Geometry worksheet focusing on medians and altitudes in triangles.

Let’s go step by step through each problem on the worksheet. We’ll solve them one at a time, carefully checking our work.

---

Problem 1:
> A median connects a ______ to the ______ of the opposite side.

✔ Answer:
A median connects a vertex to the midpoint of the opposite side.

*(This is the definition of a median in a triangle.)*

---

Problem 2:
> The point where all of the medians meet in a triangle is called the ______.

✔ Answer:
The point where all of the medians meet is called the centroid.

*(Remember: centroid = center of gravity of the triangle. All three medians intersect here.)*

---

Problem 3:
> If PT̅ is a median of △PQR, mark the triangle and find x, m∠PTQ, and m∠PTR.

Looking at the diagram:
- Triangle PQR with median from P to T on QR.
- Angles given: ∠QPT = 4x + 5°, ∠RPT = 7x - 10°
- Since PT is a median, it goes to the midpoint — but more importantly, for angles at P, if PT were also an angle bisector, those two angles would be equal. But wait — the problem doesn’t say it’s an angle bisector. However, looking at the diagram, it seems like they’re implying that PT splits angle P into two parts, and we’re to find x such that... actually, hold on — maybe there's a typo or mislabel? Let me re-read.

Wait — actually, in many textbooks, when they give you expressions for two adjacent angles formed by a line from the vertex (like PT), and ask you to find x, it often implies that those two angles are equal — meaning PT is also an angle bisector. But the problem says “PT is a median”, not necessarily an angle bisector.

BUT — look at the diagram again (even though I can't see it, based on standard problems): usually, if they give you two expressions for angles at the same vertex split by a segment, and ask you to solve for x, it’s because those two angles are meant to be equal — perhaps assuming symmetry or that it’s also an angle bisector. Alternatively, maybe the total angle is given? No, not here.

Wait — another possibility: since PT is a median, and if the triangle is isosceles with PQ = PR, then the median would also be the angle bisector. But again, not stated.

Actually — let’s think differently. Maybe the key is that points Q, T, R are colinear, so angles ∠PTQ and ∠PTR are supplementary (they form a straight line). That makes sense!

Yes! Because T is on QR, so ∠PTQ and ∠PTR are adjacent angles forming a straight line → they add up to 180°.

But the expressions given are for ∠QPT and ∠RPT — which are at vertex P, not at T.

Hold on — let me clarify notation:

In triangle PQR, with median PT (so T is midpoint of QR).

Angles labeled:
- At vertex P: ∠QPT = 4x + 5° and ∠RPT = 7x - 10° → these are the two parts of angle QPR.

If no other info, we cannot assume they are equal unless told PT is angle bisector.

But perhaps in this context, since it’s a median and they’re giving us two expressions, and asking for x, it’s likely that the intention is that PT is ALSO the angle bisector — making those two angles equal.

That’s a common setup in geometry problems: “if a median is also an angle bisector, then...”

Alternatively, maybe the diagram shows that triangle is isosceles? Without seeing it, we have to go by typical textbook problems.

Let me check both possibilities.

Possibility 1: Assume PT is angle bisector → then 4x + 5 = 7x - 10

Solve:
4x + 5 = 7x - 10
5 + 10 = 7x - 4x
15 = 3x
x = 5

Then ∠QPT = 4(5)+5 = 25°, ∠RPT = 7(5)-10 = 25° → yes, equal.

Now, what about m∠PTQ and m∠PTR?

Since T is on QR, and PT is drawn, then angles at T: ∠PTQ and ∠PTR are on a straight line → sum to 180°.

But we don’t know their individual measures yet. Unless... is there more info?

Wait — in the diagram, maybe it’s marked that PT is perpendicular? Or maybe not.

Actually, looking back at the original problem statement: “mark the triangle and find x = ___, m∠PTQ = ___, m∠PTR = ___”

And in the diagram description, it might show that PT is altitude too? But problem says only “median”.

Hmm. Perhaps after finding x=5, we need to find angles at T.

But without additional info, we can’t determine ∠PTQ and ∠PTR individually — unless...

Wait — another thought: if PT is median and also angle bisector, and if the triangle is isosceles, then PT might also be altitude → so angles at T would be 90° each.

Is that possible?

Let’s test: if x=5, and angles at P are both 25°, then total angle at P is 50°.

If triangle is isosceles with PQ=PR, then base angles at Q and R are equal.

Sum of angles in triangle = 180°, so angles at Q and R: (180 - 50)/2 = 65° each.

Now, in triangle PTQ: angles are at P: 25°, at Q: 65°, so at T: 180 - 25 - 65 = 90°.

Similarly in triangle PTR: same thing → 90°.

So yes! If PT is median and angle bisector in an isosceles triangle, it’s also altitude → so angles at T are 90°.

Therefore:

x = 5
m∠PTQ = 90°
m∠PTR = 90°

This fits perfectly.

So even though the problem didn’t explicitly say “isosceles” or “altitude”, the numbers work out that way, and it’s a standard result.

✔ Final for Problem 3:
x = 5
m∠PTQ = 90°
m∠PTR = 90°

---

Problem 4:
> If RT̅ is a median and ST = x + 7, TP = 3x - 1, mark the triangle and find x, ST, TP.

RT is a median → so T is midpoint of SP? Wait, triangle is probably SRT or something. Looking at labels: points S, T, P, R.

Diagram shows triangle SRP? With median from R to T on SP.

So RT is median → T is midpoint of SP → so ST = TP.

Given:
ST = x + 7
TP = 3x - 1

Set equal:
x + 7 = 3x - 1
7 + 1 = 3x - x
8 = 2x
x = 4

Then:
ST = 4 + 7 = 11
TP = 3(4) - 1 = 12 - 1 = 11 → checks out.

✔ Answer:
x = 4
ST = 11
TP = 11

---

Problem 5:
> What segment is a median in the triangle at the right? What kind of triangle is ΔABC?

Diagram description: triangle ABC, with point D on BC, and AD drawn. Also, BD = DC = 4, and AB = AC = 5? Wait, labels: A at top, B left, C right, D on BC. Markings: BD and DC both marked with one tick → so BD = DC → D is midpoint → so AD is median.

Also, AB and AC both marked with two ticks → so AB = AC → triangle is isosceles.

Additionally, angle at A has arc with 60°? Wait, in your text: “A 60°” — probably angle BAC = 60°.

If AB = AC and angle A = 60°, then triangle is equilateral!

Because in isosceles triangle with vertex angle 60°, base angles are (180-60)/2 = 60° each → all angles 60° → equilateral.

Also, sides: if AB = AC = 5, and BC = BD + DC = 4 + 4 = 8? Wait, that contradicts equilateral.

Wait — inconsistency?

In your problem statement: “A 60°” and “B 4 D 4 C” — so BD=4, DC=4 → BC=8.

AB and AC marked equal — say length 5? But 5,5,8 is isosceles but not equilateral.

Angle at A is 60°, sides AB=AC=5, base BC=8.

Check using Law of Cosines:

BC² = AB² + AC² - 2·AB·AC·cos(angle A)
8² = 5² + 5² - 2·5·5·cos(60°)
64 = 25 + 25 - 50·(0.5)
64 = 50 - 25 = 25 → 64 ≠ 25 → contradiction.

So probably the side lengths are not 5 — maybe the markings indicate equality, but actual lengths not given numerically except BD=DC=4.

Perhaps AB and AC are equal, angle A=60°, so it must be equilateral → so BC should equal AB=AC.

But BD=DC=4 → BC=8 → so AB=AC=8.

Then angle A=60°, sides 8,8,8 → equilateral.

Probably the "5" was a misread — in many diagrams, they mark sides with ticks, not numbers.

Assuming AB = AC (two ticks), BD = DC (one tick each), angle A = 60° → then triangle is equilateral.

Median is AD (since D is midpoint of BC).

✔ Answers:
Median: AD
Triangle type: Equilateral

*(Even if side lengths aren't specified, with AB=AC and angle A=60°, it forces equilateral.)*

---

Problems 6 and 7:
> For number 6 & 7, solve for x given the following median.

These involve centroids. Remember: centroid divides each median into segments with ratio 2:1, with the longer part toward the vertex.

So, for any median, from vertex to centroid : centroid to midpoint = 2:1

---

Problem 6:
Diagram: triangle with median from E to G, centroid at F? Labels: E to F is 4x - 12, F to G is 2x + 3.

Since F is centroid, EF : FG = 2:1

So EF = 2 * FG

Set up equation:
4x - 12 = 2*(2x + 3)

Solve:
4x - 12 = 4x + 6
Subtract 4x from both sides:
-12 = 6 → Contradiction!

That can’t be. Did I assign correctly?

EF is from vertex E to centroid F, FG is from centroid F to midpoint G → so yes, EF should be twice FG.

But 4x - 12 = 2*(2x + 3) → 4x - 12 = 4x + 6 → -12=6 → impossible.

Mistake in assignment? Perhaps the labels are different.

Maybe EF is the whole median? No, typically F is centroid.

Another possibility: perhaps 4x-12 is from E to F, and 2x+3 is from F to G, but maybe it's the other way around? But that wouldn't make sense because centroid is closer to midpoint.

Unless... perhaps the expression for EF is smaller? But 4x-12 vs 2x+3 — for positive x, 4x-12 could be less if x small.

Try setting FG = 2 * EF? But that violates centroid property.

Centroid always divides median so that vertex to centroid is twice centroid to midpoint.

So EF / FG = 2 / 1 → EF = 2 FG

But as above, leads to contradiction.

Perhaps the expressions are switched? Suppose EF = 2x + 3, FG = 4x - 12

Then 2x + 3 = 2*(4x - 12)
2x + 3 = 8x - 24
3 + 24 = 8x - 2x
27 = 6x
x = 4.5

Then EF = 2(4.5) + 3 = 9 + 3 = 12
FG = 4(4.5) - 12 = 18 - 12 = 6 → 12:6 = 2:1 → correct.

Probably the labels were misassigned in my assumption. In the diagram, likely the segment from vertex to centroid is labeled as 2x+3, and centroid to midpoint as 4x-12? But that would mean the shorter segment has larger expression, which is odd.

Perhaps in the diagram, it's written as EG is the median, F is centroid, EF = 4x-12, FG = 2x+3, but then as before, contradiction.

Another idea: perhaps "solve for x" and the expressions are for the parts, but we need to use the ratio.

Set EF / FG = 2 / 1 → (4x-12)/(2x+3) = 2/1

So 4x - 12 = 2*(2x + 3) → same as before → 4x-12=4x+6 → -12=6 → impossible.

Unless the median is from another vertex.

Perhaps F is not between E and G? Unlikely.

Or perhaps the expressions are for different segments.

Looking back at user input: "6.) EG ← 4x-12, 2x+3" — probably means on median EG, with F centroid, EF = 4x-12, FG = 2x+3.

But mathematically impossible.

Unless I have the ratio wrong. Is it possible that in some contexts it's different? No, centroid always 2:1.

Perhaps it's 1:2? But no, vertex to centroid is longer.

Another thought: maybe 4x-12 is the entire median? But then why two expressions.

Perhaps "EG" is the median, and the expressions are for EF and FG, but F is not centroid? But problem says "given the following median" and mentions centroid implicitly.

Wait, in problem 6 and 7, it says "solve for x given the following median" and diagrams show centroid.

Perhaps for problem 6, the segment from vertex to centroid is 2x+3, and from centroid to midpoint is 4x-12, but that would require 2x+3 = 2*(4x-12) as I did earlier, giving x=4.5.

And 2x+3 = 12, 4x-12=6, ratio 2:1.

Probably in the diagram, the label "4x-12" is on the shorter segment, "2x+3" on the longer, but that would be unusual labeling.

To resolve, let's assume that the expression for the part from vertex to centroid is twice the other part.

So let VtoC = 2 * CtoM

Suppose VtoC = a, CtoM = b, a=2b.

In problem 6, if we take 2x+3 as VtoC, 4x-12 as CtoM, then 2x+3 = 2*(4x-12) → x=4.5 as above.

If we take 4x-12 as VtoC, 2x+3 as CtoM, then 4x-12 = 2*(2x+3) → impossible.

So likely, the intended assignment is that the larger expression is for the longer segment, but 4x-12 > 2x+3 when x>7.5, but at x=4.5, 4*4.5-12=6, 2*4.5+3=12, so 12>6, so if 2x+3 is VtoC=12, 4x-12=CtoM=6, ratio 2:1.

So probably in the diagram, the segment from vertex to centroid is labeled "2x+3", and from centroid to midpoint is "4x-12", even though numerically at solution, 2x+3 is larger.

Perhaps the labels are placed near the segments, and "2x+3" is near the vertex part.

I think we have to go with x=4.5 for problem 6.

Similarly for problem 7.

But let's confirm with problem 7 first.

Problem 7:
Diagram: median from B to D, centroid at C? Labels: B to C is 2x+24, C to D is 4x-18.

Centroid C, so BC : CD = 2:1

So BC = 2 * CD

2x + 24 = 2*(4x - 18)

Solve:
2x + 24 = 8x - 36
24 + 36 = 8x - 2x
60 = 6x
x = 10

Then BC = 2(10)+24 = 44
CD = 4(10)-18 = 22 → 44:22 = 2:1 → perfect.

So for problem 7, x=10 works with BC=2x+24, CD=4x-18.

For problem 6, if we do similar: suppose EF = 2x+3 (vertex to centroid), FG = 4x-12 (centroid to midpoint), then 2x+3 = 2*(4x-12) → x=4.5 as before.

But in problem 6, the user wrote "EG ← 4x-12, 2x+3" — perhaps the order is from E to G, with F in between, and EF=4x-12, FG=2x+3, but then as before, impossible.

Unless the ratio is applied differently.

Another possibility: perhaps "solve for x" and the expressions are for the lengths, and we set the ratio.

But mathematically, only if we swap the assignments.

Perhaps in problem 6, the median is from G to E, and F is centroid, GF = 4x-12, FE = 2x+3, then GF should be twice FE if G is vertex.

Assume G is vertex, F centroid, E midpoint.

Then GF : FE = 2:1

So GF = 2 * FE

4x - 12 = 2*(2x + 3) → same contradiction.

If FE = 2 * GF, then 2x+3 = 2*(4x-12) → x=4.5, and FE=12, GF=6, so if F is centroid, and E is vertex, then from vertex E to centroid F is 12, F to midpoint G is 6, ratio 2:1, good.

So probably in the diagram, E is the vertex, G is the midpoint, F centroid, with EF = 2x+3, FG = 4x-12.

But the user wrote "EG ← 4x-12, 2x+3" — perhaps the arrow indicates direction, and the expressions are listed in order from E to G, so first segment EF=4x-12, second FG=2x+3, but then it doesn't work.

To avoid confusion, since problem 7 works with the first expression being the vertex-to-centroid part, for consistency, in problem 6, let's assume that the expression associated with the vertex-to-centroid segment is the one that should be twice the other.

In problem 6, if we take 2x+3 as vertex-to-centroid, 4x-12 as centroid-to-midpoint, then 2x+3 = 2*(4x-12) → x=4.5

And it works numerically.

Perhaps the "4x-12" is a distractor or mislabeled, but mathematically, x=4.5 is the only solution that satisfies the centroid ratio.

So I'll go with that.

✔ Problem 6: x = 4.5
✔ Problem 7: x = 10

---

Now, compiling all answers:

Final Answers:

1. vertex, midpoint
2. centroid
3. x = 5, m∠PTQ = 90°, m∠PTR = 90°
4. x = 4, ST = 11, TP = 11
5. Median: AD, Triangle: Equilateral
6. x = 4.5
7. x = 10

Note: For problem 6, if the diagram clearly shows which segment is which, adjust accordingly, but based on standard interpretation and mathematical consistency, x=4.5 is correct under the assumption that the vertex-to-centroid segment is twice the centroid-to-midpoint segment, and assigning 2x+3 to the former and 4x-12 to the latter.

---

Final Answer:
1. vertex, midpoint
2. centroid
3. x = 5, m∠PTQ = 90°, m∠PTR = 90°
4. x = 4, ST = 11, TP = 11
5. Median: AD, Triangle: Equilateral
6. x = 4.5
7. x = 10