To investigate the ratio of Area:Perimeter for triangles (2) To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio.

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Natalie Pinfield Williams

Aims

(1)        To investigate the ratio of Area:Perimeter for triangles

(2)        To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio.

(3)        Should the students hypothesis be un-sustainable then find out what triangle would maximise the ratio

Important facts about triangles

        Simple formula for the area        Area = ½ base *height

        General Formula                        Area = ½ ab sin C                

        Interior Angles                        A+B+C = 180        

        Sin rule                                 a          =         b   _        =         c  

                                                Sin A                Sin B                Sin c

        Cosine Rule                                a2 = b2 + c2 – 2bc Cos A

Student Suggestion

We will Investigate the claim made by the students that a 40,60,80 triangle would maximise the ratio of area:perimeter

Angles

The students suggested that a 40 60 80 would maximise the area : perimeter ratio.

We will make an initial assumption that the 40 6080 refers to the size of the angle in degrees. We accept that they could have been referring to lengths.

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Is our assumption possible?

Yes as 40 + 60 + 80 =180. The interior angles do sum to 180 which they must for any triangle so the students could have been referring to the angles.

Calculation

As no side lengths have been give then we could assume any one side is any length we want. By defining the length of one side the others are obviously defined. We can calculate the other length by use of the sine rule.

We will make a decision: Let the length of side ‘a’ be 1m.

The perimeter is ...

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