Question

1. point l is a centroid of the triangle fwd. if rl = 54 cm, what is rd?

Explanation:

Step1: Recall Centroid Property

The centroid of a triangle divides each median into a ratio of \(2:1\), with the longer segment being closer to the vertex. Also, the centroid is the midpoint of the segment connecting the midpoints of two sides (by the midline theorem and centroid properties). Wait, actually, in a triangle, if \(L\) is the centroid, and \(R\) is the midpoint of \(WF\) (since \(WR = RF\) from the marks), and \(J\) is the midpoint of \(FD\) (since \(FJ = JD\) from the marks), then \(RL\) is a midline segment? No, wait, the centroid divides the median. Wait, actually, the line connecting midpoints of two sides is parallel to the third side and half its length (midline theorem). But here, \(R\) is midpoint of \(WF\), \(J\) is midpoint of \(FD\), so \(RJ\) is midline, parallel to \(WD\) and \(RJ=\frac{1}{2}WD\). But \(L\) is the centroid, so the medians are \(WJ\) (from \(W\) to midpoint \(J\) of \(FD\)) and \(DR\) (from \(D\) to midpoint \(R\) of \(WF\))? Wait, no, the medians are from each vertex to the midpoint of the opposite side. So vertex \(D\) to midpoint \(R\) of \(WF\) is a median, and vertex \(W\) to midpoint \(J\) of \(FD\) is another median. The centroid \(L\) is the intersection of the medians. So in median \(DR\) (from \(D\) to \(R\), midpoint of \(WF\)), the centroid \(L\) divides \(DR\) into \(DL:LR = 2:1\)? Wait, no, the centroid divides each median into a ratio of \(2:1\), with the segment from the vertex to centroid being twice the segment from centroid to midpoint. Wait, so if \(R\) is the midpoint of \(WF\), then \(DR\) is a median, and \(L\) is the centroid, so \(DL = 2 \times LR\)? Wait, no, wait: the centroid is located at \(\frac{2}{3}\) of the median from the vertex, and \(\frac{1}{3}\) from the midpoint. So if \(R\) is the midpoint of \(WF\), then the median is \(DR\), with \(D\) being the vertex, \(R\) the midpoint. Then \(L\) (centroid) is \(\frac{2}{3}\) from \(D\) and \(\frac{1}{3}\) from \(R\). Wait, but the problem says \(RL = 54\) cm. Wait, maybe I got the direction wrong. Wait, \(R\) is on \(WF\), midpoint, and \(L\) is on \(DR\). So \(RL\) is the segment from \(R\) (midpoint) to \(L\) (centroid), and \(DL\) is from \(D\) to \(L\). Then, since centroid divides median into \(2:1\) (vertex to centroid : centroid to midpoint), so \(DL:LR = 2:1\)? Wait, no, vertex to centroid is twice centroid to midpoint. So \(DL = 2 \times LR\)? Wait, no, if \(R\) is midpoint, then the median is \(DR\), with length \(DL + LR\). The centroid is \(\frac{2}{3}\) from \(D\) and \(\frac{1}{3}\) from \(R\). So \(LR = \frac{1}{3}DR\), and \(DL = \frac{2}{3}DR\). Therefore, \(DR = 3 \times LR\). Wait, but the problem is asking for \(RD\), which is the same as \(DR\) (length). Wait, \(RL = 54\) cm, which is \(LR\) (from \(L\) to \(R\)). So \(DR = 3 \times RL\)? Wait, no, wait: if centroid is \(\frac{1}{3}\) from midpoint (\(R\)) and \(\frac{2}{3}\) from vertex (\(D\)), then \(LR = \frac{1}{3}DR\), so \(DR = 3 \times LR\). So \(RD = DR = 3 \times 54 = 162\)? Wait, no, wait, maybe I mixed up the median. Wait, let's re-express:

Median: from vertex \(D\) to midpoint \(R\) of \(WF\). So the median is \(DR\). Centroid \(L\) is on \(DR\), so \(DL:LR = 2:1\). Therefore, \(DL = 2 \times LR\), and \(DR = DL + LR = 2LR + LR = 3LR\). Given \(RL = LR = 54\) cm (since \(RL\) is the segment from \(R\) to \(L\), which is \(LR\)), so \(DR = 3 \times 54 = 162\) cm. Therefore, \(RD = DR = 162\) cm.

Wait, but maybe the midline is involved. Wait, \(R\) is midpoint of \(WF\), \(J\) is midpoint of \(FD\), so \(RJ\) is midline, parallel to \(WD\), and \(RJ = \frac{1}{2}WD\). But the centroid is the intersection of medians, so the medians are \(WJ\) (from \(W\) to \(J\), midpoint of \(FD\)) and \(DR\) (from \(D\) to \(R\), midpoint of \(WF\)). So their intersection is \(L\), the centroid. So in median \(DR\), \(L\) divides it into \(DL:LR = 2:1\). So \(LR = 54\) cm, so \(DL = 108\) cm, so \(DR = DL + LR = 108 + 54 = 162\) cm. Therefore, \(RD = 162\) cm.

Step1: Recall Centroid-Median Ratio

The centroid of a triangle divides each median into a ratio of \(2:1\) (vertex to centroid : centroid to midpoint).
Let the median be \(DR\) (from \(D\) to midpoint \(R\) of \(WF\)). The centroid \(L\) lies on \(DR\), so \(DL:LR = 2:1\), meaning \(DR = DL + LR = 2LR + LR = 3LR\).

Step2: Substitute \(LR = RL = 54\) cm

Given \(RL = 54\) cm (which is \(LR\), the segment from \(L\) to \(R\)), substitute into the formula:
\(RD = DR = 3 \times LR = 3 \times 54 = 162\) cm.

Answer:

\(RD = \boldsymbol{162}\) cm