Pythagorean Values & Triplets

Pythagorean Values & Triplets

Pythagorean Values & Triplets: The Theorem of Pythagoras is one of the most important theorems in geometry. Right from high school students to those appearing for competitive exams to engineers, everybody deals with this theorem as a part of their daily practice. The concept that we will study in this Appendix is not only useful for calculating Pythagorean triplets but also useful in the Apollonius theorem, coordinate geometry, and trigonometric identities like \sin ^{2} \emptyset+\cos ^{2} \emptyset=1.

Pythagorean Values & Triplets
Pythagorean Values & Triplets

Pythagorean Values & Triplets

We know that the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. For example, if the sides of the right-angled triangle are 3, 4, and 5 then the square of 5 equals the square of 3 plus the square of 4. The values 3, 4, and 5 together form a Pythagorean triplet. Other examples of triplets are (6-8-10), (9-40-41), (1160-61), etc. We are going to study a method by which we can form triplets out of a given value. This technique will be of great help to students dealing with geometric figures in competitive exams.

We will divide our study into two parts. In the first part, we will express the square of a number as the sum of two squared numbers. In the second part, we will express a given number as the difference between two squared numbers. The first part is further divided into two cases – the first case dealing with odd numbers and the second dealing with even numbers. a

(A) Expressing the square of a number as the sum of two squared numbers

Case 1: Odd Numbers

Rule: The square of an odd number is also odd. This square is the sum of two consecutive middle digits.

Example: (a) 3^2 = 9 = 4 + 5 (b) 5^2 = 25 = 12 + 13 (c) 9^2 = 81 = 40 +41

We see that the square of 3 (odd) is 9 (odd), which is the sum of two consecutive numbers, viz. 4 and 5. Similarly, 25 is the sum of two consecutive numbers, 12 and 13.

But how do we find the consecutive numbers? The answer is very simple. To find the consecutive numbers, just divide the square by 2 and round it off.

In the first example, the square of 3 is 9. When 9 is divided by 2 the answer is 4.5. When 4.5 is rounded off to its next higher and lower numbers we have 4 and 5! Thus, (3-4-5) form a triplet where the square of the highest number is equal to the sum of the square of the other two numbers.

In the second example, the square of 5 is 25. When 25 is divided by 2 we get 12.5. When 12.5 is rounded off to its next higher and lower numbers, we have 12 and 13. Thus, (5-12-13) forms a triplet where the square of 13 equals the square of 5 plus the square of 12.

In the last example, the square of 9 is 81. When 81 is divided by 2 and rounded off it gives 40 and 41. Thus, (9-40-41) is a triplet where the square of the biggest number (41) is equal to the sum of the square of the other two numbers (9 and 40)

Similarly, 35^2= 1225 (1225/2 = 612.5)

Thus, 35, 612, and 613 form a triplet where 613 units are the length of the hypotenuse and 35 units and 612 units are the lengths of the other two sides of the triangle.

Case 2: Even Numbers

The square of an even number is an even number. Therefore, we cannot have two middle digits on dividing it by 2. Thus, we divide the even number by 2, 4, 8, 16, etc. (powers of 2) until we get an odd number. Once we get an odd number we continue the procedure as done in the case of odd numbers.

If you have divided the even number by 2, 4, 8, etc. the final answer will have to be multiplied by the same number to get the triplet.

Q. One value of a Pythagorean triplet is 6; find the other two values.

We divide 6 by 2 to get an odd number 3. Next, we follow the same procedure as explained in the case of odd numbers to form a triplet (3-4-5). Now, since we have divided the number by 2 we multiply all the values of (3-4-5) by 2 to form the triplet (6-8-10).

Thus, the other two values are 8 and 10.

Q. One value of a triplet is 20; find the other two values.

We divide 20 by 4 to get an odd number 5. Next, we follow the same procedure as explained before to form a triplet (5-1213). Now, since we have divided the number by 4, we multiply all values of the triplet (5-12-13) by 4 and make it (20-48-52).

Thus the other two values are 48 and 52

(B) Expressing a given number as a difference of two squares

We express a given number ‘n’ as a product of two numbers ‘a’ and ‘b’ and then express it as:

\mathrm{n}=[(\mathrm{a}+\mathrm{b}) / 2]^{2}-[(\mathrm{a}-\mathrm{b}) / 2]^{2}

Q. Express 15 as a difference of two squared numbers. We know that 15 = 5 x 3. Thus, the value of ‘a’ is 5 and ‘b’ is 3.

Thus,
\begin{aligned} 15 &=[(5+3) / 2]^{2}-[(5-3) / 2]^{2} \\ &=(8 / 2)^{2}-(2 / 2)^{2} \\ &=(4)^{2}-(1)^{2} \end{aligned}

(We can verify that 16 – 1 = 15)

From the above, it can be proved that if one side of the triangle is \sqrt{15} then the other side is 1 unit and the hypotenuse is 4 units. This can be checked with the expression \sqrt{15}^2 + 12 = 42

Q. In A ABC, the value of side angle B is 90 degrees and side BC is 6 units. Find a few possible values of side AB and side AC.

Since the triangle is a right-angled triangle, BC = ACP – AB2. The value of BC2 = 36. Thus, 36 = AC2 – AB?. In other words, we have to express 36 as a difference between the two squares. We will apply the same rule that we used in the previous example.

We have to express the number 36 as a product of 2 numbers ‘a’ and ‘b’.

(a) 36 = 18 x2 = [(18+2)/2]^2[(18-2)/2]^2 = [(20)/2]^2[(16)/2]^2 = 10^28^2 = 10,8

(b) 36 = 9 x4 = [(13)/2]^2[(5)/2]^2 = 6.5^22.5^2 = 6.5, 2.5

(c) 36 = 12 x 3 = [(15)/2]^2[(9)/2]^2 = 7.5^24.5^2 = 7.5, 4.5

In example (a) the value of sides AC and AB is 10 units and 8 units.
In example (b) the value of sides AC and AB is 6.5 units and 2.5 units.
In example (c) the value of sides AC and AB is 7.5 units and 4.5 units.

We can verify:

6^2= 10^28^2
6^2= 6.5^22.5^2
6^2= 7.5^24.5^2

From the examples, we can conclude that if the length of side BC is 6 units, then the lengths of sides AC and AB can be either 10 and 8 or 6.5 and 2.5 or 7.5 and 4.5 units. We have thus formed three triplets (6,8,10) (2.5, 6, 6.5) and (4.5, 6, 7.5).

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