Question

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\

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.