Classify triangles by angles and sides with this geometry worksheet.

Let’s go step by step to classify each triangle correctly.

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Part 1: Classify by Angles (acute, equiangular, right, or obtuse)

Remember:
- Acute = all angles < 90°
- Right = one angle = 90°
- Obtuse = one angle > 90°
- Equiangular = all angles equal (each 60° — also acute!)

#### Problem 1:
All angles marked with same arc → likely all equal? But no numbers. Wait — in geometry, if all three angles have the same mark, it means they’re equal → so 60° each → equiangular

But let’s check others first.

Actually, looking again — problem 1 has three arcs, but not labeled. In many worksheets, this means “all angles equal” → equiangular.

But wait — maybe we should calculate from other problems where angles are given.

Let’s do ones with numbers first.

#### Problem 2:
Angles: 150°, 20°, 10° → sum = 180° ✓
One angle is 150° > 90° → obtuse

#### Problem 3:
Angles: 40°, 60°, 80° → all < 90° → acute

#### Problem 4:
Has a square corner → that’s 90° → right

#### Problem 5:
Angles: 55°, 55°, 70° → all < 90° → acute

#### Problem 6:
Has square corner → 90° → right

#### Problem 7:
Angles: 140°, 15°, 25° → 140° > 90° → obtuse

#### Problem 8:
Angles: 60°, 60°, 60° → all equal → equiangular

Now back to Problem 1: Since no numbers, but all angles marked same → must be equiangular → equiangular

Wait — actually, in some systems, if all angles are marked with same symbol, it implies equality → so yes, equiangular.

So:

1.) equiangular
2.) obtuse
3.) acute
4.) right
5.) acute
6.) right
7.) obtuse
8.) equiangular

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Part 2: Classify by Sides (scalene, isosceles, equilateral)

Remember:
- Scalene = all sides different
- Isosceles = at least two sides equal
- Equilateral = all three sides equal

Look for tick marks on sides — same number of ticks = same length.

#### Problem 9:
Two sides have one tick → those two equal → isosceles

#### Problem 10:
All three sides have different ticks (one, two, three) → all different → scalene

Wait — actually, look: side with one tick, another with two, another with three → all different → scalene

#### Problem 11:
Two sides have one tick → those two equal → isosceles

#### Problem 12:
All three sides have same tick (three ticks each?) Wait — looks like all three sides have identical markings → probably equilateral? Let me see.

In image: triangle 12 — all three sides have same double-tick? Or triple? Actually, in standard notation, if all sides have same marking, it’s equilateral.

Looking carefully: in problem 12, all three sides have the same pattern (looks like three small lines) → so equilateral

#### Problem 13:
Two sides have one tick → those two equal → isosceles

#### Problem 14:
Two sides have one tick → those two equal → isosceles

Wait — problem 14: left and bottom sides have one tick, right side has none? No — actually, in the image, two sides have single tick, one side has no tick? That would mean only two equal → isosceles.

Yes.

So:

9.) isosceles
10.) scalene
11.) isosceles
12.) equilateral
13.) isosceles
14.) isosceles

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Part 3: Complete Charts (Classify by Angles AND Sides)

We need to find angles and side lengths for each sub-triangle.

#### Problem 15:

Triangle ABD:
- Angles: At A: 55°, at B: 70°, so at D: 180 - 55 - 70 = 55° → angles: 55°, 70°, 55° → all < 90° → acute
- Sides: AB = 7, AD = 13, BD = ? Not given directly, but since angles at A and D are both 55°, sides opposite them should be equal → side opposite A is BD, opposite D is AB → so BD = AB = 7 → so sides: AB=7, BD=7, AD=13 → two sides equal → isosceles

Wait — let's confirm:

In triangle ABD:
∠A = 55°, ∠B = 70°, ∠D = 55° → yes, so sides opposite equal angles are equal:
- Side opposite ∠A (55°) is BD
- Side opposite ∠D (55°) is AB
→ So BD = AB = 7
Side AD = 13 → so two sides equal → isosceles

So ΔABD: acute, isosceles

Triangle DBC:
Points D, B, C
Angles: At D: we know from before, in big triangle ABC, point D is on AC.
From diagram: ∠DBC = ? We have ∠ABC = 70° + ? Wait, better to use given angles.

At point B: total angle is split into 70° (for ABD) and then angle DBC.

Also, at point C: 35°, at point D in triangle DBC: angle at D is adjacent to angle ADB which was 55°, so angle BDC = 180 - 55 = 125°? Because they form straight line at D on AC.

Yes! Point D is on AC, so angles ADB and BDC are supplementary → ∠ADB = 55° → ∠BDC = 125°

Then in triangle DBC:
∠D = 125°, ∠C = 35°, so ∠B = 180 - 125 - 35 = 20°

So angles: 125°, 35°, 20° → one >90° → obtuse

Sides: DB = 7 (from earlier), BC = 24, DC = 12 → all different → scalene

So ΔDBC: obtuse, scalene

Triangle ABC:
Whole triangle: points A, B, C
Angles: ∠A = 55°, ∠C = 35°, so ∠B = 180 - 55 - 35 = 90° → right

Sides: AB = 7, BC = 24, AC = AD + DC = 13 + 12 = 25
Check: 7-24-25 is Pythagorean triple? 7²+24²=49+576=625=25² → yes! So right triangle, and all sides different → scalene

So ΔABC: right, scalene

Summary for 15:
ΔABD: acute, isosceles
ΔDBC: obtuse, scalene
ΔABC: right, scalene

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#### Problem 16:

Triangle LMO:
Given: LM = 5, MO perpendicular to LN (since right angle at O), LO = ? ON = 3, MN = 5, angle at M is 20°? Wait, arrow shows angle between LM and MN is 20°? Actually, label says "20°" near M, between LM and MN.

But triangle LMO: points L, M, O
We know: MO ⊥ LN → so ∠MOL = 90°
LM = 5, LO = ? But we can find angles.

Since MO is altitude, and LM = MN = 5, and LO and ON may not be equal.

Angle at M for whole triangle LMN is 20°, but that’s angle LMN.

In triangle LMO:
∠O = 90°
LM = 5
LO = ? Not given, but perhaps we can find angles.

Note: since LM = MN = 5, triangle LMN is isosceles with LM=MN, so base angles at L and N are equal.

Angle at M is 20°, so angles at L and N: (180-20)/2 = 80° each.

So in triangle LMO:
∠L = 80° (same as in big triangle)
∠O = 90°
So ∠M = 180 - 80 - 90 = 10°

Angles: 80°, 90°, 10° → has 90° → right

Sides: LM=5, MO=?, LO=?
But we don’t need exact lengths — since angles are all different, sides are all different → scalene

Wait — is that true? In right triangle, if angles are 10,80,90, all different, so sides all different → scalene.

Yes.

So ΔLMO: right, scalene

Triangle OMN:
Similarly, ∠O = 90°, ∠N = 80° (from above), so ∠M = 10°

Same as above → angles 90°,80°,10° → right, and sides: MN=5, MO same as before, ON=3 → but LO and ON may differ.

Actually, since triangle LMN is isosceles with LM=MN=5, and MO is altitude to base LN, then it should also be median, so LO = ON.

Is that true? Only if it’s isosceles with apex M, which it is — LM=MN, so yes, altitude from M to LN bisects LN → so LO = ON = 3? But ON is given as 3, so LO=3.

Then in triangle OMN:
Sides: OM (common), ON=3, MN=5
Angles: 90° at O, 80° at N, 10° at M → same as LMO.

But sides: ON=3, MN=5, OM = sqrt(MN² - ON²) = sqrt(25-9)=sqrt(16)=4

Similarly, in LMO: LO=3, LM=5, OM=4 → same sides!

So triangle LMO and OMN are congruent? Both have sides 3,4,5 → right triangles.

So for ΔOMN: angles 90°,80°,10° → right, sides 3,4,5 → all different → scalene

But wait — sides are 3,4,5 — all different → scalene.

Yes.

Triangle LMN:
Points L,M,N
Sides: LM=5, MN=5, LN= LO+ON=3+3=6
So two sides equal → isosceles

Angles: we already said ∠M=20°, ∠L=80°, ∠N=80° → all <90° → acute

So ΔLMN: acute, isosceles

Summary for 16:
ΔLMO: right, scalene
ΔOMN: right, scalene
ΔLMN: acute, isosceles

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#### Problem 17:

Triangle EFH:
Given: right angle at E, so ∠E=90°
FH has tick marks — looks like EH and FH have same ticks? Let's see.

Diagram: F to H has one tick, H to G has one tick, E to H has two ticks? And E to F has no tick? Also, angle at H in triangle EFH is 60°.

Actually, in triangle EFH:
∠E = 90° (given by square)
∠H = 60° (labeled)
So ∠F = 180 - 90 - 60 = 30°

Angles: 90°,60°,30° → right

Sides: need to see markings.
EF: no tick
EH: two ticks
FH: one tick
All different → scalene

But wait — is there any equality? Probably not, since different ticks.

So ΔEFH: right, scalene

Triangle EHG:
Points E,H,G
∠E = 90° (same right angle)
∠G = 30° (labeled)
So ∠H = 180 - 90 - 30 = 60°

Angles: 90°,30°,60° → right

Sides: EH has two ticks, HG has one tick, EG has no tick? All different → scalene

But note: in diagram, FH and HG both have one tick → so FH = HG
And EH is common.

But for triangle EHG, sides are EH, HG, EG — all seem different.

Yes.

So ΔEHG: right, scalene

Triangle EFG:
Whole triangle: E,F,G
∠E = 90°
∠F = 30° (from earlier, in EFH)
∠G = 30° (given)
So angles: 90°,30°,30° → right

Sides: EF and EG — are they equal? Angles at F and G are both 30°, so sides opposite should be equal: side opposite F is EG, opposite G is EF → so EF = EG → isosceles

Also, FG is hypotenuse.

So ΔEFG: right, isosceles

Summary for 17:
ΔEFH: right, scalene
ΔEHG: right, scalene
ΔEFG: right, isosceles

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#### Problem 18:

This one is complex. Let's break down.

First, identify triangles: XYZ, TUV, TSZ

Given:
- Triangle XYZ: XY=14, YZ=10, XZ=6+T Z? Points X,T,Z,V on line.
X to T:6, T to Z:? , Z to V:10
Also, angle at U is 75°, at V is 45°, etc.

Perhaps better to find angles first.

Start with triangle TSZ:
Points T,S,Z
Given: angle at S is 120°, angle at Z is ?
Also, SZ=6, TZ=? , ST=7? Labels: ST=7, SZ=6, and angle at S is 120°.

In triangle TSZ:
Sides: ST=7, SZ=6, TZ=?
Angle at S: 120°
So we can find other angles, but perhaps classify by angles first.

Angle at S is 120° >90° → so obtuse

Sides: 7,6, and TZ — probably all different → scalene

To confirm, use law of cosines, but for classification, since angle is obtuse, and no indication of equal sides, assume scalene.

Now triangle TUV:
Points T,U,V
Given: TU=7, UV=14, angle at U=75°, angle at V=45°
So angle at T: 180 - 75 - 45 = 60°
All angles <90° → acute

Sides: TU=7, UV=14, TV=?
TV = TZ + ZV, but ZV=10, TZ unknown.

But sides: 7,14, and TV — likely all different → scalene

Unless TV equals one, but 7 and 14, TV probably not equal.

Now triangle XYZ:
Points X,Y,Z
XY=14, YZ=10, XZ= XT + TZ =6 + TZ
But we need angles or more info.

Note: there is a right angle at X? Diagram shows square at X for triangle XYZ? Looking back: in problem 18, at point X, there is a square, so ∠X=90° for triangle XYZ.

Yes! So in ΔXYZ, ∠X=90°

Sides: XY=14, XZ=6 + TZ, YZ=10
But YZ=10 is hypotenuse? If ∠X=90°, then hypotenuse is YZ=10, but legs XY=14 and XZ — but 14>10, impossible for leg to be longer than hypotenuse.

Contradiction.

Perhaps I misread.

Look: in diagram, for triangle XYZ, points X,Y,Z.
At X, there is a right angle symbol, so ∠YXZ=90°
Sides: XY=14, XZ= distance from X to Z.
X to T is 6, T to Z is ? Let's call TZ=x, so XZ=6+x
YZ=10

By Pythagoras: XY² + XZ² = YZ²? But 14² + (6+x)² = 10²? 196 + ... = 100? Impossible.

Perhaps YZ is not the hypotenuse. If ∠X=90°, then hypotenuse is YZ, so YZ should be longest side, but XY=14 > YZ=10, contradiction.

Unless the right angle is not in triangle XYZ? But the square is at X, and lines are XY and XZ, so it should be for triangle XYZ.

Perhaps I have the points wrong.

Another possibility: the right angle is for triangle XTY or something, but the label is at X for the corner.

Let's read the diagram carefully.

In problem 18, there is a large figure with points X,Y,Z,V,U,T,S.

At point X, there is a right angle between XY and XV (the horizontal line).

Triangle XYZ has vertices X,Y,Z, with Z on XV.

So in triangle XYZ, sides are XY, YZ, XZ.

Angle at X is between XY and XZ, and since XZ is part of XV, and angle between XY and XV is 90°, so yes, ∠YXZ=90°.

But then, by Pythagoras, XY² + XZ² = YZ²

14² + XZ² = 10² → 196 + XZ² = 100 → XZ² = -96 impossible.

So mistake.

Perhaps YZ is not 10; let's see the labels.

In the diagram: "10" is written next to YZ, but perhaps it's the length of YU or something.

Look: "10" is on YZ, "14" on XY, "6" on XT, "10" on ZV, "7" on ST and TU, "6" on SZ, "14" on UV, "75°" at U, "45°" at V, "120°" at S.

Also, there is a point S connected to T and Z, with ST=7, SZ=6, angle at S=120°.

For triangle XYZ, perhaps the right angle is not included, or I need to calculate.

Another idea: perhaps the 10 on YZ is not the length, but no, it's likely length.

Let's calculate using coordinates or other way.

Notice that in triangle TSZ, we have sides ST=7, SZ=6, angle S=120°, so we can find TZ.

By law of cosines:

TZ² = ST² + SZ² - 2*ST*Sz*cos(angle S)

Angle S is 120°, cos120°= -0.5

So TZ² = 7² + 6² - 2*7*6*(-0.5) = 49 + 36 - 84*(-0.5) = 85 + 42 = 127

So TZ = sqrt(127) ≈11.27

Then XZ = XT + TZ = 6 + sqrt(127)

In triangle XYZ, XY=14, XZ=6+sqrt(127)≈17.27, YZ=10

But then check if right-angled at X: XY² + XZ² = 14² + (6+sqrt(127))² = 196 + [36 + 12sqrt(127) + 127] = 196 + 163 + 12sqrt(127) = 359 + 12sqrt(127) > 100, while YZ²=100, so not equal.

But the diagram shows a right angle at X, so it must be that for triangle XYZ, angle at X is 90 degrees, so perhaps YZ is not 10; maybe the "10" is for something else.

Let's look back at the user's image description or standard interpretation.

Perhaps "10" on YZ is a typo or I misread.

Another possibility: the "10" is the length of YU or UV, but UV is labeled 14.

Let's list all given lengths:

- XY = 14
- YZ = 10? But that causes problem.
- XT = 6
- ZV = 10
- ST = 7
- SZ = 6
- TU = 7
- UV = 14
- Angles: at U 75°, at V 45°, at S 120°

Also, there is a point S, and lines to T and Z.

For triangle XYZ, perhaps the right angle is at X, and we need to find if it's right, but the symbol is there, so it is given as right angle.

Perhaps YZ is not 10; let's see the position.

In the diagram, "10" is written along YZ, but perhaps it's the length of the segment from Y to the intersection or something.

To resolve, let's assume that the right angle at X is for triangle XYZ, so ∠X=90°, and then YZ must be the hypotenuse, so YZ > XY and YZ > XZ.

But XY=14, so YZ >14, but it's labeled 10, which is less, so contradiction.

Unless the "10" is not for YZ. Let's read the diagram description again.

In the user's input, for problem 18: "10" is on YZ, "14" on XY, etc.

Perhaps it's a different configuration.

Another idea: perhaps the right angle at X is for the large figure, but for triangle XYZ, it's not necessarily the angle at X for the triangle, but that doesn't make sense.

Let's calculate the angles for triangle TSZ first, as it's clearer.

Triangle TSZ:
Sides: ST = 7, SZ = 6, angle at S = 120°
As above, TZ² = 7² + 6² - 2*7*6*cos(120°) = 49 + 36 - 84*(-0.5) = 85 + 42 = 127, so TZ = sqrt(127)

Angles: we have angle S=120°, so obtuse.

Sides: 7,6,sqrt(127)≈11.27, all different → scalene

So ΔTSZ: obtuse, scalene

Triangle TUV:
Sides: TU = 7, UV = 14, angle at U = 75°, angle at V = 45°, so angle at T = 60° as calculated.

Sides: TU=7, UV=14, TV = TZ + ZV = sqrt(127) + 10 ≈11.27+10=21.27

All sides different: 7,14,21.27 → scalene

Angles all <90° → acute

So ΔTUV: acute, scalene

Triangle XYZ:
Vertices X,Y,Z
Sides: XY = 14, XZ = XT + TZ = 6 + sqrt(127) ≈6+11.27=17.27, YZ = 10

But then check the angles.

Use law of cosines to find angles.

First, find angle at X.

Cos angle X = (XY² + XZ² - YZ²) / (2 * XY * XZ) = (14² + (6+sqrt(127))² - 10²) / (2*14*(6+sqrt(127)))

Calculate numerically:

sqrt(127) ≈ 11.2694

XZ = 6 + 11.2694 = 17.2694

XY = 14

YZ = 10

XY² = 196

XZ² = (17.2694)^2 ≈ 298.22

YZ² = 100

So numerator = 196 + 298.22 - 100 = 394.22

Denominator = 2*14*17.2694 = 28*17.2694 ≈ 483.5432

Cos angle X = 394.22 / 483.5432 ≈ 0.8153

Angle X ≈ cos^{-1}(0.8153) ≈ 35.4°

Not 90°.

But the diagram shows a right angle at X. This is confusing.

Perhaps the right angle is for triangle XTY or something else.

Let's look at the diagram description: "at X, there is a square", and lines are XY and the horizontal, so for triangle XYZ, if Z is on the horizontal, then angle at X between XY and XZ is the same as between XY and the horizontal, which is 90 degrees, so it should be 90 degrees.

But calculation shows otherwise, so perhaps the "10" on YZ is incorrect or for a different segment.

Another possibility: "10" is the length of YU, not YZ.

In the diagram, "10" might be on YU, but the user said "10" on YZ.

Perhaps in the original, YZ is not 10; let's assume that for triangle XYZ, with right angle at X, and XY=14, and XZ= let's say b, YZ=c, with c^2 = 14^2 + b^2.

But we have other information.

Notice that there is a point S, and triangle TSZ with ST=7, SZ=6, angle S=120°, so TZ=sqrt(127) as before.

Then XZ = XT + TZ = 6 + sqrt(127)

Then for triangle XYZ, if angle at X is 90°, then YZ^2 = XY^2 + XZ^2 = 14^2 + (6+sqrt(127))^2 = 196 + 36 + 12sqrt(127) + 127 = 359 + 12sqrt(127) ≈ 359 + 12*11.2694 = 359 + 135.2328 = 494.2328, so YZ ≈ sqrt(494.2328) ≈ 22.23

But in the diagram, it's labeled 10, which is not matching.

Perhaps the "10" is for ZV, and YZ is not given, but in the diagram, "10" is on YZ.

I think there might be a mistake in my assumption or in the diagram interpretation.

Let's try to use the given angles for other triangles.

For example, in triangle TUV, we have angles 75° at U, 45° at V, 60° at T, sides TU=7, UV=14, TV= TZ + ZV = sqrt(127) + 10

But for classification, we can do without exact values.

For triangle XYZ, perhaps the right angle is given, so we take it as right, and ignore the length conflict, or perhaps the "10" is a distractor.

Maybe "10" is the length of the entire XV or something.

Another idea: perhaps "10" on YZ is the length, but for triangle XYZ, with right angle at X, then it must be that YZ is hypotenuse, so YZ > XY=14, so "10" cannot be YZ; perhaps it's a different number.

Let's check the user's input: "10" is written next to YZ, but in some fonts, it might be 20 or something, but unlikely.

Perhaps it's 10 for ZV, and YZ is not labeled, but in the text, it's "10" on YZ.

To proceed, let's assume that for triangle XYZ, angle at X is 90 degrees as per diagram, and we'll classify based on that, and for sides, we'll see.

So for ΔXYZ:
- Angle at X = 90° → right
- Sides: XY=14, XZ=6 + TZ, YZ= ? but from right triangle, YZ = sqrt(XY^2 + XZ^2) > XY=14, and XZ >6, so all sides different unless specified, so scalene

Similarly, for others.

For ΔTUV: as above, angles 60°,75°,45° all <90° → acute, sides 7,14, TV>10+11=21, all different → scalene

For ΔTSZ: angle at S=120° >90° → obtuse, sides 7,6, TZ=sqrt(127)≈11.27, all different → scalene

So summary for 18:
ΔXYZ: right, scalene
ΔTUV: acute, scalene
ΔTSZ: obtuse, scalene

Even though there's a length discrepancy, for classification purposes, we go by the given right angle and the side markings or calculations.

So final answers.

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Now compile all.

Final Answer:

1.) equiangular
2.) obtuse
3.) acute
4.) right
5.) acute
6.) right
7.) obtuse
8.) equiangular

9.) isosceles
10.) scalene
11.) isosceles
12.) equilateral
13.) isosceles
14.) isosceles

15.)
ΔABD: acute, isosceles
ΔDBC: obtuse, scalene
ΔABC: right, scalene

16.)
ΔLMO: right, scalene
ΔOMN: right, scalene
ΔLMN: acute, isosceles

17.)
ΔEFH: right, scalene
ΔEHG: right, scalene
ΔEFG: right, isosceles

18.)
ΔXYZ: right, scalene
ΔTUV: acute, scalene
ΔTSZ: obtuse, scalene