Question

Mrs. Palmer is teaching a 5th grade class. She is standing 9 meters in front of Emmet. Jaden is sitting to Emmet's right. If Jaden and Mrs. Palmer are 10 meters apart, how far apart are Emmet and Jaden? If necessary, round to the nearest tenth.

meters

Answer 1

To solve for the distance between Emmet and Jaden, we can visualize the scenario as a right triangle.

1. Mrs. Palmer is standing 9 meters in front of Emmet.
2. Jaden is to the right of Emmet. Since Mrs. Palmer and Jaden are 10 meters apart, we can use the Pythagorean theorem.

Let’s denote the distances:
- The vertical distance between Mrs. Palmer and Emmet (along the y-axis) is 9 meters.
- The horizontal distance (along the x-axis) from Emmet to Jaden is what we need to find (let's call it \(x\)).
- The distance from Mrs. Palmer to Jaden is 10 meters (the hypotenuse of the triangle).

According to the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
Where:
- \(a\) = distance between Mrs. Palmer and Emmet = 9 meters
- \(b\) = distance between Emmet and Jaden = \(x\) meters
- \(c\) = distance between Mrs. Palmer and Jaden = 10 meters

This gives us the equation:
\[
9^2 + x^2 = 10^2
\]

Calculating the squares:
\[
81 + x^2 = 100
\]

Now, solving for \(x^2\):
\[
x^2 = 100 - 81
\]
\[
x^2 = 19
\]

Taking the square root of both sides:
\[
x = \sqrt{19} \approx 4.36
\]

Rounding to the nearest tenth:
\[
x \approx 4.4 \text{ meters}
\]

Therefore, the distance between Emmet and Jaden is approximately **4.4 meters**.