PYTHAGORAS' THEOREM

 

“The square on the Hypotenuse is equal to the sum of the squares on the other two sides.”

Contents:

 

Theorem: There is a relationship between the sides of a right angled triangle. There are no angles to be considered.

The square on the hypotenuse is equal to the sum of the squares on the other two sides.

As in Trigonometry the hypotenuse is the side opposite the right angle. If the lengths of any two sides are known then by using the theorem the third can be calculated.

c = a + b.

 

 

 

 

 

 

 

 

 

 

 

 

The diagram above shows the square on the hypotenuse: 6 = the square on the other 2 sides:

6 = 33 + 5.


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Two Dimensional Examples:

Example (i): Two sides of a right angled triangle are of length 4 and 5. What is the length of the hypotenuse ?

Answer: Hyp = 4+ 5; Hyp = 16 + 25; Hyp = 41; Hyp = 41;

Hyp = 64.


Example (ii): P is a point 8 distance from the centre of a triangle of 3 radius. A tangent from the circle meets at P. Calculate the distance from T to P.

Answer: This time the side 8 is the hypotenuse: 8 = 3 + TP. Rearrange for TP: TP = 8 - 3; TP = 55;

Therefore TP = 55; = 74.


Exercise:


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Three Dimensional Examples:

Example (iii). The diagram shows a cube with all dimensions 5. What is the length of the line from the front corner A to the rear corner G.

Answer: As the diagram is three dimensional, the first step is to make it two dimentional. This is done by drawing a line A to C which is in the same direction as A to G, but on the base where a calculation can be done to find it's length. (AC is the hypotenuse). AC = AB + BC. AC = 50; = 71. Now there is a two dimensional triangle ACG of which two lengths are known. The missing length is also a hypotenuse, therefore AG = AC + CG.

AG = 71 + 5; = 866.


Example: (iv). The diagram shows a triangular prism with some lengths given. Find the lengths B to F and A to F also their angle of elevation.

Answer: BF is the hypotenuse of the triangle BFD. Therefore BF = BD + DF. BF = 205; = 143.

The angle that B rises from F is tan(314); = 12.

To find AF, first the line CF has to be calculated (see example: (iii)), CF is a hypotenuse, CF = CD + DF; CF = 1220; = 349. Now two sides are known AC and CF; AF is the hypotenuse of this right angled triangle. AF = AC + CF; AF = 1227; = 35.

There is a short cut as the right angle triangle ABF is in the same plain, and length BF has already been found. AF is the hypotenuse, AF = AB + BF; AF = 1228; = 35. (check).

The angle that A rises from F is tan(3349); = 49.

Using short cut method (length CF is unknown) sin(335); = 49.


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