Triangle medians worksheet with problems to find segment lengths given other median lengths.

Problem Analysis:

The task involves solving for various segment lengths in triangles where the medians are drawn. Recall that:
- A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
- The medians of a triangle intersect at a point called the centroid.
- The centroid divides each median into two segments in the ratio 2:1, with the longer segment being closer to the vertex.

General Approach:

For each problem, we will use the property that the centroid divides each median in the ratio 2:1. This means:
- If a median is divided by the centroid into two segments, the segment closer to the vertex is twice as long as the segment closer to the midpoint of the opposite side.

Solutions:

#### 1) Find YN if UN = 10.
- In the triangle, \( UY \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( UY \) in the ratio 2:1.
- Let \( UC = x \) and \( CY = 2x \). Then \( UY = UC + CY = x + 2x = 3x \).
- Given \( UN = 10 \), and since \( N \) is the midpoint of \( UY \), \( UN = \frac{1}{2} UY \).
- Therefore, \( UY = 2 \times UN = 2 \times 10 = 20 \).
- Since \( UY = 3x \), we have \( 3x = 20 \), so \( x = \frac{20}{3} \).
- The segment \( YN \) is the same as \( CY \), which is \( 2x = 2 \times \frac{20}{3} = \frac{40}{3} \).

Answer:
\[
\boxed{\frac{40}{3}}
\]

#### 2) Find JQ if KQ = 16.
- In the triangle, \( JQ \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( JQ \) in the ratio 2:1.
- Let \( JC = x \) and \( CQ = 2x \). Then \( JQ = JC + CQ = x + 2x = 3x \).
- Given \( KQ = 16 \), and since \( K \) is the midpoint of \( JQ \), \( KQ = \frac{1}{2} JQ \).
- Therefore, \( JQ = 2 \times KQ = 2 \times 16 = 32 \).

Answer:
\[
\boxed{32}
\]

#### 3) SP = 18. Find CS.
- In the triangle, \( SP \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( SP \) in the ratio 2:1.
- Let \( SC = x \) and \( CP = 2x \). Then \( SP = SC + CP = x + 2x = 3x \).
- Given \( SP = 18 \), we have \( 3x = 18 \), so \( x = 6 \).
- The segment \( CS \) is the same as \( SC \), which is \( x = 6 \).

Answer:
\[
\boxed{6}
\]

#### 4) CH = 17. Find JH.
- In the triangle, \( JS \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( JS \) in the ratio 2:1.
- Let \( JC = x \) and \( CS = 2x \). Then \( JS = JC + CS = x + 2x = 3x \).
- Given \( CH = 17 \), and since \( H \) is the midpoint of \( JS \), \( CH = \frac{1}{2} JS \).
- Therefore, \( JS = 2 \times CH = 2 \times 17 = 34 \).
- Since \( JH = JC \) and \( JC = x \), and \( JS = 3x = 34 \), we have \( x = \frac{34}{3} \).
- Thus, \( JH = x = \frac{34}{3} \).

Answer:
\[
\boxed{\frac{34}{3}}
\]

#### 5) BY = 51. Find CY.
- In the triangle, \( BS \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( BS \) in the ratio 2:1.
- Let \( BC = x \) and \( CS = 2x \). Then \( BS = BC + CS = x + 2x = 3x \).
- Given \( BY = 51 \), and since \( Y \) is the midpoint of \( BS \), \( BY = \frac{1}{2} BS \).
- Therefore, \( BS = 2 \times BY = 2 \times 51 = 102 \).
- Since \( CY = CS \) and \( CS = 2x \), and \( BS = 3x = 102 \), we have \( x = \frac{102}{3} = 34 \).
- Thus, \( CY = 2x = 2 \times 34 = 68 \).

Answer:
\[
\boxed{68}
\]

#### 6) Find OZ if RZ = 14.
- In the triangle, \( RZ \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( RZ \) in the ratio 2:1.
- Let \( RO = x \) and \( OZ = 2x \). Then \( RZ = RO + OZ = x + 2x = 3x \).
- Given \( RZ = 14 \), we have \( 3x = 14 \), so \( x = \frac{14}{3} \).
- The segment \( OZ \) is \( 2x = 2 \times \frac{14}{3} = \frac{28}{3} \).

Answer:
\[
\boxed{\frac{28}{3}}
\]

#### 7) Find JA if TA = 12.
- In the triangle, \( JA \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( JA \) in the ratio 2:1.
- Let \( JC = x \) and \( CA = 2x \). Then \( JA = JC + CA = x + 2x = 3x \).
- Given \( TA = 12 \), and since \( T \) is the midpoint of \( JA \), \( TA = \frac{1}{2} JA \).
- Therefore, \( JA = 2 \times TA = 2 \times 12 = 24 \).

Answer:
\[
\boxed{24}
\]

#### 8) HA = 12. Find CH.
- In the triangle, \( HL \) is a median, and \( C \) is the centroid.
- The centroid divides the median \( HL \) in the ratio 2:1.
- Let \( HC = x \) and \( CL = 2x \). Then \( HL = HC + CL = x + 2x = 3x \).
- Given \( HA = 12 \), and since \( A \) is the midpoint of \( HL \), \( HA = \frac{1}{2} HL \).
- Therefore, \( HL = 2 \times HA = 2 \times 12 = 24 \).
- Since \( CH = HC \) and \( HC = x \), and \( HL = 3x = 24 \), we have \( x = \frac{24}{3} = 8 \).
- Thus, \( CH = x = 8 \).

Answer:
\[
\boxed{8}
\]

Final Answers:

\[
\boxed{\frac{40}{3}, 32, 6, \frac{34}{3}, 68, \frac{28}{3}, 24, 8}
\]