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What Are Some Real-Life Examples of the Pythagorean Theorem?

Answer – The Pythagorean theorem is used in architecture (for building right-angled structures), surveying, and navigation (for calculating the distance between two points on a map).

Explanation: 
The Pythagorean theorem is a basic rule in geometry for right-angled triangles. It states that the square of the longest side (opposite to the right angle) is equal to the sum of the squares of the other two sides. This can be written as a2 + b2 = c2, if a and b are two sides of a right-angle triangle with a hypotenuse “c.” This theorem is named after the ancient Greek mathematician Pythagoras.

A diagram of a labeled right-angled triangle along with the Pythagorean theorem
The Pythagorean theorem is used to calculate the relationship between the sides of a right-angled triangle.
Image credits: Andre Engels / Wikimedia Commons (licensed under CC BY-SA 3.0)

Let’s say there’s a roof with a pitch angle of 30 degrees and one needs to calculate the length of the roof beams. This can be done using the Pythagorean theorem if the values for the length of the base and the vertical rise are provided. For this example, let’s consider the length of the horizontal base (a) to be 10 meters and the vertical rise [height (b)] to be 5 meters. Now, by applying the Pythagorean theorem, the length of the roof beam (hypotenuse) can be calculated as follows:

[Length of the roof beam (c)]2  = [base (a)2 + height (b)2 ] =  (10)2 + (5)2 = 100 + 25 = 125

Length of the roof beam (c)      =  125

                                                  = 11.18 meters

Another instance where the Pythagoras theorem can be used is while calculating the distance between two points on a map. For example, let’s say the coordinates for two points (X1, Y1) and (X2, Y2) are (3, 4) and (7, 9) respectively, on a plane of the x- and y-axis. So, the distance (D) between these two points will be the hypotenuse. Therefore, as per the theorem, D is calculated as follows:

D = (X2  X1)2+(Y2  Y1)2

D = (7  3)2+(9  4)2

D = 16 +25

D 6.40 units.

Therefore, the distance between the two points is approximately 6.40 units.


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