How to find sides of a right triangle?
Pythagorean theorem
a^2 + b^2 = c^2
a and b are the legs of the triangle, c is the side opposite the right angle (the hypotenuse).
History records that Pythagoras and Diophantus were probably the two most well known mathematicians that had anything to do with right triangles having integer sides. The Pythagoreans were known for their intense interest in relationships that could be expressed in whole numbers. The famous Pythagorean Theorem states that in any right triangle, the area of the square constructed on the hypotenuse is equal to the sum of the areas of the squares constructed on the two legs. Stated in its more familiar form, x^2 + y^2 = z^2, x and y being the two sides forming the right angle and z being the hypotenuse of the triangle, the side opposite the right angle. Not too surprisingly, such integer sided triangles were ultimately referred to as Pythagorean triangles. The Pythagoreans were thought to have been one of the first to provide a proof.
Almost everyone has heard of the famous 3-4-5 right triangle where 3^2 + 4^2 = 5^2, the simplest, and most fundamental triangle based on the Pythagorean theorem and the only one with the three sides being consecutive integers. Not so surprisingly, fewer people know of the 5, 12, and 13 right triangle, the 7, 24, and 25 right triangle, the 20-21-29 right triangle, and infinitely more, which also satisfy the relationship without any proportional relationship to the 3, 4, and 5 triangle. These infinitely many sets of three integers, all satisfying the Pythagorean theorem relationship, are traditionally referred to as Pythagorean Triples.
According to many history sources, several formulas were developed to define certain types of Pythagorean triangles. One set that was attributed to Pythagoras took the form of x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 where n is any integer. It was ultimately discovered that the formulas created only triangles where the hypotenuse exceeded the larger leg by one.
All Pythagorean Triples of the form x^2 + y^2 = z^2 derive from x = k(m^2 - n^2), y = k(2mn), and z = k(m^2 + n^2) where k is any positive integer and m and n are arbitrary positive integers, m greater than n. Pythagorean Triples that have no common factor, or a greatest common divisor of 1, are called primitive. Those with a common factor other than 1 are called non-primitive triples. Primitive Pythagorean Triples are obtained only when k = 1, m and n are of opposite parity (one odd one even) and have no common factor, and m is greater than n. (For x, y, & z to be a primitive solution, m and n cannot have common factors and cannot both be even or odd. Violation of these two limitations will produce non-primitive Pythagorean Triples.)
Pythagorean Triples can also be derived from
1--x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 where n is any integer. The formulas create only triangles where the hypotenuse exceeds the larger leg by one.
2--x = n^2, y = (n^2 - 1)^2/2, and z = (n^2 + 1)^2/2.
3--x = 2m, y = m^2 - 1, and z = m^2 + 1 for any positive value of m greater than 1.
4--Others can be derived by taking any consecutive odd or even numbers and adding their reciprocals as with x
and y, we get 1/x + 1/y = (x + y)/xy. It follows that (x + y)^2 + (xy)^2 = N^2.
Example, from 5 and 7 we get 1/5 + 1/7 = 12/35 from which 12^2 + 35^2 = 37^2 or from 6 and 8 we get 1/6 + 1/8 = 7/24 and 7^2 + 24^2 = 25^2.
5--Some Pythagorean Triples can be derived using the odd squares. Using any odd square, express it as the sum of two consecutive integers, as in x + (x + 1) = N = n^2 and n^2 + x^2 = (x + 1)^2. For example, 25 = 12 + 13 = 5^2 and 5^2 + 12^2 = 13^2 or 121 = 60 + 61 = 11^2 and 11^2 + 60^2 = 61^2.
6--Another series of equations often used for generating triples are x = pq, y = (p^2 - q^2)/2 and z = (p^2 + q^2)/2 where p and q are odd integers with no common factors and p > q > 1.
Primitive Pythagorean Triples always have one leg odd, one leg even, and the hypotenuse odd (x and y cannot both be odd or even).
In every primitive Pythagorean triple, x, y and z, either x or y is divisible by 2 or 4, either x or y is divisible by 3, and either x, y, or z is divisible by 5. The area is always divisible by 6 and the product of all three sides is always divisible by 60.
Why?
1--Since either x or y is equal to 2mn or 4mn, the one that is equal to 2mn, or 4nm, is obviously divisible by 2 or 4.
2--If either m or n is divisible by 3, then x or y = 2mn is divisible by 3. If neither m or n is divisible by 3, then y = m^2 - n^2 is as the squares of both m and n are of the form 3k + 1.
3--If the product of the 3 sides is divisible by 5, then one of the sides is divisible by 5. The product is 2mn(m^2 - n^2)(m^2 + n^2) = 2mn(m^4 - n^4). If m or n is a multiple of 5, so is 2mn. If neither m or n is a multipe of 5, then (m^4 - n^4) is as m is then of the form 5k+/- 1 or 5k+/-2. Expanding the fourth power of each of them by the binomial theorem, it is found that m^4 is of the form 5h + 1 and similarly for n^4.
5--With x divisible by 4, x or y divisible by 3 and x, y or z divisible by 5, the area , xy/2, is divisible by 4x4/2 and xyz is divisible by 4x3x5.
(Ref: Mathematical Recreations by Maurice Kraitchik, Dover Publications, Inc., 1953.)
Quite surprisingly, the perimeter and area of any Pythagorean triangle are even.
The perimeter of any Pythagorean triangle is P = 2m(m + n).
The area of any Pythagorean triangle is A = mn(m + n)(m - n) = mn(m^2 - n^2).
Triples with all sides even are non-primitive but can be reduced to a primitive triple by dividing through by a constant factor.
Triples with all sides odd are impossible.
The sum and difference of the hypotenuse and the even side of a primitive Pythagorean triple are squares.
Only one side of a Pythagorean triangle can be a square.
If "m" and "n" have no factor in common but mn is a square, then both "m" and "n" are squares.
All primitive hypotenuses are primes of the form 4n + 1.
A number may be a primitive hypotenuse only if all of its prime factors are of the form 4n + 1. No primitive hypotenuse has a prime factor of the form 4n - 1.
The product of the two legs of a right triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse.
The area of a Pythagorean triangle can never be a square number or twice a square number.
The determination of two integers, the sum of whose squares equals another square integer, has always been, and remains today, a principal subject in the popular field of recreational mathematics. It is always possible to find at least one square which, when added to a given square, will result in a sum which is a square. The only numbers this does not work for are the squares of 1 and 2. This is not true for the hypotenuse of a right triangle
however where an integer must be of a certain type to be the hypotenuse of an integral right triangle.
There is much more to be observed and learned about Pythagorean triangles which would take much too much room to present here. There are many problems dealing with Pythagorean triangles that have enticed mathematicians and puzzle lovers for centuries, such as, "Given an integer A, how many primitive Pythagrorean triangles can you find with A as a leg?" or "Can you find more than two Pythagorean triangles having equal areas?" Then there are problems involving Pythagorean triangles whose areas have a given ratio; whose area equals its perimeter; whose perimeter is a square; the sum of whose area and perimeter is a cube; whose area equals the hypotenuse; whose perimeter is a given number; and so on, ad infinatum.
To find the sides of a right triangle, you can use a few different methods depending on the information you have. Here are a few approaches:
1. Pythagorean Theorem: If you know the lengths of two sides of a right triangle, you can use the Pythagorean theorem to find the third side. The theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. So, if you have side lengths A and B, and you want to find side length C (the hypotenuse), the equation would be C^2 = A^2 + B^2. You can then take the square root of C^2 to get the length of C.
2. Trigonometric Ratios: If you know one angle and one side length in a right triangle, you can use trigonometric ratios to find the remaining sides. The three primary trigonometric ratios used in right triangles are sine (sin), cosine (cos), and tangent (tan). For example, if you know the length of one of the non-hypotenuse sides and one of the acute angles, you can use sine or cosine to find the lengths of the other sides. The sine of an angle is equal to the length of the side opposite the angle divided by the length of the hypotenuse. The cosine of an angle is equal to the length of the side adjacent to the angle divided by the length of the hypotenuse. By rearranging these formulas, you can solve for the missing side lengths.
3. Special Right Triangle Properties: There are two special right triangles with easily calculable side lengths. The first is the 45-45-90 triangle, where the two acute angles are each 45 degrees. In this triangle, the two legs are congruent and the hypotenuse is the length of one leg multiplied by the square root of 2. The second special triangle is the 30-60-90 triangle, where one acute angle is 30 degrees and the other is 60 degrees. In this triangle, the lengths of the sides follow specific ratios: the side opposite the 30-degree angle is half the length of the hypotenuse, and the side opposite the 60-degree angle is equal to the length of the hypotenuse multiplied by the square root of 3.
By using one or a combination of these methods, you can find the sides of a right triangle.