Which equation could be used to find the length of the hypotenuse?

A. [tex]6^2 + 11^2 = c^2[/tex]
B. [tex]6^2 + c^2 = 11^2[/tex]
C. [tex]n^2 + k^2 = 11^2[/tex]



Answer :

To determine which equation could be used to find the length of the hypotenuse in a right-angled triangle, let's recall the Pythagorean theorem. The Pythagorean theorem states that in a right-angled 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. This can be written as:

[tex]\[ a^2 + b^2 = c^2 \][/tex]

where:
- [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the two legs of the triangle,
- [tex]\(c\)[/tex] is the length of the hypotenuse.

Now, let's analyze each given equation to determine if it follows the form of the Pythagorean theorem:

1. Equation [tex]\(6^2 + 11^2 = c^2\)[/tex]

This equation takes the form [tex]\(a^2 + b^2 = c^2\)[/tex], where:
- [tex]\(a = 6\)[/tex],
- [tex]\(b = 11\)[/tex],
- [tex]\(c\)[/tex] is the length of the hypotenuse.

This matches the Pythagorean theorem perfectly. Therefore, this equation is correct and can be used to find the length of the hypotenuse.

2. Equation [tex]\(6^2 + c^2 = 11^2\)[/tex]

This equation suggests that one of the legs of the triangle is 6, the hypotenuse is 11, and another leg's length is represented by [tex]\(c\)[/tex]. This does not match the form of the Pythagorean theorem, where the hypotenuse squared should be on one side of the equation:

[tex]\[ a^2 + b^2 = c^2 \][/tex]

Here, it mistakenly implies that one of the sides is actually the hypotenuse, which is incorrect.

3. Equation [tex]\(n^2 + k^2 - 11^2\)[/tex]

This equation does not even form a proper equation because it lacks an equals sign and clear structure. It is not recognizable under the Pythagorean theorem and therefore cannot be used to find the length of the hypotenuse.

In conclusion, the correct equation that can be used to find the length of the hypotenuse is:

[tex]\[ 6^2 + 11^2 = c^2 \][/tex]