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Question

An obtuse triangle is drawn with vertices upper A upper B upper C.

Suppose a midsegment line XY¯¯¯¯¯¯¯¯
was drawn within △ABC
and it intersected lines CB¯¯¯¯¯¯¯¯
and AB¯¯¯¯¯¯¯¯
. If XY¯¯¯¯¯¯¯¯=16
units, what would be the length of AC¯¯¯¯¯¯¯¯
?

(1 point)
AC¯¯¯¯¯¯¯¯ =
units

Answer 1

In triangle geometry, a midsegment connects the midpoints of two sides of a triangle and is parallel to the third side. The length of the midsegment is always half the length of the third side.

In your case, the midsegment \( \overline{XY} \) has a length of 16 units. According to the properties of midsegments:

\[ \overline{XY} = \frac{1}{2} \cdot \overline{AC} \]

To find the length of \( \overline{AC} \), we can rearrange the equation:

\[ \overline{AC} = 2 \cdot \overline{XY} \]

Now substitute the length of the midsegment:

\[ \overline{AC} = 2 \cdot 16 = 32 \]

Thus, the length of \( \overline{AC} \) is:

\[ AC = 32 \text{ units} \]