Mathematics

By the end of this unit, I will be able to solve problems of lengths in right angled triangle by using Pythagoras' theorem.

• State Pythagoras' theorem

• Identify hypotenuse in three sided of a right angled triangle

• Demonstration of Pythagoras' theorem

• Application of Pythagoras' theorem in calculations

Any triangle that contains a right angle is called a right triangle or right-angled triangle. Right triangles occur frequently in everyday life situations. For example, right triangles are formed when you lean a ladder against a wall or you brace a book shelf.

Fig 6.1 shows a right triangle. The longest side in any right triangle is called the hypotenuse.

                     

For a very long time, it has been known, as a matter of fact, that a triangle with sides 3, 4 and 5 units is right-angled.  However, this is only one isolated case and does not provide an answer to the question above.  The lengths of the sides of a right triangle are related in a special way as you will discover from the following activities

           

Activity 6.1

Consider a floor that is tiled with tiles of the same size, each is a right-angled isosceles triangle, as in Fig. 6.2.

1. Make a copy of this figure on a larger scale.

2. Pick any of the small triangular tiles (e.g. A) and shade the three squares standing on its sides.  The squares on the shorter sides are equal (each comprising of two triangular tiles).

3. How many tiles make the square on the longest side (called the hypotenuse)?

4. Shade a triangle composed of two triangular tiles (e.g. B).  This triangle is similar to the previous one. 

5. Count the number of tiles that make the squares on its sides.

6. Continue shading triangles of the same shape but of increasing size and record your results as in Table 6.1.

What is the relationship between the values of a, b and c? Suppose the area of a triangle represents 1sq. unit, find the areas of the squares on the three sides of the triangle. What is the relationship between the three areas? Suppose also that the area of triangle B in the same figure represents 1sq.unit state the areas of the three squares on the side of B.

What is the relationship between the three areas?

In both cases, the sum of the areas of the square on the two shorter sides is equal to the area of the square on the hypotenuse. Is this fact true if the right-angled triangle is not isosceles?  Verify this by carrying out Activities 6.2 and 6.3.

1. Draw ∆ABC, right-angled at B on a stiff paper, such that BC is longer than AB. Construct the three squares on its sides, as in Fig. 6.3.

2. Locate the centre P of the square on side BC. Through P, construct a line perpendicular to AC and another line parallel to AC, to subdivide the square into four pieces as in Fig. 6.4.

3. Cut out the squares on AB and BC. Cut out the square on BC into the four pieces indicated.  Arrange the pieces to cover completely the square on AC (See Fig. 6.5). Note that the pieces can be moved into their new positions without rotating any of them or turning them over.  (This is like a jigsaw puzzle and is referred to as Perigal’s dissection).

What can you say about the areas of the squares on the two shorter sides of the triangle

1. (a)  Construct a right triangle with sides of 3 cm,4 cm and 5 cm respectively on a piece of grid paper Fig 6.6

          

(b) Draw a square on each side of the triangle. Calculate the area of each square.

(c) How does the area of the square on the hypotenuse seem to relate to the areas of the squares drawn on the other two sides?

2. Repeat question 1 by constructing right triangles

(a) 6 cm, 8 cm, 10 cm

(b) 5 cm, 12 cm, 13 cm

3. The side of various right triangles are shown in the table 6.2.

(a) Draw each right triangle using the sides given measure the hypotenuse.

(b) Draw a square on each side of the right triangle. Find the area of each square.

(c) How does the area of the square drawn on the hypotenuse seem to relate to the areas of the other two sides?

              

From activities 6.1 to 6.3, the area of the square on the hypotenuse is equal to the sum of the areas on the other two sides of any right triangle.

                               

ABC is a right-angled triangle. The following relationship is obtained

           

This relationship is called the Pythagorean relation or Pythagoras theorem. It states that “in aright triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides”.

The Pythagoras relation can be used to help you find the missing side in a right triangle.

A right triangle has a hypotenuse 12 cm long. Find the length of the third side if one of the two shorter sides is 8 cm long. Give your answer to the nearest centimetres.

             

1. Lengths of the sides of four triangles are shown. Identify which of the triangles are right angled, showing your method.

(a) AB = 24 cm,  BC = 10 cm,  AC = 26 cm

(b) DE = 7 cm,  EF = 8 cm,   FD = 13 cm

(c) GH = 10.6 cm, HF = 5.6 cm,  IG = 9.0 cm

(d) JK = 16 mm,  KL = 34 mm,   LJ = 30 mm

2. Use Fig 6.9 to copy and complete the following, given that A,B,C

represent areas of the three squares.

             

For pythagoras’ theorem to be completely general, think of a triangle T whose sides are of lengths a, b and c units, as in Fig. 6.10.

               

Note that line CBC′ is straight and C′B′ is at right angles to it.

2. Rotate triangle T′ in a clockwise direction about O twice through 90° in each case, to positions T′′ and T′′′ as in Fig. 6.12.
 

 Note that the figure CC′C′′C′′′ obtained after the three rotations, is a square of side (a + b) units.

                          

As we have seen, Pythagoras’ theorem concerns areas of the square on the sides of a right angled triangle. Its main use, however, is in calculating lengths.  It also provides us with a test for a right-angled triangle.

A triangle is right-angled, whenever the square of the length of the longest side equals the sum of the squares of the lengths of the other two sides.

          

1. Fig. 6.16 is a right-angled triangle with squares A, B and C on its sides.

7. The sides of a rectangle are 7.8 cm and 6.4 cm long.  Find the length of the diagonal of the rectangle.

8. The length of the diagonal of a rectangle is 23.7 cm and the length of one side is 18.8 cm. Find its perimeter.

CD = 20 cm. Calculate the length of a diagonal of the rectangle giving your answer to the nearest a whole number.

Revisit your answers to question 5 of Exercise 6.1 (b).  You notice that the values give right-angled triangles. Such sets of values which give right-angled triangles are known as Pythagorean triples or Pythagorean numbers

A Pythagorean triple (a, b, c,) is a group of three numbers which give the respective lengths of the sides and the hypotenuse of a right-angled triangle and are related thus:  a² + b²  =  c²

The group (3, 4, 5) is the most famous and most commonly used Pythagorean triple.  The triple was known even before the time of Pythagoras.  It was, and is still, used for setting out the base lines on tennis courts and other sports pitches.

In general if we donate a Pythagorean triple as a, b, c, such that:

(i) a < b < c and a, b and c are positive integers.

(ii) b and c are consecutive numbers i.e. b + 1 = c, then b + c = a²

Suppose 11, b, c are Pythagorean triple, find b and c.

b + c = 11² ⇒ b + c = 121

We need two consecutive numbers whose sum = 121

(121 – 1) /2 = 60

The two numbers are therefore 60 and 61

Check for the Pythagorean property

in 11, 60, and 61 112 + 602 

= 612 LHS 112 = 121  

602 = 3 600 112 + 602 = 121 + 3600    

= 3 721 RHS = 612 = 3 721

Therefore, 11, 60 and 61 is a pythagorean triple

Discuss with your classmate the following working from numbers 1-4 then work out numbers 5-10 in your exercise books.

In general,

if a, b, c is a Pythagorean triple then

(i) a < b < c  (ii) b + 2 = c

(iii)  b + c = 1/2 a² Note that a is always an even number

so that  1/2a² can make sense.

Suppose that 14, b, c is a Pythagorean triple.

Find the value of b and c.

Examples 6.5 and 6.6 show two ways of finding Pythagorean triples.  But they do not seem to give all the possible sets of such triples.  Is  there a general way of finding Pythagorean triples?

Are these all Pythagorean triples? In general:

Given any two positive integers m and n, where m > n, we always obtain the Pythagorean triple:  (m² – n²,  2mn,  m² + n²)

We can use Pythagorean theorem to determine if a given triangle is rightangled or not. If a triangle is right-angled, then the sum of the squares of the shorter sides equals the square of the longer side. Example 6.7 illustrates the process.

Find out whether a triangle with sides 11, 15 and 18 cm is right-angled.

Solution

The two shorter sides are 11 cm and 15 cm in length. The sum of the squares of their lengths is

  11² + 15²  =  121 +225   =  346

The square of the length of the longest side is

 18²  =  324 Now  

11² + 15²  ≠  182

∴  the triangle is not right-angled.

1. (5, 12, 13) is a Pythagorean triple.

(a) Write down four multiples of it.

(b) Are all the four multiples in (a) Pythagorean triples?

(c) Using a multiplier n and any Pythagorean triple (a, b, c,) state the general result for such multiples as in (a).

2. Find out if the following are Pythagorean triples.

(a) (7 , 24 , 25)

(b) (8 , 15 , 17)

(c) (15 , 22 , 27)

(d) (28 , 43 , 53)

(e) (11 , 60 , 61)

(f) (20 , 21 , 29)

3. The following are the dimensions of two triangles.  Which one of them is a         right-angled triangle?

(a) 15 cm,  30 cm,  35 cm

(b) 33 cm,  56 cm,  65 cm

4. Use the following numbers to generate Pythagorean triples.

(a) 1 and 4 

(b)  1 and 5

(c) 6 and 2 

(d)  3 and 8

5. Complete the following Pythagorean triples.

(a) (25 , … , …) 

(b) (31 , … , …)

(c) (43 , … , …)

(d) (49 , … , …)

(e) (30 , … , …)

(f) (38 , … , …)

(g) (44 , … , …)

(64 , … , …)

6. Which of the following measurements would give a right-angled triangle?

(a) 6 cm by 8 cm by 10 cm

(b) 5 cm by 12 cm by 13 cm

(c) 4 cm by 16 cm by 17 cm

(d) 7(1/2) cm by 10 cm by 12(1/2) cm

(e) 9 cm by 30 cm by 35 cm

(f) 12 cm by 35 cm by 37 cm

(g) 12 m  by 60 m by 61 m

21 m by 90 m by 101 m

(i) 20 m by 21 m by 28 m

(j) 28 m by 45 m by 53 m

(k) 27 m by 35 m by 50 m

(l) 33 m by 44 m by 55 m

(m) 4 m by 7(1/2) m by 8(1/2) m

14 m by 48 m by 50 m

(o) 2.7 m by 36.4 m by 36.5 m

(p) 2.9 m by 42 m by 42.1 m

As we have seen, Pythagoras’ theorem connects the areas of actual squares.  Its main use, however, is in calculating lengths without  having to draw any squares.  The theorem also acts as a test for right-angled triangles

There are many real life situations which require the use of Pythagoras’ theorem. A pedestrian may take shortcut along the diagonal rather than walk along street M and street A. How much shorter is the shortcut?

Working in groups, identify five real life situations in which you find application of Pythagoras' theorem. Describe the situations either in words or by use of clearly labeled diagrams.

The following are some of the situations that represent the application of pythagoras' theorem in real life. Did you capture all of them in Activity 6.7? List those that are not captured below in your exercise book:

In this case we can find the distance between the wall and the foot of the ladder, the vertical height of the top of the ladder from the ground or the length of the ladder depending on the information given or known.

2. Imagine a guy wire or chain used to steady an electric post. The post is upright i.e. perpendicular to the ground and the guy chain fixed to the ground on one end and to the post on the other. (see Fig. 6.24

                 

In this case you can find the distance of the wire from the post, the height of the point at which the wire is secured on the post or even the length of the wire depending on what information is given or known. The wire, the post and the ground form a right angled triangle.

3. Imagine a hawk on a branch of a tree, and observed a chick on the ground. Some distance from the foot of the tree. (see Fig. 6.25)

                      

The view line of the hawk to the chick represents the hypotenuse of a right angled triangle connecting the hawk, the chick and the foot of the tree.

4. Imagine a man standing at the edge of a cliff and using binoculars observes a ship on the sea at a distance. The point on the cliff where the man stands, the position of the ship and the height of the cliff above the water connected would form a right angled triangle whose dimensions can be calculated given appropriate information make a sketch of this situation.   (see Fig. 6.26)
                             

These are just a few examples where Pythagoras' theorem can be applied to calculate distances. I believe of many more which you are ready to share with your class.

A ladder, 3.9 m long, leans against a wall.  If its foot is 1.2 m from the wall, how high up the wall does it reach?

Fig. 6.27 is an illustration of the situation.  Note that the ground must be assumed to be horizontal and level and hence at right angles to the wall.

                

1. Fig. 6.28 shows a television antenna.  Find the length of the wire AB holding the antenna.

                            

2. A ladder reaches the top of a wall of height 6 m when the end on the ground is 2.5 m from the wall.  What is the length of the ladder?

3. The length of a diagonal of a rectangular flower bed is 24.6 m and the length of one side is 18.9 m.  Find the perimeter and the area of the flower bed.

4. A piece of rope with 12 knots that are equally spaced has been laid out and pinned down on the ground as in Fig. 6.29.

                          

(a) What can you say about the triangle whose corners the stakes mark?

(b) Does it matter how great the distance between the knots is?

5. A rectangular chalkboard in a classroom measures 2.2 m by 1.2 m.  What is the length of the longest straight line that can be drawn on it?

6. Fig. 6.30 shows a road that turns through a right angle to go round a rectangular recreational garden in a town.  To save time, people on foot cut off the  corner, thus making a path that meets the road at 45°.  If the path is 48 m long, find the distance that the people save?

                     

7. A hall is 16 m long, 14 m wide and 9 m high. Find the length of the diagonal of the floor.

8. Fig. 6.31 represents a roof truss which is symmetrical about QS.  Beam PQ is 5 m long, strut TS 2.4 m long and the distance TQ is 1.8 m.

(a) Find the height QS.

(b) Hence, find the span PR of the roof.

                        

9. Three ships A, B and C leave and sailed in different directions. A sails due west, B sails due East and C sails due south. In 1 hour, A had sailed 23 km, B 18 km and C 12 km. Draw a sketch to show the relative positions at this instance. Hence calculate how far apart are

(a) B and C

(b) A and C

(c) A and B

10. Fig. 6.32 gives an example of the network of streets in a modern city,  a pedestrain may make a shortcut along the diagonal rather than walk along street M and street A. How much shorter is the shortcut.
          

Unit Summary

1. Any triangle that contain a right angle is called a right triangle or right- angled triangle.

2. The longest side of any right angled triangle is called the hypotenuse.

3. Pythagoras' theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

4. A group of three numbers (a, b, c) which give the respective lengths of the sides and the hypotenuse of a right-angled triangle and are related is called a pythagorean triple.

5. A set of values which give right-angled triangles is called Pythagorean numbers.

1. Find AB if AC = 18 cm and   BC = 24 cm in Fig. 6.33.

                            

2. A rectangle measures 3 cm by 4 cm. Calculate the length of its diagonal. 

3. A string is firmly tied onto the top of a flag post 10 m tall and supported on the ground by pegs. If the string is pegged 7.5 m away from the base of the flag post, find the length of the string. 

4. A ladder 22 m long leans against a vertical wall. If it is 16 m away from the vertical wall, calculate the height of the wall.

 5. In  Fig. 6.34, PQ is parallel to SR and PQRS is a trapezium.  Angle QPS = Angle RSP = 90°,  PS = 6 cm, SR = 19 cm and  PQ = 11 cm.

Calculate: (a)  the length of QR.

                  (b) the length of diagonal PR.

                   

6. The diagonal of a rectangle measures 7.5 cm. If one of the sides measures 4 cm, calculate the perimeter of the rectangle

7. In Fig. 6.35, QR = 32 cm, PTS = TRS = TQP = PUS = 90º. Given that T divides QR in the ratio 1:1 and S divides RU in the ratio 3:1, find the length of PS.