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look at the diagram. which term describes \\( \\overline{fi} \\)? \\( \…

Question

look at the diagram.
which term describes \\( \overline{fi} \\)?
\\( \quad \\) altitude \\( \quad \\) perpendicular bisector
\\( \quad \\) median \\( \quad \\) none of these

Explanation:

Step1: Recall Definitions

  • Altitude: A segment from a vertex perpendicular to the opposite side (or its extension).
  • Perpendicular Bisector: A line/segment that is perpendicular to a segment and bisects it (divides into two equal parts).
  • Median: A segment from a vertex to the midpoint of the opposite side.

Step2: Analyze the Diagram

  • \( HI = 80 \), \( IE = 46 \), so \( I \) is not the midpoint of \( HE \) (since \( 80

eq 46 \)). Thus, \( \overline{FI} \) is not a perpendicular bisector (needs to bisect \( HE \)) or a median (needs to connect to midpoint).

  • \( \overline{FI} \) is perpendicular to \( HE \) (right angle symbol), but since it doesn't bisect \( HE \) or connect to a midpoint, it's an altitude (as it's perpendicular to the opposite side, even if not bisecting or connecting to midpoint in this case? Wait, no—wait, altitude is just perpendicular to the opposite side (or its line). Wait, but let's check again. Wait, the length from \( H \) to \( I \) is 80, \( I \) to \( E \) is 46, so \( I \) is not the midpoint. So perpendicular bisector requires bisecting (so \( HI = IE \)), which is not true here. Median requires \( I \) to be midpoint, which it's not. But \( \overline{FI} \) is perpendicular to \( HE \), so it's an altitude (since altitude is a perpendicular segment from a vertex to the line containing the opposite side). Wait, but the options: let's re-express.

Wait, maybe I made a mistake. Wait, the altitude is a perpendicular segment from a vertex to the opposite side (or its extension). So \( F \) is a point on \( GE \), and \( FI \perp HE \). So \( \overline{FI} \) is an altitude of triangle \( FHE \) (or the larger triangle? Wait, the diagram: triangle \( GHE \), with \( F \) on \( GE \), and \( FI \perp HE \) at \( I \), with \( HI = 80 \), \( IE = 46 \). So \( I \) is not the midpoint, so perpendicular bisector is out (needs to bisect \( HE \), so \( HI = IE \)), median is out (needs \( I \) to be midpoint). But \( FI \) is perpendicular to \( HE \), so it's an altitude. Wait, but let's confirm the definitions again.

Altitude: A perpendicular segment from a vertex to the line containing the opposite side. So even if \( I \) is not the midpoint, as long as it's perpendicular, it's an altitude. Perpendicular bisector: must bisect the segment (so \( HI = IE \)) and be perpendicular. Since \( HI
eq IE \), perpendicular bisector is out. Median: connects vertex to midpoint, \( I \) is not midpoint, so median is out. So \( \overline{FI} \) is an altitude. Wait, but wait—the problem is about triangle \( GHE \) or \( FHE \)? Let's see: \( F \) is on \( GE \), so \( \overline{FI} \) is from \( F \) to \( HE \), perpendicular. So yes, altitude.

Wait, but maybe the answer is "altitude". Let's check the options again. The options are altitude, perpendicular bisector, median, none of these. So since \( FI \) is perpendicular to \( HE \), it's an altitude (even though it doesn't bisect \( HE \), because altitude doesn't require bisecting, just being perpendicular to the opposite side's line). Perpendicular bisector requires bisecting (so \( HI = IE \)), which is not true here (80 ≠ 46). Median requires \( I \) to be midpoint, which it's not. So the correct term is altitude.

Answer:

altitude