Question

if $qu = 70$, what is $rt$?
$rt = $

Explanation:

Step1: Identify Midsegment

From the diagram, \( RT \) is a midsegment of \( \triangle QSU \), and \( R \) is the midpoint of \( QS \), \( T \) is the midpoint of \( SU \). Also, \( QS \) is divided into two equal parts by \( R \), so \( QR = RT \) (wait, actually, since \( RT \) is a midsegment, by the Midline Theorem, \( RT=\frac{1}{2}QU \)? Wait no, wait the segments on \( QS \): the marks show \( QR = RS \), so \( R \) is the midpoint of \( QS \), and \( T \) is the midpoint of \( SU \), so \( RT \) is parallel to \( QU \) and \( RT=\frac{1}{2}QU \)? Wait no, wait \( QU \) is a side, and \( QS \) is another side. Wait, the length \( QU = 70 \), but \( RT \) is a midsegment? Wait, no, maybe I got the segments wrong. Wait, the diagram: \( Q \) to \( R \) to \( S \) is a vertical segment, with \( QR = RS \) (the two marks), so \( R \) is the midpoint of \( QS \). Then \( T \) is the midpoint of \( SU \) (since \( ST = TU \), the two marks on the horizontal segment). So by the Midline Theorem (Midsegment Theorem) in a triangle, the segment connecting the midpoints of two sides is parallel to the third side and half its length. So here, \( RT \) connects midpoints of \( QS \) and \( SU \), so \( RT \parallel QU \) and \( RT=\frac{1}{2}QU \)? Wait, no, \( QU \) is the third side? Wait, \( Q \) to \( U \) is a side, \( Q \) to \( S \) is a side, \( S \) to \( U \) is a side. So midpoints of \( QS \) (R) and \( SU \) (T), so the midsegment \( RT \) should be parallel to \( QU \) and half its length. Wait, but \( QU = 70 \), so \( RT=\frac{1}{2} \times 70 = 35 \)? Wait, no, wait maybe I mixed up the sides. Wait, \( QS \) is the vertical side, with \( QR = RS \), so \( QS = QR + RS = 2QR \). Then \( RT \) is parallel to \( QU \), and \( RT \) is half of \( QU \)? Wait, let's re-examine. The problem is: If \( QU = 70 \), what is \( RT \)? Wait, maybe \( RT \) is a midsegment, so \( RT = \frac{1}{2} QU \)? Wait, no, maybe \( QU \) is a side, and \( RT \) is half of \( QU \). Wait, let's check the Midsegment Theorem: In a triangle, the segment connecting the midpoints of two sides is parallel to the third side and half as long. So if \( R \) is the midpoint of \( QS \) and \( T \) is the midpoint of \( SU \), then \( RT \parallel QU \) and \( RT = \frac{1}{2} QU \). So \( RT = \frac{70}{2} = 35 \). Wait, but maybe I made a mistake. Wait, no, the Midsegment Theorem: midsegment length is half the length of the third side. So here, third side is \( QU \), so \( RT = \frac{1}{2} QU \). So \( RT = \frac{70}{2} = 35 \).

Wait, but let's confirm. The diagram: \( Q \) at top, \( S \) at bottom left, \( U \) at bottom right. So triangle \( QSU \), with \( QS \) vertical, \( SU \) horizontal, \( QU \) slant. Midpoints \( R \) (of \( QS \)) and \( T \) (of \( SU \)), so \( RT \) is midsegment, so \( RT \parallel QU \) and \( RT = \frac{1}{2} QU \). Therefore, \( RT = \frac{70}{2} = 35 \).

Step1: Apply Midsegment Theorem

The Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. Here, \( R \) is the midpoint of \( QS \) (since \( QR = RS \)) and \( T \) is the midpoint of \( SU \) (since \( ST = TU \)). Thus, \( RT \) is the midsegment of \( \triangle QSU \), so \( RT = \frac{1}{2} QU \).

Step2: Calculate \( RT \)

Given \( QU = 70 \), substitute into the formula:
\( RT = \frac{1}{2} \times 70 \)
\( RT = 35 \)

Answer:

\( 35 \)