Use the formulas [tex]a = x^2 - y^2[/tex], [tex]b = 2xy[/tex], [tex]c = x^2 + y^2[/tex] to answer this question.

One leg of a right triangle has length 11, and all sides are whole numbers. Find the lengths of the other two sides.

The other leg [tex]= \ \square[/tex] and the hypotenuse [tex]= \ \square[/tex]



Answer :

To solve the problem, let's start by using the formulas for the sides of a right triangle where all sides are whole numbers (forming a Pythagorean triple):

1. One leg of the triangle is represented by [tex]\( a = x^2 - y^2 \)[/tex]
2. The other leg of the triangle is represented by [tex]\( b = 2xy \)[/tex]
3. The hypotenuse of the triangle is represented by [tex]\( c = x^2 + y^2 \)[/tex]

We know one leg of the triangle is 11. Therefore, either [tex]\( a = 11 \)[/tex] or [tex]\( b = 11 \)[/tex]. We need to find suitable whole numbers [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that will satisfy these equations.

### Step-by-Step Solution:

1. Let's consider the possibility that [tex]\( a = 11 \)[/tex].
[tex]\[ a = x^2 - y^2 = 11 \][/tex]

2. For the equation [tex]\( x^2 - y^2 = 11 \)[/tex] to hold, we need to find whole numbers [tex]\( x \)[/tex] and [tex]\( y \)[/tex] where [tex]\( y < x \)[/tex].

3. By the formulas of the Pythagorean triple, iterate through possible pairs [tex]\( (x, y) \)[/tex]:

4. After solving, we find the pair [tex]\( (x, y) = (6, 5) \)[/tex] satisfies [tex]\( x^2 - y^2 = 11 \)[/tex]:
[tex]\[ 6^2 - 5^2 = 36 - 25 = 11 \implies a = 11 \][/tex]

5. With [tex]\( x = 6 \)[/tex] and [tex]\( y = 5 \)[/tex], calculate the other leg [tex]\( b \)[/tex]:
[tex]\[ b = 2xy = 2 \cdot 6 \cdot 5 = 60 \][/tex]

6. Finally, calculate the hypotenuse [tex]\( c \)[/tex]:
[tex]\[ c = x^2 + y^2 = 6^2 + 5^2 = 36 + 25 = 61 \][/tex]

Therefore, the lengths of the other two sides of the right triangle are:
- The other leg [tex]\( b = 60 \)[/tex]
- The hypotenuse [tex]\( c = 61 \)[/tex]

So, the complete side lengths of the right triangle are:
- One leg [tex]\( = 11 \)[/tex]
- The other leg [tex]\( = 60 \)[/tex]
- The hypotenuse [tex]\( = 61 \)[/tex]