Question

Amy is trying to prove that a triangle with the vertices X(-6,2), Y(-4,6), and Z(-2,2) is an isosceles triangle. She found that side XY has a length of approximately 4.5 and that side overline XZ has a length of 4. What is the length of side overline YZ ? Round your answer to one decimal place.

Answer 1

To find the length of side \( \overline{YZ} \), we can use the distance formula, which is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

We have the coordinates of the points \( Y(-4, 6) \) and \( Z(-2, 2) \).

Substituting these coordinates into the distance formula, we get:

\[
d_{YZ} = \sqrt{((-2) - (-4))^2 + (2 - 6)^2}
\]

Now, calculate the differences:

\[
(-2) - (-4) = -2 + 4 = 2
\]

\[
2 - 6 = -4
\]

Substituting these values back into the formula gives:

\[
d_{YZ} = \sqrt{(2)^2 + (-4)^2}
\]

Calculating the squares:

\[
d_{YZ} = \sqrt{4 + 16} = \sqrt{20}
\]

Now, simplify \( \sqrt{20} \):

\[
d_{YZ} = \sqrt{4 \cdot 5} = 2\sqrt{5}
\]

To find the approximate value of \( 2\sqrt{5} \):

Using \( \sqrt{5} \approx 2.236 \):

\[
d_{YZ} \approx 2 \cdot 2.236 \approx 4.472
\]

Rounding this to one decimal place:

\[
d_{YZ} \approx 4.5
\]

Thus, the length of side \( \overline{YZ} \) is approximately \( \boxed{4.5} \).