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Solve for the missing side using the Pythagorean Theorem.

Use the decoder to find out what the numbers 3, 6, 10, and 15 have in common.

Decoder Table:
[tex]
\[
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
A & B & E & G & H & I & L & M & N & R & S & T & U & Y \\
\hline
3 & 4\sqrt{2} & 3\sqrt{3} & 8 & 9 & 10 & 12 & 13 & 15 & 17 & 20 & 3\sqrt{5} & 25 & 26 \\
\hline
\end{array}
\]
[/tex]

Code:
[tex]
\[
\overline{10} \overline{1} \overline{9} \quad \overline{8} \overline{14} \overline{9} \quad \overline{2} \overline{3} \\
\frac{10}{10} \frac{14}{5} \frac{1}{2} \overline{12} \frac{11}{11} \frac{}{3} \frac{1}{2} \frac{14}{14}
\]
[/tex]

Note: The task involves using the decoder table and the given code, but further context or instructions are required to fully understand how to decode the message or apply it to solving for the missing side.

Answer :

To solve this problem, we need to use the Pythagorean Theorem. The Pythagorean Theorem is a fundamental principle in geometry that relates the sides of a right triangle. It states that in a right 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.

In this case, we're given the lengths of two sides of a right triangle: the height (let's call it [tex]\(a\)[/tex]) is 10 units, and the base (let's call it [tex]\(b\)[/tex]) is 17 units. We're trying to find the length of the hypotenuse (let's call it [tex]\(c\)[/tex]).

Here's the step-by-step solution:

1. Apply the Pythagorean Theorem:
[tex]\[
a^2 + b^2 = c^2
\][/tex]
Substitute the given values into the equation:
[tex]\[
10^2 + 17^2 = c^2
\][/tex]

2. Calculate the squares:
[tex]\[
10^2 = 100
\][/tex]
[tex]\[
17^2 = 289
\][/tex]

3. Add the squares:
[tex]\[
100 + 289 = 389
\][/tex]

4. Find the hypotenuse by taking the square root:
[tex]\[
c = \sqrt{389}
\][/tex]

5. Calculate the hypotenuse:
[tex]\[
c \approx 19.723
\][/tex]

So, the hypotenuse of the triangle is approximately 19.723 units.

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Rewritten by : Barada