Given a right triangle whose side lengths are integral multiples of 7, how many units are in the smallest possible perimeter
Let's consider a right triangle with side lengths that are integral multiples of 7. In a right triangle, the two shorter sides are called the legs, and the longest side is called the hypotenuse.
Since the side lengths are integral multiples of 7, we can represent them as 7x, 7y, and 7z, where x, y, and z are positive integers.
Now, using the Pythagorean theorem, we know that the sum of the squares of the legs is equal to the square of the hypotenuse:
(7x)^2 + (7y)^2 = (7z)^2
Simplifying this equation:
49x^2 + 49y^2 = 49z^2
Dividing both sides by 49:
x^2 + y^2 = z^2
Now, let's find the smallest possible values for x, y, and z that satisfy this equation. To minimize the perimeter, we want to minimize the values of x, y, and z.
The smallest integral values that satisfy the equation x^2 + y^2 = z^2 are x = 1, y = 1, and z = 1. This means that the smallest possible side lengths of the triangle are 7, 7, and 7.
To find the perimeter, we add up the three side lengths:
Perimeter = 7 + 7 + 7 = 21 units
Therefore, the smallest possible perimeter of the right triangle with side lengths that are integral multiples of 7 is 21 units.
To find the smallest possible perimeter of a right triangle whose side lengths are integral multiples of 7, we can use the Pythagorean theorem. According to the theorem, in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Let's assume the side lengths of the right triangle are 7a, 7b, and 7c, where a, b, and c are positive integers.
Applying the Pythagorean theorem, we have:
(7a)^2 + (7b)^2 = (7c)^2
49a^2 + 49b^2 = 49c^2
Dividing both sides of the equation by 49:
a^2 + b^2 = c^2
Now we need to find integral values of a, b, and c that satisfy this equation.
To minimize the perimeter, we want to find the smallest possible values that satisfy the equation. Since the side lengths are integral multiples of 7, let's start by assigning the smallest possible values to a and b, which are both 1.
Plugging in these values into the equation:
1^2 + 1^2 = c^2
2 = c^2
c = √2
However, c cannot be a decimal value since the side lengths are required to be integral multiples of 7.
Let's try with a = 2 and b = 1:
2^2 + 1^2 = c^2
5 = c^2
This again gives a non-integral value for c.
By continuing this process, we can see that for any combination of integer values for a and b, the resulting value of c^2 is always a non-integral value.
Therefore, there are no integral solutions for a, b, and c in this case, and thus there is no smallest possible perimeter for a right triangle with side lengths that are integral multiples of 7.
21 + [(square root of 5) x 7]
The smallest possible perimeter is obtained when we have the smallest possible sides.
But the sides are supposed to be multiples of 7
So they have to be 7 and 14
Hypotenuse^2 = 7^2 + 14^2
Hypotenuse = √245
so smallest possible perimter = 7+14+√245
= 21 + √245