Geometry Worksheets | Triangle Worksheets - Free Printable

To solve these problems, we need to use a special rule about medians in a triangle.

The Rule:
1. The three medians of a triangle meet at a single point called the centroid (labeled as $C$ in your diagrams).
2. The centroid divides each median into two parts with a specific ratio: 2:1.
* The part from the vertex (corner) to the centroid is the longer part ($\frac{2}{3}$ of the total length).
* The part from the centroid to the midpoint of the opposite side is the shorter part ($\frac{1}{3}$ of the total length).

So, mathematically:
* $\text{Vertex to Centroid} = 2 \times (\text{Centroid to Midpoint})$
* $\text{Total Median Length} = 3 \times (\text{Centroid to Midpoint})$
* $\text{Vertex to Centroid} = \frac{2}{3} \times (\text{Total Median Length})$

Let's solve each problem step-by-step using this rule.

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1) Find $YN$ if $UN = 10$.
* Look at triangle $QUN$. The line segment $NY$ is a median starting from vertex $N$, going through centroid $C$, to midpoint $Y$. Wait, looking at the diagram, $U-Y-N$ is the bottom side. The median is drawn from vertex $Q$ to side $UN$? No, let's look closer.
* In diagram 1, the vertices are $Q, U, N$. The medians are $QY$, $NV$, and $UO$. They intersect at $C$.
* The question asks for $YN$. $Y$ is on the side $UN$. Since $QY$ is a median, $Y$ is the midpoint of side $UN$.
* Therefore, $UY = YN$.
* We are given the total length of side $UN = 10$.
* Since $Y$ is the midpoint, $YN$ is half of $UN$.
* $YN = 10 / 2 = 5$.

2) Find $JQ$ if $KQ = 16$.
* Look at triangle $DJQ$. The vertices are $D, J, Q$. The medians are $DK$, $JM$, and $QE$. They intersect at $C$.
* The question asks for $JQ$. $K$ is on the side $JQ$. Since $DK$ is a median, $K$ is the midpoint of side $JQ$.
* Therefore, $JK = KQ$.
* We are given $KQ = 16$.
* The total length $JQ = JK + KQ = 16 + 16 = 32$.

3) $SP = 18$. Find $CS$.
* Look at triangle $USY$. Vertices $U, S, Y$. Medians are $UK$, $SP$, and $YQ$. Intersection is $C$.
* $SP$ is a median connecting vertex $S$ to midpoint $P$ on side $UY$.
* $C$ is the centroid on median $SP$.
* The distance from the vertex to the centroid ($SC$ or $CS$) is $\frac{2}{3}$ of the total median length ($SP$).
* $CS = \frac{2}{3} \times SP$
* $CS = \frac{2}{3} \times 18$
* $CS = 2 \times (18 / 3) = 2 \times 6 = 12$.

4) $CH = 17$. Find $JH$.
* Look at triangle $PJS$. Vertices $P, J, S$. Medians are $PM$, $JH$, and $SB$. Intersection is $C$.
* $JH$ is the median from vertex $J$ to midpoint $H$ on side $PS$.
* $C$ is the centroid.
* The segment $CH$ is the shorter part (from centroid to midpoint). It is $\frac{1}{3}$ of the total median $JH$.
* So, $JH = 3 \times CH$.
* Given $CH = 17$.
* $JH = 3 \times 17$.
* $3 \times 10 = 30$, $3 \times 7 = 21$. $30 + 21 = 51$.

5) $BY = 51$. Find $CY$.
* Look at triangle $PBS$. Vertices $P, B, S$. Medians are $PZ$, $BY$, and $SF$. Intersection is $C$.
* $BY$ is the median from vertex $B$ to midpoint $Y$ on side $PS$.
* $C$ is the centroid.
* The segment $CY$ is the shorter part (from centroid to midpoint). It is $\frac{1}{3}$ of the total median $BY$.
* $CY = \frac{1}{3} \times BY$.
* Given $BY = 51$.
* $CY = 51 / 3$.
* $3 \times 17 = 51$, so $CY = 17$.

6) Find $OZ$ if $RZ = 14$.
* Look at triangle $TRZ$? No, vertices are $T, R, Z$. Medians are $TO$, $RI$, and $ZY$. Intersection is $C$.
* Wait, let's re-read the diagram labels carefully.
* Triangle vertices: $T, R, Z$.
* Median from $T$ goes to $O$ on side $RZ$. So $O$ is the midpoint of $RZ$.
* The question asks for $OZ$.
* Since $O$ is the midpoint of side $RZ$, $RO = OZ$.
* We are given the total length of side $RZ = 14$.
* $OZ$ is half of $RZ$.
* $OZ = 14 / 2 = 7$.
*(Note: This question tests if you know what a median does to the side it touches, not just the centroid ratio. $TO$ is the median, but $RZ$ is the side being bisected.)*

7) Find $JA$ if $TA = 12$.
* Look at triangle $LJA$. Vertices $L, J, A$. Medians are $LT$, $JZ$, and $AQ$. Intersection is $C$.
* $LT$ is the median from vertex $L$ to side $JA$. So $T$ is the midpoint of side $JA$.
* The question asks for the total length of side $JA$.
* We are given $TA = 12$.
* Since $T$ is the midpoint, $JT = TA = 12$.
* Total length $JA = JT + TA = 12 + 12 = 24$.

8) $HA = 12$. Find $CH$.
* Look at triangle $DHL$. Vertices $D, H, L$. Medians are $DE$, $HL$? No.
* Let's trace the lines. Line from $D$ goes to $E$ on $HL$. Line from $H$ goes to $A$ on $DL$. Line from $L$ goes to $P$ on $DH$.
* So, $HA$ is a median from vertex $H$ to midpoint $A$ on side $DL$.
* The intersection is $C$.
* We need to find $CH$. $CH$ is the part of the median from the vertex $H$ to the centroid $C$.
* The rule is: Vertex-to-Centroid ($CH$) is $\frac{2}{3}$ of the total median ($HA$).
* $CH = \frac{2}{3} \times HA$.
* Given $HA = 12$.
* $CH = \frac{2}{3} \times 12$.
* $12 / 3 = 4$.
* $2 \times 4 = 8$.

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Summary of Answers:

1) $YN = 5$
2) $JQ = 32$
3) $CS = 12$
4) $JH = 51$
5) $CY = 17$
6) $OZ = 7$
7) $JA = 24$
8) $CH = 8$

Final Answer:
1) 5
2) 32
3) 12
4) 51
5) 17
6) 7
7) 24
8) 8