- Level: AS and A Level
- Subject: Maths
- Word count: 1576
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.
Task 1:
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 |
Task3:
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
- , 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
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