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# Investigate the relationships between the lengths of the 3 sides of the right angled triangles and the perimeters and areas of these triangles.

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Introduction

Kamran Ali 10.1                GCSE Maths Coursework

Aim: To investigate the relationships between the lengths of the 3 sides of the right angled triangles and the perimeters and areas of these triangles.

a)

The numbers 5, 12, 13 satisfy the condition.

5² + 12² = 13²

Because 5² = 5x5 = 25

12² = 12x12 = 144

13² = 13x13 = 169

And so

5² + 12² = 25 + 144 = 169 = 132

b)

The Numbers 7, 24, 25 also satisfy the condition.

7² + 24² =25²

Because  7² = 7x7 = 49

24² = 24x24 = 576

25² = 25x25 = 525

And so

7² + 24² = 49+ 576 = 625 = 25²

Task2: The perimeter and area of the triangle are:

a)    b)     Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 12 6 5 12 13 30 30 7 24 25 84 84

Length of short side is going to be in fixed steps meaning that this is a linear sequence and the length of middle side and longest side is actually a quadratic sequence because they are not in fixed steps and in geometric sequence.

4     , 12     , 24     , 40      8     , 12     , 16   4     , 4

5     , 13     , 25     , 41      8     , 12     , 16 1. , 4 Length of shortest side:

 Term no 1 2 3 4 5 Sequence 3 5 7 9 11 Sequence 2n 1 4 6 8 10 Sequence 1 1 1 1 1

Middle

I am doing this to eliminate C from these equations

9a + 3b + c = 24 –eqn3

-  4a + 2b + c = 12 –eqn2 5a + b          = 12 –eqn4

Equation 2 – Equation 1

I am doing this to eliminate C and form a fifth equation that I will subtract with equation 4.

4a + 2b + c = 12 –eqn2

-  a + b + c = 4 –eqn1 3a + b          = 8 –eqn5

Equation 4 – Equation 5

I am doing this to eliminate B and finally work out what A is worth.

5a + b = 12 –eqn4

-  3a + b = 8 –eqn5 2a          = 4

A = 4/2

A = 2

Substitute A =2 into equation 5

I am doing this to find what B is worth.

3a + b = 8

3 x 2 + b = 8

6 + b = 8

B = - 6

B = 2

Substitute A =2 and B = 2 into equation 1.

I am doing this to find what C is worth.

A + B + C = 4

2 + 2 + C = 4

4 + C       = 4

C = 0

F (n) = an² + bn + c

F (n) = 2n² + 2n

Try n = 1

F (1) = 2 x 1² + 2 x 1

= 2 + 2

= 4

Try n = 2

F (2) = 2 x 2² + 2 x 2

= 8 + 4

= 12

Try n = 100

F (100) = 2 x 100² + 2 x 100

= 10000 + 200

= 20200

Length of longest side:

F (n) = an² + bn +c

F= (1) = a x 1² + b x 1 + c

= a + b + c = 5 – eqn1

F= (2)

Conclusion

-  a + b + c + d = 6 –eqn1 7a + 3b + c      = 24- eqn7

Equation 6 – Equation 7

I am doing this to eliminate C.

19a + 5b + c = 54 –eqn6

-  7a + 3b + c = 24- eqn7 12a + 2b       = 30- eqn8

Equation 5 – Equation 6

I am doing this to eliminate C.

37a + 7b + c = 96 –eqn5

-  19a + 5b + c = 54 –eqn6 18a + 2b       = 42- eqn9

Equation 9 – Equation 8

I am doing this to eliminate B and to find the value of A is worth.

18a + 2b       = 42- eqn9

-    12a + 2b       = 30- eqn8 6a                      = 12

A = 12/6

A = 2

Substitute A = 2 into equation 9.

I am doing this to find what B is worth.

12a + 26 = 30

12 x 2 + 2b = 30

24 + 2b = 30

2b = 30 – 24

2b = 6

b = 6/2

B=3

Substitute A = 2 and B = 3 into equation 6.

I am doing this to find what C is worth.

19a + 5b + c = 54

19 x 2 + 5 x 3 + c = 54

38 + 15 + c = 54

53 + 15 + c = 54

c = 54 – 53

C = 1

Substitute A = 2 and B = 3 and C = 1 into equation 1.

I am doing this to find what D is worth.

A + B + C + D = 6

2 + 3 + 1 + D = 6

6 + D = 6

D = 6 – 6

D = 0

F (n) = a³ + bn² + cn + d

F (n) = 2n³ + 3n² + n

Try n = 1

F (1) = 2 x 1³ + 3 x 1² + 1 x 1

= 2 + 3 + 1

= 6

Try n = 2

F (2) = 2 x 2³ + 3 x 2² + 1 x 2

= 16 + 12 + 2

= 30

By:

Kamran Ali 10.1

Kamran Ali 10.1                GCSE Maths Coursework

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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