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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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)

     

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    

  1. , 4

Length of shortest side:

2n + 1

Length of the middle side:

F (n) = an² + bn +c

F= (1) = a x 1² + b x 1 + c

  = a + b + c = 4 – eqn1

F= (2) = a x 2² + b x 2 + c

  = 4a + 2b + c = 12 – eqn2

F= (3) = a x 3² + b x 3 + c

  = 9a + 3b + c = 24 – eqn3

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Equation 3 - Equation 2

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

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