Question

jk is a midsegment of △fgh. find the values of x and y.
(there is a triangle fgh with midsegment jk. segment fj is labeled x, jh is labeled 10, segment fk is labeled y, kg is labeled 28, and fg is labeled 28. there are boxes to fill in for x = and y =)

Explanation:

Step1: Recall Midsegment Theorem

The Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle (midsegment) is parallel to the third side and half as long. Also, the midsegment divides the other two sides into segments of equal length. So, \( FJ = JH \) (since \( JK \) is a midsegment, \( J \) is the midpoint of \( FH \)) and \( y=\frac{1}{2}FG \), \( x = JH = FJ = 10\) (wait, no, let's correct. Wait, \( J \) is on \( FH \), \( K \) is on \( GH \), and \( JK \) is midsegment. So \( FJ = JH \), so \( FH = FJ + JH = 10 + x \)? Wait, no, the length from \( F \) to \( J \) is \( x \)? Wait, no, looking at the diagram: \( FJ \) is labeled \( x \)? Wait, no, the diagram: \( F \) to \( J \) is \( x \)? Wait, no, the given length: \( FJ \) is \( x \)? Wait, no, \( HJ \) is 10? Wait, the diagram shows \( HJ = 10 \), \( FJ = x \), \( FG = 28 \), \( JK = y \). Since \( JK \) is a midsegment, \( J \) is the midpoint of \( FH \), so \( FJ = JH \). So \( FJ = JH = 10 \)? Wait, no, \( JH \) is 10, so \( FJ = x = 10 \)? Wait, no, maybe \( FJ = x \), \( JH = 10 \), so since \( J \) is midpoint, \( FJ = JH \), so \( x = 10 \)? Wait, no, that can't be. Wait, no, the midsegment theorem: the midsegment is parallel to the third side and half its length, and the midsegment divides the two sides into equal segments. So \( J \) is the midpoint of \( FH \), so \( FJ = JH \). So if \( JH = 10 \), then \( FJ = x = 10 \). Then, \( JK \) is midsegment, so \( JK = \frac{1}{2}FG \). \( FG = 28 \), so \( y = \frac{28}{2} = 14 \). Wait, let's re-express:

Midsegment Theorem: In a triangle, the segment connecting the midpoints of two sides (midsegment) is parallel to the third side and half as long. Also, the midsegment divides the two sides into segments of equal length. So:

1. \( J \) is the midpoint of \( FH \), so \( FJ = JH \). Given \( JH = 10 \), so \( FJ = x = 10 \).

1. \( K \) is the midpoint of \( GH \), so \( JK \parallel FG \) and \( JK = \frac{1}{2}FG \). Given \( FG = 28 \), so \( y = \frac{28}{2} = 14 \).

Wait, that makes sense. So \( x = 10 \) (since \( FJ = JH = 10 \), so \( x = 10 \)), and \( y = 14 \) (since midsegment is half of \( FG \)).

Step2: Calculate x

Since \( J \) is the midpoint of \( FH \), \( FJ = JH \). Given \( JH = 10 \), so \( x = FJ = 10 \).

Step3: Calculate y

Since \( JK \) is the midsegment, \( JK = \frac{1}{2}FG \). Given \( FG = 28 \), so \( y = \frac{28}{2} = 14 \).

Answer:

\( x = 10 \), \( y = 14 \)