the maximum number of machines which is 11. Therefore the greatest number of machines that can be bought by the school is 11 machines. Exercise 3 1. Show on a graph the feasible region for which the restrictions are: y ≤ 2x, x≥ 6, y≥2 and 2x + 3y ≤30 From the graph at which point does: y – x take a maximum value? x + y take a maximum value? y – x take a maximum value? 2. With only 20,000 shillings to spend on fish, John had the choice of buying two types of fish. The price of a single fish type 1 was 2,500shillings and each fish of type 2 was sold at 2,000 shillings. He wanted to buy at least four of type 1. What is the greatest number of fish did John buy? How many of each type could he buy? 3. How many corner points does the feasible region restricted by the inequalities? x≥0, y ≥ 0, 3x + 2y ≤ 18 and 2x + 4y ≤16 have? Which corner point maximizes the objective function f (x, y) = 2x + 5y?, Matrices and Transformation, Topic 6: Vectors - Basic Mathematics Form Four, Trigonometry, Probability, Three Dimensional Figures, Area and Perimeter, Coordinate Geometry, Statistics, Similarity
   

Topic 2: Area and Perimeter – Basic Mathematics Form Four

Area of any Triangle

The Formula for the Area of any Triangle

Derive the formula for the area of any triangle

Area of triangle is given by½bh, whereby b is the base of the triangle and h is the height of the given triangle. Consider the illustrations below:

040523 1524 TOPIC2AREAA1

From the figure above, we see where the base and height are located.

Applying the Formula to find the Area of any Triangle

Apply the formula to find the area of any triangle

Example 1

The base of a triangle is 12cm long. If the corresponding height is 7cm, find the area of the triangle.

Solution

Consider the figure below:

040523 1524 TOPIC2AREAA2

The area of a triangle is given by½bh.

Area =½×12cm×7cm

Area = 42cm2

Therefore area of a triangle is 42cm2

Example 2

The lengths of two sides of a triangle are 6cm and 8cm. Find the area of a triangle if the included is

Solution:

Consider the triangle below, name it triangle ABC.

040523 1524 TOPIC2AREAA3

The area of a triangle above is given by½b×h

So, Area = ½× 8cm× 6cm × sin 45°

=24cm2× sin45°

= 16.97cm2

Therefore the area of ABC = 16.97cm2

Area of a Rhombus

The Formula for Finding the Area of Rhombi in Terms of the Diagonals

Derive the formula for finding the area of rhombi in terms of the diagonals.

The area of a rhombus is the same as the area of a parallelogram because rhombus is a special kind of parallelogram. Rhombus is a parallelogram with equal sides. Consider the figure below of a rhombus with base b and height h.

040523 1524 TOPIC2AREAA4

So, Area of rhombus = bh

Another formula for finding the area of rhombus can be obtained by using the diagonals. Consider the rhombus below:

040523 1524 TOPIC2AREAA5

Diagonals of a rhombus bisect each other at right angles (means the diagonal lines are half equally), so the area of a rhombus ABCD can be found as follows:

Area of a triangle ABC = area of a triangle ADC

040523 1524 TOPIC2AREAA6

Since the triangles are equal, the area of ABCD:

040523 1524 TOPIC2AREAA7

Therefore, the area of a rhombus is equal to half the product of the length of the diagonals.

The Formula to Find the Areas of Quadrilaterals

Apply the formula to find the areas of quadrilaterals

Area of a Trapezium

Consider the trapezium with constructed lines as shown in the figure below:

040523 1524 TOPIC2AREAA8

In order to find the area of a trapezium, first let us find the area of the triangles ABD and BDF with the same height h.

The base of the triangle ABD is b1

The base of the triangle BDF is b2

The area of the triangle ABD = ½b1h

The area of the triangle ABE = ½b1h

Area of ABDF =½b1h +½b2h

Area of ABDF=½ (b1 + b2)h

Generally, the area of the trapezium is given by½ (b1+ b2)hor is the product of half the sum of the parallel sides (bases) and the perpendicular distance between them (height)

Example 3

Find the height of the trapezium with area 90 square units and bases of 6 units and 14 units

Solution:

Consider the trapezium below:

040523 1524 TOPIC2AREAA9

Therefore the height of the trapezium is 9 units.

Area of a Parallelogram

Consider parallelogram below with constructed lines as shown in the figure.

040523 1524 TOPIC2AREAA10

The area of the parallelogram can be formed from the formula for the area of the trapezium. The important thing to note is that the bases for a parallelogram are equal.

Now, the area of the parallelogram ABCD=½ (AB+DC)hsquare units.Since AB = DC

Then, the area of parallelogram ABCD =½ × 2AB × h = AB× h

If AB =b, then Area of the parallelogram ABCD = bh, where b is the base and h is the height of a parallelogram.

Therefore; the area of a parallelogram is equal to the product of the base and the perpendicular height.

Area of a Rectangle

Consider the rectangle below:

040523 1524 TOPIC2AREAA11

The rectangle ABCD is divided into two congruent triangles, which are triangle ABD and triangle ACD by the diagonal AD.

The area of ABCD = area of triangle ABD + area of triangle ACD

040523 1524 TOPIC2AREAA12

Since the triangles are equal, the area of ABCD is equals to double the area of one of the triangles.The area of ABCD

040523 1524 TOPIC2AREAA13

If CD is the length ‘l’of the Rectangle and AC is the width ‘w‘ of the Rectangle, then, the area of ABCD = l×w or lw

Therefore, the area of the Rectangle is the product of the length and width.

Area of a Square

A Square is a special rectangle with equal sides. Therefore the area of the square is the product of its lengths. i.e. Area of a square = l × l=l2.

We can also find the area of a square by using the length of the diagonals. Consider the square below with diagonals AC and DB:

040523 1524 TOPIC2AREAA14

Each of the diagonals of a square bisect at a right angle.Area of a triangle ABC = Area of ADC

040523 1524 TOPIC2AREAA15

But the area of ABCD

040523 1524 TOPIC2AREAA16

Since the length of the diagonals are equal, then AC = BD.So, the area of ABCD = ½ (AC)2

Therefore the area of a Square is equals to the half of the product of the lengths of the diagonals.

Example 4

Find the area of a parallelogram ABCD if AC = 7cm, AB = 9cm and the angle ZWX =.

040523 1524 TOPIC2AREAA17

Solution

Area of a parallelogram = base×height

Consider the figure below:

040523 1524 TOPIC2AREAA18

Consider the triangle ACE to find h

040523 1524 TOPIC2AREAA19

Therefore the area of the parallelogram ABCD is 53.427cm2.

Perimeter of a Regular Polygon

The Formula for Finding the Length of a Side of a Regular Polygon

Derive the formula for finding the length of a side of a regular polygon

When we sum up the lengths of the sides of the polygon we obtain what is called perimeter of a polygon. Therefore, perimeter of a regular polygon is the sum of the lengths of the sides of the polygon.

How to find the perimeter of a Regular Polygon inscribed in a circle?

An inscribed polygon is the one whose vertices lie on the circle. If the lengths of the sides of the polygon are the same we say that the polygon is an inscribed Regular Polygon.

A Regular Polygon with number of sides larger than 2 say n sides can be inscribed in a circle as follows:

For example, if you want to construct an inscribed regular hexagon (6 sides), first draw a circle and locate the center of the circle. Then draw rays that intersect the circle in six points from the center of the circle. Each angle at the center will measure 360°/6 = 60°. Connect the points of intersection on the circle by line segments. The figure formed is an inscribed regular polygon. See the figure below:

040523 1524 TOPIC2AREAA20

Now, to obtain the formula of finding the perimeter of a regular polygon inscribed in a circle with radius r and center O, let AB be the side of the polygon and OC the perpendicular from O to AB as shown in the figure below:

040523 1524 TOPIC2AREAA21

The angle AOB = 360°/n, since the polygon has n sides.

The angle AOC = the angle AOB and the angle AOB = angle AOC + BOC

Therefore, the angle AOC = ½(360°/n) = 180°/n

Let the length of the side of the regular polygon ABbe S.

Then,

040523 1524 TOPIC2AREAA22

Therefore the length of a regular polygon with n sides inscribed in a circle is given by

040523 1524 TOPIC2AREAA23

The Perimeter of a Regular Polygon

The formula to determine the perimeter of a regular polygon

If we let 2r = d

040523 1524 TOPIC2AREAA24

From the concept of perimeter that perimeter of a regular polygon is the sum of the lengths of the sides of the polygon , if we have n sides each with length ‘S’ then the sum of the lengths of these sides will be nS. Therefore, Perimeter P of a regular polygon of n sides each with length S is given by:

040523 1524 TOPIC2AREAA25

Example 5

Find the length of one side of eight-sided regular polygon inscribed in a circle with radius 7cm.

Solution

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Therefore the length of one side of eight-sided regular polygon with radius of 7cm is 5.358cm

Area of a Regular Polygon

The Formula for Finding the Area of a Regular Polygon

Derive the formula for finding the area of a regular polygon

Consider the regular polygon with n sides inscribed in a circle of radius r and center O as shown below:

040523 1524 TOPIC2AREAA27

040523 1524 TOPIC2AREAA28

But since each vertex of the polygon is connected to O, the polygon region is divided into n triangles which are equal.

Now,

040523 1524 TOPIC2AREAA29

Therefore the area of a polygon of n sides inscribed in a circle of radius r is given by:

040523 1524 TOPIC2AREAA30

The Formula to Calculate the Area of a Regular Polygon

Apply the formula to calculate the area of a regular polygon

Example 6

Find the area of twelve-sided regular polygon inscribed in a circle of radius 14 cm

Solution

040523 1524 TOPIC2AREAA31

Circumference and area of a circle

Circumference of a circle is the distance around it. Circumference of a circle can be estimated by using a regular polygon with many sides inscribed in a circle with radius r.

We know perimeter of the regular polygon is given by:

040523 1524 TOPIC2AREAA32

Here we see that, as n increases the value of nsin 180°/napproaches the value ofπ.When n is very large the perimeter of a regular polygon approaches the circumference of the circle. The value ofnsin 180°/ncan be replaced byπbecause it approaches the value ofπwhen n is very large.

Therefore, circumference of the circle C, is given by C = 2πr

Area of a Circle

In similar way we can generate the formula of calculating the area of a circle by considering area of a regular polygon inscribed in a circle of radius r.

We know that, area of a regular polygon is given by:

040523 1524 TOPIC2AREAA33

Alternatively we can write it as:

040523 1524 TOPIC2AREAA34

040523 1524 TOPIC2AREAA35

Example 7

Find the circumference of a circle of radius 21cm. (takeπ= 3.14).

Solution

Circumference of a circle, C = 2πr

= 2× 3.14 × 21cm = 131.88cm

Area of Similar Polygons

The Ratio of Areas of Similar Polygons

Find the ratio of areas of similar polygons

Let ABC and A’ B’ C’ be two similar triangles:

040523 1524 TOPIC2AREAA36

Area and Perimeter

If we find the ratio of their sides we get;

040523 1524 TOPIC2AREAA37

Generally, if the ratio of the lengths of the corresponding sides of two similar polygons is k, then the ratio of their areas is k2.

Problems Related to Ratio for Areas of Similar Polygons

Solve problems related to ratio for areas of similar polygons

Example 8

We are given two triangles which are similar. The length of one side is 8cm and the length of the corresponding side is 14cm. if the area of a smaller triangle is 24cm2find the area of the other triangle.

Solution

040523 1524 TOPIC2AREAA38

Therefore the area of the other triangle is 73.5cm2.

Example 9

The ratio of the areas of two similar polygons is 36:48. The length of a side of the smaller polygon is 10cm. find the length of the corresponding side of the other polygon.

Solution

040523 1524 TOPIC2AREAA39

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