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Joseph and Isabelle left Omyra's house at the same time. Joseph jogged north at 8 kilometers per hour, while Isabelle rode her bike west at 12 kilometers per hour. Omyra tried to figure out how far apart they were after 15 hours. Her work is shown below. Which statements describe her errors? Check all that apply.

[tex]$
\begin{aligned}
8^2 + 12^2 & = a^2 \\
64 + 24 & = a^2 \\
88 & = a^2
\end{aligned}
$[/tex]


Sagot :

Let's review the problem step-by-step and identify where Omyra made errors in her calculations.

### Step-by-Step Solution:

1. Determine the speeds and time:
- Joseph jogs at 8 kilometers per hour.
- Isabelle rides at 12 kilometers per hour.
- Both are traveling for 15 hours.

2. Calculate the distance traveled by each:
- Joseph's distance: [tex]\( 8 \, \text{km/h} \times 15 \, \text{hours} = 120 \, \text{km} \)[/tex].
- Isabelle's distance: [tex]\( 12 \, \text{km/h} \times 15 \, \text{hours} = 180 \, \text{km} \)[/tex].

3. Apply the Pythagorean theorem to find the distance between Joseph and Isabelle:
- They are traveling on perpendicular paths (north and west), forming a right-angled triangle.
- Let the distances traveled by Joseph and Isabelle be the legs of the triangle, a and b, respectively:
[tex]\[ a = 120 \, \text{km} \quad \text{and} \quad b = 180 \, \text{km} \][/tex]
- The distance apart (hypotenuse, c) is found using:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
[tex]\[ c^2 = 120^2 + 180^2 \][/tex]
[tex]\[ c^2 = 14400 + 32400 \][/tex]
[tex]\[ c^2 = 46800 \][/tex]

4. Solve for c:
- Take the square root of both sides to find the hypotenuse:
[tex]\[ c = \sqrt{46800} \approx 216.33 \, \text{km} \][/tex]

### Review of Omyra's Work:
Let's compare this correct solution to Omyra's work:
[tex]\[ \begin{aligned} 8^2+12^2 & -a^2 \\ 64+24 & =a^2 \\ 88 & =a^2 \end{aligned} \][/tex]

Here are Omyra's errors:

1. Error in Squaring the Speeds:
- The squares should be [tex]\(8^2 = 64\)[/tex] and [tex]\(12^2 = 144\)[/tex], not [tex]\(64 + 24\)[/tex].
2. Error in Adding the Squares:
- Omyra incorrectly added the squared values as [tex]\(64 + 24 = 88\)[/tex]. The correct sum is [tex]\(64 + 144 = 208 \, \text{(if calculating speeds after 1 hour only, which she did wrong here)}\)[/tex].

### Corrected Calculations:
- Correct speeds and distances after 15 hours squared:
[tex]\[ 120^2 = 14400 \, \text{km}^2 \quad \text{and} \quad 180^2 = 32400 \, \text{km}^2 \][/tex]
- Correct sum:
[tex]\[ 14400 + 32400 = 46800 \, \text{km}^2 \][/tex]
- Distance between Joseph and Isabelle:
[tex]\[ c = \sqrt{46800} \approx 216.33 \, \text{km} \][/tex]


To sum up, the correct results lead to the following errors:
- Error in Basic Squaring:
[tex]\(8^2\)[/tex] should equal 64 and [tex]\(12^2\)[/tex] should equal 144.
- Error in Summation:
Sum [tex]\(64\)[/tex] and [tex]\(144\)[/tex] correctly rather than the false [tex]\( 64 + 24 = 88 \)[/tex].

By identifying these key mistakes, we can see why her result was inaccurate.