do these measurements form a right triangle?
side 1: $8\sqrt{17}$
side 2: $42\sqrt{17}$
side 3: $32\sqrt{17}$
show your work here
hint: to add the square root symbol ($\sqrt{\square}$), type
oot\

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Accepted Answer

# Explanation:
## Step1: Recall Pythagorean theorem
For a right triangle, $a^2 + b^2 = c^2$, where $c$ is the longest side. First, identify the longest side. Among $8\sqrt{17}$, $42\sqrt{17}$, $32\sqrt{17}$, the longest is $42\sqrt{17}$ (since $42>32>8$). So $c = 42\sqrt{17}$, $a = 8\sqrt{17}$, $b = 32\sqrt{17}$.

## Step2: Calculate $a^2 + b^2$
$$
\begin{align*}
a^2&=(8\sqrt{17})^2 = 8^2\times(\sqrt{17})^2 = 64\times17 = 1088\
b^2&=(32\sqrt{17})^2 = 32^2\times(\sqrt{17})^2 = 1024\times17 = 17408\
a^2 + b^2&= 1088 + 17408 = 18496
\end{align*}
$$

## Step3: Calculate $c^2$
$$
c^2=(42\sqrt{17})^2 = 42^2\times(\sqrt{17})^2 = 1764\times17 = 30008
$$

## Step4: Compare $a^2 + b^2$ and $c^2$
Since $18496
eq30008$, $a^2 + b^2
eq c^2$.

# Answer:
These measurements do not form a right triangle.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Pythagorean theorem

Step2: Calculate \(a^2 + b^2\)

Step3: Calculate \(c^2\)

Step4: Compare \(a^2 + b^2\) and \(c^2\)

Answer: