Year 11 GCSE Maths - Intermediate Triangles and Interior and Exterior Angles In this lesson you will learn: How to prove that the angles of a triangle.

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Presentation transcript:

Year 11 GCSE Maths - Intermediate Triangles and Interior and Exterior Angles In this lesson you will learn: How to prove that the angles of a triangle will always add up to 180º ; How to use angle notation – the way we refer to angles in complicated diagrams ; How to work out the total of the interior (inside) angles of any polygon.

a b c In the diagram above we have a triangle in- between two parallel lines. At the top of the triangle there are three angles: a, b and c. Because these three angles make a straight line: a + b + c = 180º How to prove the angles of a triangle = 180º

a b c c Because of the Z-rule, we see that this angle here is also equal to c Remember: this means the two angles marked c are ALTERNATE angles!!

a b c a Because of the Z-rule again, we see that this angle here is equal to a Remember: this means the two angles marked a are also ALTERNATE angles!! c

a b c a Now we have a, b and c as the three angles in the triangle…….. …. And we already know that a + b + c = 180º so this proves the angles in a triangle add up to 180º !! c

Using Angle notation Often we can get away with referring to an angle as just a, or b, or c or even just x or y. But sometimes this can be a little unclear. Copy the diagram on the next slide…..

Just saying ‘the angle F’ could actually be referring to one of ten possible angles at the point F. If we actually mean angle 1, then we give a three-letter code which starts at one end of the angle, goes to F, and finishes at the other end of the angle we want A B C D E F

So for angle 1 we start at B, then go to F and finish at A, and we write: Angle 1 = BFA (sometimes you write this as BFA) A B C D E F

BUT notice we could go the other way round and start at A, then go to F and finish at B, and we write: Angle 1 = AFB instead. Either answer is correct!! A B C D E F

Also for angle 4 we start at D, then go to F and finish at E, and we write: Angle 4 = DFE (or EFD) (sometimes you write this as DFE) A B C D E F

And for angle 9 we start at F, then go to C and finish at D, and we write: Angle 9 = FCD (or DCF) (sometimes you write this as FCD) A B C D E F

A B C D E F Now you have a go at writing the three-letter coding for the following angles: Angle 2Angle 4Angle 10 Angle 6Angle 12Angle 3+4

A B C D E F The answers are: Angle 2 = BFC or CFBAngle 4 = DFE or EFD Angle 10 = CDF or FDCAngle 6 = ABF or FBA Angle 12 = FED or DEFAngle 3+4 = CFE or EFC

Interior Angles of a Polygon A polygon is any shape with straight lines for sides, so a circle is NOT a polygon. A pentagon

Interior Angles of a Polygon To find the total of the angles inside any polygon, just pick a vertex (corner) and divide the polygon into triangles, starting at that vertex: VERTEX

Interior Angles of a Polygon Now each triangle has a total of 180º, so with three triangles, the pentagon has total interior angles of 3 x 180º = 540º

Interior Angles of a Polygon What about a heptagon? This has 7 sides. Copy the one below into your book and label the vertex shown: VERTEX Now divide it into triangles…

Interior Angles of a Polygon You can see now that the heptagon has been divided into 5 triangles. That means the interior angles of a heptagon must add up to 5 x 180º = 900º.

Interior Angles of a Polygon Now copy this table and fill it in for the 2 polygons we have looked at so far: Name of Polygon Number of sides Number of triangles Working out Total of Interior angles Triangle31 1 x º Quadrilateral Pentagon Hexagon Heptagon75 5 x º Octagon8 Decagon10

Interior Angles of a Polygon Now complete your table – here’s a hint: look for patterns in the numbers!! Name of Polygon Number of sides Number of triangles Working out Total of Interior angles Triangle31 1 x º Quadrilateral Pentagon Hexagon Heptagon75 5 x º Octagon8 Decagon10

Interior Angles of a Polygon Challenge Question: What would be the total of the Interior angles of a 42- sided polygon? Answer: The number of triangles that can be drawn is always two less than the number of sides in the polygon, so: 40 x 180 = 7200º !!