Similar Triangles Worksheets - Math Monks - Free Printable

Let’s go through each problem one by one. We’re checking if the triangles are similar — meaning their angles are equal and their sides are in proportion (same ratio).

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Problem 1:

Triangle PQR: sides 72, 84, 36
Triangle ABC: sides 14, 12, 6

Let’s match smallest to smallest, middle to middle, largest to largest.

PQR sorted: 36, 72, 84
ABC sorted: 6, 12, 14

Check ratios:

36 ÷ 6 = 6
72 ÷ 12 = 6
84 ÷ 14 = 6 → All same! So yes, similar.

Now match vertices by side order:

In PQR:
- Side 36 is QR → opposite P
- Side 72 is PQ → opposite R
- Side 84 is PR → opposite Q

Wait — better to match by corresponding sides.

Actually, let’s label properly:

Triangle PQR:
PQ = 72, PR = 84, QR = 36

Triangle ABC:
AB = 6, AC = 14, BC = 12? Wait — diagram shows:

Looking at image description:

Triangle ABC: A—B is 6, A—C is 14, B—C is 12? But that doesn’t make sense with labeling.

Actually, from standard notation:

In triangle ABC, side AB = 6, BC = 12, AC = 14? Let me recheck.

From your image description:

“Triangle ABC: A 6 B, then down to C, with 14 on left (A to C), 12 on right (B to C)”

So:

AC = 14, BC = 12, AB = 6

Triangle PQR: PQ = 72, PR = 84, QR = 36

So let’s pair:

Smallest side: QR=36 and AB=6 → ratio 6
Next: PQ=72 and BC=12 → 72/12=6
Largest: PR=84 and AC=14 → 84/14=6

So correspondence:

Q ↔ B (since QR and AB are smallest)
R ↔ C (since PQ and BC are next — wait no)

Better: Match sides:

QR (36) corresponds to AB (6) → so Q↔A, R↔B? Not quite.

Actually, since all ratios are 6, we can say:

ΔPQR ~ ΔCBA? Let's see:

If we map:

P → C (because PR=84 corresponds to CA=14? 84/14=6)
Q → B (PQ=72 corresponds to CB=12? 72/12=6)
R → A (QR=36 corresponds to BA=6? 36/6=6)

Yes! So:

P ↔ C
Q ↔ B
R ↔ A

Thus: ΔPQR ~ ΔCBA

But usually we write in order of matching vertices.

Since QR corresponds to BA, and PQ to CB, etc., it’s easier to write:

ΔPQR ~ ΔCBA

But let’s confirm angle correspondence isn't needed here — since SSS similarity, just side ratios.

Answer for #1: ΔPQR ~ ΔCBA

Wait — but in the blank it says “ΔPQR ~ _____” — probably expects the name of the second triangle in correct vertex order.

Given sides:

In PQR: PQ=72, QR=36, RP=84
In ABC: AB=6, BC=12, CA=14

Ratio: 72/12=6, 36/6=6, 84/14=6 → so PQ corresponds to BC, QR to AB, RP to CA

So:

P ↔ B (since PQ and BC)
Q ↔ C (since QR and CA? No)

This is messy. Better to sort sides:

PQR sides: 36, 72, 84 → assign to vertices opposite? Or just match numerically.

Standard way: List sides in increasing order and match.

PQR: QR=36, PQ=72, PR=84
ABC: AB=6, BC=12, AC=14

So:

QR : AB = 36:6 = 6
PQ : BC = 72:12 = 6
PR : AC = 84:14 = 6

So correspondence:

Q ↔ A (since QR and AB share Q and A?) Actually, vertex between two sides.

Perhaps easier: Since all ratios equal, triangles are similar, and we can write:

ΔPQR ~ ΔBCA? Let's try:

If P→B, Q→C, R→A

Then PQ→BC (72→12), QR→CA (36→14? No, 36≠14*6? 14*6=84, not 36)

I think I made a mistake.

Let me recalculate:

Triangle ABC: from image — A to B is 6, A to C is 14, B to C is 12.

So sides: AB=6, AC=14, BC=12

Triangle PQR: PQ=72, PR=84, QR=36

Now, let's find which side corresponds to which.

Divide each side of PQR by 6:

72/6=12, 84/6=14, 36/6=6

So PQR sides divided by 6 give: 12, 14, 6 — which matches BC=12, AC=14, AB=6

So:

PQ=72 corresponds to BC=12 → so P↔B, Q↔C
PR=84 corresponds to AC=14 → P↔A, R↔C — conflict.

Better: The side lengths correspond as:

The side of length 36 in PQR corresponds to side of length 6 in ABC → QR corresponds to AB
Side 72 in PQR corresponds to side 12 in ABC → PQ corresponds to BC
Side 84 in PQR corresponds to side 14 in ABC → PR corresponds to AC

So:

QR ↔ AB → Q↔A, R↔B
PQ ↔ BC → P↔B, Q↔C — again conflict.

Ah, I see the issue — when we say "corresponds", we mean the sides are proportional, but the vertex mapping must be consistent.

Let’s use the fact that in similar triangles, the order of vertices matters for correspondence.

Let me list the sides with endpoints:

For PQR:
- Side PQ = 72
- Side QR = 36
- Side RP = 84

For ABC:
- Side AB = 6
- Side BC = 12
- Side CA = 14

Now, notice:

PQ / BC = 72 / 12 = 6
QR / AB = 36 / 6 = 6
RP / CA = 84 / 14 = 6

So:

PQ corresponds to BC → so P↔B, Q↔C
QR corresponds to AB → Q↔A, R↔B — contradiction because Q can't be both C and A.

Unless... perhaps the correspondence is different.

Maybe:

Let’s assume ΔPQR ~ ΔXYZ where X,Y,Z are A,B,C in some order.

Suppose P corresponds to C, Q to B, R to A.

Then:

PQ should correspond to CB = BC = 12 → 72 / 12 = 6
QR should correspond to BA = AB = 6 → 36 / 6 = 6
RP should correspond to AC = 14 → 84 / 14 = 6

Perfect! And no conflict.

So:

P ↔ C
Q ↔ B
R ↔ A

Therefore, ΔPQR ~ ΔCBA

Yes! So answer is ΔCBA

But typically we write the letters in order, so ΔPQR ~ ΔCBA means P->C, Q->B, R->A

And sides match as above.

So for #1: ΔPQR ~ ΔCBA

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Problem 2:

Triangle XYZ: XY=6, YZ=5, XZ=8
Triangle EFQ: EF=36, FQ=30, EQ=48

Check ratios:

Sort sides:

XYZ: 5,6,8
EFQ: 30,36,48

5/30 = 1/6
6/36 = 1/6
8/48 = 1/6 → all same ratio 1/6, so similar.

Correspondence:

Smallest: YZ=5 ↔ FQ=30 → Y↔F, Z↔Q
Middle: XY=6 ↔ EF=36 → X↔E, Y↔F
Largest: XZ=8 ↔ EQ=48 → X↔E, Z↔Q

Consistent: X↔E, Y↔F, Z↔Q

So ΔXYZ ~ ΔEFQ

Answer: ΔEFQ

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Problem 3:

Triangle KLM: KL=3, LM=3, KM=3 → equilateral
Triangle NPQ: NP=27, PQ=27, NQ=27 → also equilateral

All angles 60°, so definitely similar.

Ratio: 3/27 = 1/9

Correspondence: any order, but usually match vertices as labeled.

K↔N, L↔P, M↔Q? But sides: KL=3, NP=27; LM=3, PQ=27; KM=3, NQ=27

So yes, K↔N, L↔P, M↔Q

So ΔKLM ~ ΔNPQ

But the question asks for ΔNPQ ~ _____

So ΔNPQ ~ ΔKLM

Answer: ΔKLM

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Problem 4:

Triangles WQP and SML — marked with tick marks.

WQP:
- WQ has one tick, WP has two ticks, QP has three ticks? From image:

"Triangle WQP: W to Q has one arc, W to P has two arcs, Q to P has three lines"

Actually, from description:

"WQP: Q has one arc, P has two arcs, W has three arcs? No"

Looking back:

"Triangle WQP: at Q: one arc, at P: two arcs, at W: three arcs? But usually arcs indicate angles.

In the image description:

"Triangle WQP: Q has one curved mark, P has two curved marks, W has three curved marks" — this likely indicates angles.

Similarly, triangle SML: S has one curved mark, M has two, L has three.

So angles:

In WQP: ∠Q = one mark, ∠P = two marks, ∠W = three marks
In SML: ∠S = one mark, ∠M = two marks, ∠L = three marks

So corresponding angles equal: ∠Q = ∠S, ∠P = ∠M, ∠W = ∠L

Therefore, similar by AAA.

Correspondence: Q↔S, P↔M, W↔L

So ΔWQP ~ ΔLSM? Let's see:

W↔L, Q↔S, P↔M

So ΔWQP ~ ΔLSM

But the blank is for ΔWQP ~ _____

So ΔLSM

But let's write in order: since W↔L, Q↔S, P↔M, so ΔWQP ~ ΔLSM

Yes.

Answer: ΔLSM

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Problem 5:

Triangles EFG and RPQ — marked with ticks.

EFG:
- E to F: two ticks, F to G: three ticks, G to E: one tick? From description:

"Triangle EFG: E has one arc, F has two arcs, G has three arcs" — angles.

Similarly, RPQ: R has one arc, P has two arcs, Q has three arcs.

So angles:

∠E = one mark, ∠F = two marks, ∠G = three marks
∠R = one mark, ∠P = two marks, ∠Q = three marks

So ∠E = ∠R, ∠F = ∠P, ∠G = ∠Q

Correspondence: E↔R, F↔P, G↔Q

So ΔEFG ~ ΔRPQ

Answer: ΔRPQ

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Problem 6:

Triangle ABC: angles given: ∠A=21°, ∠B=105°, so ∠C=180-21-105=54°
Triangle PQR: ∠P=54°, ∠R=21°, so ∠Q=180-54-21=105°

So angles:

ABC: A=21°, B=105°, C=54°
PQR: P=54°, Q=105°, R=21°

So corresponding angles:

∠A = R = 21°
∠B = ∠Q = 105°
∠C = ∠P = 54°

So correspondence: A↔R, B↔Q, C↔P

Thus ΔABC ~ ΔRQP

Answer: ΔRQP

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Problem 7:

Triangle MNP: MN=8, NP=14, MP=? Not given — wait, only two sides? In image: "M to N is 8, N to P is 14, M to P not given" — but for similarity, need all sides or angles.

Looking back: "Triangle MNP: M to N is 8, N to P is 14, and M to P is not labeled? But in the other triangle XYZ: X to Y not given, Y to Z is 28, X to Z is 49.

Perhaps I missed.

From description: "Triangle MNP: M-N=8, N-P=14, and M-P is not specified? But for similarity, we need three sides or angles.

Wait, in the image, perhaps only two sides are given, but that might not be enough.

Let me read carefully: "Triangle MNP: M to N is 8, N to P is 14" — and no third side? But in XYZ: Y to Z is 28, X to Z is 49, and X to Y not given.

This is problematic. Perhaps the third side is implied or we assume it's included.

Maybe it's a typo, or perhaps we can check ratio of given sides.

Assume that the sides correspond as labeled.

Suppose MN corresponds to YZ, NP to ZX, etc.

MN=8, YZ=28 → ratio 8/28=2/7
NP=14, ZX=49 → 14/49=2/7

Same ratio. Now, if the included angle is the same, or if the third side also matches.

But we don't have third side. However, in many such problems, if two sides are proportional and the included angle is equal, but here no angles given.

Perhaps from the diagram, the triangles are oriented similarly, and we can assume the third side will match.

Since two pairs of sides are proportional with same ratio, and if the triangles are both scalene, likely similar, but strictly speaking, we need SAS or SSS.

But in this case, since only two sides are given, and ratio is same, and no other info, probably intended to be similar.

Moreover, in the answer format, it's expected to fill.

So ratio 2/7.

Correspondence: M↔Y, N↔Z, P↔X? Let's see:

MN=8 corresponds to YZ=28 → M↔Y, N↔Z
NP=14 corresponds to ZX=49 → N↔Z, P↔X

So M↔Y, N↔Z, P↔X

Thus ΔMNP ~ ΔYZX

But the blank is for MNP ~ _____

So ΔYZX

Answer: ΔYZX

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Problem 8:

Triangle TVU: TU=70, UV=42, VT=84
Triangle PQR: PQ=25, QR=30, RP=15

Sort sides:

TVU: 42,70,84
PQR: 15,25,30

Check ratios:

42/15 = 2.8
70/25 = 2.8
84/30 = 2.8 → all same! 42÷15=2.8, 70÷25=2.8, 84÷30=2.8

Yes, 42/15 = 14/5 = 2.8, 70/25=14/5, 84/30=14/5.

So similar.

Correspondence:

Smallest: UV=42 ↔ RP=15 → U↔R, V↔P
Middle: TU=70 ↔ PQ=25 → T↔P, U↔Q — conflict.

Better:

UV=42 corresponds to RP=15 → so U↔R, V↔P
TU=70 corresponds to PQ=25 → T↔P, U↔Q — U can't be both R and Q.

List:

TVU sides: TU=70, UV=42, VT=84
PQR sides: PQ=25, QR=30, RP=15

Now:

VT=84 corresponds to QR=30? 84/30=2.8
TU=70 corresponds to PQ=25? 70/25=2.8
UV=42 corresponds to RP=15? 42/15=2.8

So:

VT ↔ QR → V↔Q, T↔R
TU ↔ PQ → T↔P, U↔Q — conflict.

Set:

Let me map:

Suppose T↔P, V↔Q, U↔R

Then TV ↔ PQ? TV is not a side; sides are TU, UV, VT.

VT is from V to T, which would correspond to Q to P, i.e., PQ.

VT=84, PQ=25 → 84/25=3.36, not 2.8.

Earlier calculation: 84/30=2.8, and 30 is QR.

So VT=84 corresponds to QR=30 → so V↔Q, T↔R
TU=70 corresponds to RP=15? 70/15≈4.66, no.

TU=70, and 70/25=2.8, 25 is PQ.

So TU=70 ↔ PQ=25 → T↔P, U↔Q
UV=42 ↔ RP=15 → U↔R, V↔P — conflict.

Notice that in TVU, sides are 42,70,84
In PQR, 15,25,30

42/15=2.8, 70/25=2.8, 84/30=2.8

So the correspondence of sides is:

42 (UV) ↔ 15 (RP)
70 (TU) ↔ 25 (PQ)
84 (VT) ↔ 30 (QR)

So vertices:

UV connects U and V, RP connects R and P → so U↔R, V↔P or U↔P, V↔R

Similarly, TU connects T and U, PQ connects P and Q → T↔P, U↔Q or T↔Q, U↔P

VT connects V and T, QR connects Q and R → V↔Q, T↔R or V↔R, T↔Q

Let's solve:

From UV ↔ RP: say U↔R, V↔P
From TU ↔ PQ: T↔P, U↔Q — but U can't be both R and Q.

If U↔P, V↔R from UV↔RP
Then TU↔PQ: T↔Q, U↔P — consistent if U↔P
Then VT↔QR: V↔R, T↔Q — consistent.

Yes!

So:

U ↔ P
V ↔ R
T ↔ Q

Thus ΔTVU ~ ΔQRP

Because T↔Q, V↔R, U↔P

So ΔTVU ~ ΔQRP

Answer: ΔQRP

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Now, compiling all answers:

1. ΔPQR ~ ΔCBA
2. ΔXYZ ~ ΔEFQ
3. ΔNPQ ~ ΔKLM
4. ΔWQP ~ ΔLSM
5. ΔEFG ~ ΔRPQ
6. ΔABC ~ ΔRQP
7. ΔMNP ~ ΔYZX
8. ΔTVU ~ ΔQRP

For problem 7, I assumed similarity based on two sides proportional, but technically, without third side or included angle, it's not sufficient. However, in context of worksheet, likely intended.

Similarly, others seem solid.

Final Answers:

1. CBA
2. EFQ
3. KLM
4. LSM
5. RPQ
6. RQP
7. YZX
8. QRP

But for the blanks, it's the name of the triangle.

In problem 1, it's "ΔPQR ~ _____" so we put "ΔCBA" but probably just "CBA" as per format.

Looking at the worksheet, it says "complete the similarity statement", so likely just the letters.

In the blank, it's after "~", so probably "CBA" for first one.

Similarly.

So:

Final Answer:
1. CBA
2. EFQ
3. KLM
4. LSM
5. RPQ
6. RQP
7. YZX
8. QRP