Question

elinor determined that a triangle with side lengths 6, 10, and 8 does not form a right triangle. elinors work $6^{2}+10^{2}=8^{2}$ $36 + 100 = 64$ $136
eq64$ is her answer correct? yes, she is correct. no, she should have subtracted $10^{2}$ from $8^{2}$ and compared that to $6^{2}$. no, she should have added $8^{2}$ and $10^{2}$ and compared that to $6^{2}$. no, she should have added $6^{2}$ and $8^{2}$ and compared that to $10^{2}$.

Explanation:

Step1: Recall the Pythagorean theorem

For a right - triangle with sides \(a\), \(b\), \(c\) (\(c\) is the hypotenuse), \(a^{2}+b^{2}=c^{2}\). The hypotenuse is the longest side. Here, the sides are \(6\), \(8\), \(10\), and \(10\) is the longest side.

Step2: Apply the Pythagorean theorem correctly

We should check if \(6^{2}+8^{2}=10^{2}\). Calculate \(6^{2}=36\), \(8^{2}=64\), \(10^{2} = 100\). Then \(6^{2}+8^{2}=36 + 64=100\) and \(10^{2}=100\). Elinor wrongly took \(8\) as the hypotenuse instead of \(10\).

Answer:

No, she should have added \(6^{2}\) and \(8^{2}\) and compared that to \(10^{2}\).