Question

Use the Converse of the Pythagorean Theorem to determine whether a right triangle can be formed given sides a , b , and c , where a=6 , b=10 , and c=12 .(1 point) Responses No, a right triangle cannot be formed because 62+102≠122. No, a right triangle cannot be formed because 6 squared plus 10 squared is not equal to 12 squared . No, a right triangle cannot be formed because 62+102=122. No, a right triangle cannot be formed because 6 squared plus 10 squared equals 12 squared . Yes, a right triangle can be formed because 62+102=122. Yes, a right triangle can be formed because 6 squared plus 10 squared equals 12 squared . Yes, a right triangle can be formed because the Pythagorean Theorem produced a false statement. Yes, a right triangle can be formed because the Pythagorean Theorem produced a false statement. Skip to navigation page 17 of 17

Answer 1

To determine whether a right triangle can be formed with sides \( a = 6 \), \( b = 10 \), and \( c = 12 \), we will use the Converse of the Pythagorean Theorem, which states:

If \( c \) is the longest side, then a triangle is a right triangle if \( a^2 + b^2 = c^2 \).

In this case, we have:

- \( a = 6 \)
- \( b = 10 \)
- \( c = 12 \)

First, we calculate \( a^2 \) and \( b^2 \):

\[
a^2 = 6^2 = 36
\]
\[
b^2 = 10^2 = 100
\]

Now, we sum \( a^2 \) and \( b^2 \):

\[
a^2 + b^2 = 36 + 100 = 136
\]

Next, we calculate \( c^2 \):

\[
c^2 = 12^2 = 144
\]

Now, we compare \( a^2 + b^2 \) with \( c^2 \):

\[
a^2 + b^2 = 136 \quad \text{and} \quad c^2 = 144
\]

Since \( 136 \neq 144 \), we find that \( a^2 + b^2 \neq c^2 \).

Thus, the correct response is:

**No, a right triangle cannot be formed because \( 6^2 + 10^2 \neq 12^2 \).**