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Medians and the Centroid: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A side half and a median portion

In nondegenerate triangle ABC, D is the midpoint of BC and G is the centroid. The side BC is 18 cm long. The portion GD is 6.4 cm long.

Find BD, AG, and AD. Explain why BD and GD need not be equal.

The side and median contain different requested lengths.

Text description of this figure

D is the midpoint of BC and the given centroid G lies on median AD. Equal side halves are marked. The diagram has the given side and median proportions, with no computed length labels.
1. Hint 1
  Identify which lengths lie on the side and which lie on the median.
2. Hint 2
  The midpoint halves BC, while the centroid makes two vertex parts for each midpoint part.
Answer

$\begin{aligned}BD&=9\text{ cm}\\AG&=12.8\text{ cm}\\AD&=19.2\text{ cm}.\end{aligned}$
BD and GD are portions of different segments and need not be equal.

Full solution

D is a midpoint, so
$\begin{aligned}BD&=18\div2\\BD&=9\text{ cm}.\end{aligned}$

The centroid gives the vertex portion twice the shorter median portion.
$\begin{aligned}AG&=2(6.4)\\AG&=12.8\text{ cm}.\end{aligned}$

The whole median contains both portions.
$\begin{aligned}AD&=12.8+6.4\\AD&=19.2\text{ cm}.\end{aligned}$

BD lies on the triangle side BC.
GD lies on median AD.
The midpoint and centroid theorems govern different segments, so they do not make these lengths equal.

Answer

$\begin{aligned}BD&=9\text{ cm}\\AG&=12.8\text{ cm}\\AD&=19.2\text{ cm}.\end{aligned}$
BD and GD are portions of different segments and need not be equal.

Key idea

Side bisection and median division are different length relationships.
Problem 2 Compatibility on two medians

In nondegenerate triangle ABC, AD and BE are medians. Can a common point G on both medians be the centroid if all four of the following measurements are exact?

The whole median AD is 30 cm. Its portion AG is 20 cm.

The whole median BE is 24 cm. Its portion BG is 15 cm.

Justify your decision and give the required value of BG if the whole lengths stay fixed.
1. Hint 1
  The same division theorem must hold on each median separately.
2. Hint 2
  Compare the given vertex portion with two thirds of each whole median.
Answer

The four measurements are incompatible with a centroid. The required value of BG is 16 cm.

Full solution

On the first median, the vertex portion required by the theorem is
$\frac23(30)=20\text{ cm}.$
The supplied AG is compatible with that median.

On the second median, the required vertex portion is
$\frac23(24)=16\text{ cm}.$
This differs from the supplied 15 cm.
Since the theorem must hold on both medians, the measurements cannot all be exact for a centroid.

With the whole lengths fixed, BG must be 16 cm.
The midpoint portion GE would then be 8 cm.

Answer

The four measurements are incompatible with a centroid. The required value of BG is 16 cm.

Key idea

A candidate centroid must satisfy the division theorem on every median.
Problem 3 Link lengths on different medians

In nondegenerate triangle ABC, AD and BE are medians meeting at centroid G. E is the midpoint of CA. The length GE is 9 cm.

The vertex portions satisfy the ratio AG to BG of 4 to 5. Find BG, AG, and AD.

Connect the given comparison to two different medians.

Text description of this figure

D and E are midpoints of BC and CA. The given centroid G lies on both medians. The drawing places AG and BG in the stated four-to-five proportion, without labeling the requested numerical lengths.
1. Hint 1
  Start on the median whose shorter portion is known.
2. Hint 2
  Find one ratio part from BG, then use the four-to-five comparison to obtain AG.
Answer

$\begin{aligned}BG&=18\text{ cm}\\AG&=14.4\text{ cm}\\AD&=21.6\text{ cm}.\end{aligned}$

Full solution

On BE, the vertex portion is twice GE.
$\begin{aligned}BG&=2(9)\\BG&=18\text{ cm}.\end{aligned}$

The given ratio compares vertex portions on different medians.
$\begin{aligned}AG&=\frac45(18)\\AG&=14.4\text{ cm}.\end{aligned}$

AG contains two of the three parts of AD.
$\begin{aligned}AD&=\frac32(14.4)\\AD&=21.6\text{ cm}.\end{aligned}$

The corresponding shorter portion GD is 7.2 cm, which checks both the whole length and the two-to-one division.

Answer

$\begin{aligned}BG&=18\text{ cm}\\AG&=14.4\text{ cm}\\AD&=21.6\text{ cm}.\end{aligned}$

Key idea

Use the centroid theorem within each median and a supplied ratio to connect different medians.
Problem 4 Make a construction that records its reasons

Draw a scalene nondegenerate triangle ABC on paper. Using a compass and unmarked straightedge, construct all three medians and label their common point G.

Keep the midpoint-construction arcs. Mark each equal side-half pair with its own tick pattern. Label the vertex and midpoint portions of one median with their ratio. Explain how your construction and the theorem justify the labels.
1. Hint 1
  A median is determined once its opposite side midpoint is found.
2. Hint 2
  Use equal-radius arcs with radius greater than half of each side to construct its perpendicular bisector. Then join the opposite vertex to the midpoint.
Answer

A valid drawing has AD, BE, and CF concurrent at G, with D, E, F the midpoints of BC, CA, AB. Each side-half pair is marked separately; on AD, the order A, G, D has ratio two to one.

Full solution

For BC, use equal-radius circles centered at B and C, with a radius greater than half of BC.
Join their two distinct intersections.
This perpendicular bisector meets BC at midpoint D.
Draw AD.

Repeat on CA to obtain E and draw BE.
Repeat on AB to obtain F and draw CF.
The midpoint constructions justify the three pairs of equal side halves.

The concurrence theorem identifies the common point G.
It also gives two ratio parts for AG and one for GD.
Do not put matching equality ticks on those unequal median portions.

The completed segments are shown here; retain your own compass arcs on paper.

Answer

A valid drawing has AD, BE, and CF concurrent at G, with D, E, F the midpoints of BC, CA, AB. Each side-half pair is marked separately; on AD, the order A, G, D has ratio two to one.

Key idea

A construction should record the midpoint evidence as well as the final medians.
Problem 5 Derive a ratio without assuming the centroid theorem

In nondegenerate triangle ABC, D is the midpoint of BC and E is the midpoint of CA. AD and BE meet at the interior point G.

Prove that AG is twice GD using earlier midsegment and similarity results. Do not assume the centroid division theorem itself.

Only given midpoint equalities are marked.

Text description of this figure

Triangle ABC has midpoints D and E on BC and CA. Medians AD and BE cross at G. DE connects the given midpoints.
1. Hint 1
  Look for a segment connecting the two supplied midpoints.
2. Hint 2
  Compare triangles AGB and DGE. Identify an angle pair from parallel lines and another at the crossing.
Answer

$AG=2GD$ .

Full solution

DE joins two side midpoints, so the midsegment theorem gives $DE\parallel AB$ .
It also gives $AB=2DE$ .

Parallel lines give congruent angles GAB and GDE.
Angles AGB and DGE are vertical angles.
Thus AA establishes $\triangle AGB\sim\triangle DGE$ .

The correspondence matches AG with GD and AB with DE.
$\frac{AG}{GD}=2.$
G is strictly inside AD, so GD is positive and multiplication gives the result.
$AG=2GD.$

Match the side opposite each corresponding vertex.

Answer

$AG=2GD$ .

Key idea

The factor of two comes from a midsegment and transfers to median portions by similarity.
Problem 6 The endpoint condition

In nondegenerate triangle ABC, D is the midpoint of BC. P lies strictly inside AD, and AP is twice PD. Prove that P is the centroid.

If D is instead any point strictly inside BC, does the same ratio alone identify the centroid? Explain, or give a labeled counterexample sketch.
1. Hint 1
  Locate the actual centroid on the known median and compare its position with P.
2. Hint 2
  A ratio on a fixed median selects one point. Ask what changes when the endpoint no longer identifies a median.
Answer

With D the midpoint, P is the centroid. Without the midpoint condition, the ratio alone is insufficient.

Full solution

Let G be the centroid.
Since AD is a median, its division gives $GD=AD/3$ .

For P, segment addition and the given ratio give
$\begin{aligned}AD&=AP+PD\\AD&=3PD\\PD&=AD/3.\end{aligned}$

Both P and G lie the same distance from D along the same segment DA.
They are the same point, so P is the centroid.

For the second part, take a triangle with side midpoint M and choose a different point D inside BC.
The distinct segments AM and AD have no common point apart from A.
The centroid lies strictly inside AM, while the point P two thirds along AD lies strictly inside AD.
Hence P is not that centroid.

The second case places the same ratio on a segment that is not a median.

Answer

With D the midpoint, P is the centroid. Without the midpoint condition, the ratio alone is insufficient.

Key idea

A two-to-one division identifies the centroid only when the segment is known to be a median.
Problem 7 Two drawings with one fixed segment

Draw a segment UV 12 cm long. Mark P on UV, 8 cm from U.

Draw two different nondegenerate triangles that each have U as a vertex and UV as a median. Use different lines through V for their opposite sides.

Mark the required side midpoints, and explain why P is the centroid of both triangles even though their outlines differ.
1. Hint 1
  For UV to be a median, V must be the midpoint of the opposite side.
2. Hint 2
  On each chosen line through V, put one endpoint on either side of V at equal distances. Keep the line away from U.
Answer

Any two valid triangles with opposite-side midpoint V work. P is the centroid of both; it is two thirds of their common median from U.

Full solution

Choose a line through V that does not contain U.
Mark B and C on opposite sides of V at equal distances from V.
Then UBC is nondegenerate and UV is a median.

Choose a different line through V that also avoids U.
Mark X and Y on opposite sides of V at equal distances.
Then UXY is a second nondegenerate triangle with the same median UV.

The marked point P is 8 cm from U on the 12 cm segment.
Thus
$UP=\frac23 UV.$
Each triangle has its centroid at the unique point two thirds of that median from U, so both centroids are P.

Different outlines can have the same centroid on a shared median.

Answer

Any two valid triangles with opposite-side midpoint V work. P is the centroid of both; it is two thirds of their common median from U.

Key idea

The centroid position on one fixed median is independent of the other side choices.
Problem 8 The interior condition survives an obtuse angle

Prove that the centroid of any nondegenerate triangle lies strictly inside the triangle, including when the triangle is obtuse.

Your explanation must identify why the centroid lies inside a median and why that part of the median is inside the triangle. A drawing alone is not a proof.
1. Hint 1
  Use a median from one vertex to the midpoint of the opposite side.
2. Hint 2
  The ratio gives positive portions, and the opposite midpoint lies strictly between the two side endpoints.
Answer

The centroid lies strictly inside every nondegenerate triangle, whether its angles are acute, right, or obtuse.

Full solution

Take median AD with D strictly between B and C.
The two-median proof places G strictly between A and D and gives positive lengths AG and GD.

The open segment from a triangle vertex to an interior point of its opposite side lies inside the triangle.
To see why it is not on a boundary side, note that the line AD differs from AB and AC, and meets BC only at D.
Thus its interior points are strictly inside the triangular region.

This argument does not depend on an acute-angle condition.
It therefore applies to obtuse and right triangles as well.

The obtuse case illustrates the general interior-segment argument.

Answer

The centroid lies strictly inside every nondegenerate triangle, whether its angles are acute, right, or obtuse.

Key idea

The centroid location follows from midpoint placement and interior segment division, not angle type.
Problem 9 Which change preserves the balance model?

A uniform thin triangular sheet has geometric centroid G. Compare two separate changes to this object.

Change (a) rotates the entire sheet without changing its material.

Change (b) replaces the filled sheet by uniform wire along the same scalene triangular boundary, keeping the total mass the same.

For each change, decide whether the uniform-sheet interpretation still guarantees that the object balances at its geometric centroid. Explain why equal total mass is not enough in case (b).

A filled region and its boundary distribute material differently.

Text description of this figure

The blue solid figure represents a filled uniform triangular sheet. A translated red dashed copy shows wire only on the same boundary shape. No balance points are marked.
1. Hint 1
  The balance interpretation depends on where the mass is distributed.
2. Hint 2
  A rotation moves the same material together. Replacing the interior with boundary material changes the distribution.
Answer

(a) The centroid balance guarantee is preserved under rotation. (b) The uniform-sheet guarantee does not apply to the wire frame; equal total mass does not fix where that mass lies.

Full solution

(a) A rotation carries the whole uniform sheet and its medians together.
Its geometric centroid moves with it, and the object is still a uniform triangular sheet.
The same balance conclusion applies.

(b) A wire frame has material only on the boundary.
The filled-sheet strip explanation does not describe it.
Keeping the total mass equal does not keep the spatial distribution of that mass equal, so balance at the geometric centroid is not guaranteed.

Answer

(a) The centroid balance guarantee is preserved under rotation. (b) The uniform-sheet guarantee does not apply to the wire frame; equal total mass does not fix where that mass lies.

Key idea

Rigidly moving the same material preserves the model; redistributing it requires a new model.
Problem 10 Build a triangle around a prescribed point

You are given two distinct points A and G on paper. Construct a nondegenerate triangle ABC having A as one vertex and G as its centroid. Use a compass and unmarked straightedge.

Give the construction, justify why its centroid is G, and explain whether the triangle is unique.
1. Hint 1
  Which special segment through a vertex must contain the centroid?
2. Hint 2
  The missing portion of that segment must be half as long as AG.
3. Hint 3
  After locating its other endpoint, consider how the opposite side can have that point as its midpoint.
Answer

Construct D beyond G on ray AG with GD half of AG. Choose B and C on any line through D avoiding A, on opposite sides of D with equal positive distances from D. The resulting triangle has centroid G, and the triangle is not unique.

Full solution

Construct midpoint M of AG using its perpendicular bisector.
The given points are distinct, so the half-length GM is positive.
Extend ray AG beyond G, and copy GM from G onto that extension to locate D.

The order is A, G, D, and the copied length gives
$AG=2GD.$

Draw any line through D that does not contain A.
Choose a point B on it distinct from D, then use the compass to place C on the other side of D with DC equal to DB.
D is the midpoint of BC, and A is off its line, so ABC is a nondegenerate triangle with median AD.

The centroid is the unique interior two-to-one division point of AD.
The prescribed G has that position, so it is the centroid.
Different choices of the line and the positive distance DB give different triangles, so the triangle is not unique.

Use equal double ticks for the copied half-lengths, and single ticks for the side halves.

Answer

Construct D beyond G on ray AG with GD half of AG. Choose B and C on any line through D avoiding A, on opposite sides of D with equal positive distances from D. The resulting triangle has centroid G, and the triangle is not unique.

Key idea

A midpoint construction and a unique median division can create a triangle with a prescribed centroid.

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