Math 7 worksheet on solving for missing side lengths in similar figures, with six problems involving triangles and proportional relationships.

Let’s solve each problem one by one. Since the triangles are similar, their corresponding sides are proportional. That means we can set up ratios (fractions) of matching sides and solve for the missing length.

---

Problem 1: ΔABC ~ ΔXYZ

We’re told triangle ABC is similar to triangle XYZ.

Looking at the right angles:
- In ΔABC, right angle is at C → so side AC and BC are legs, AB is hypotenuse.
- In ΔXYZ, right angle is at Z → so side XZ and YZ are legs, XY is hypotenuse.

So correspondence:
- A ↔ X
- B ↔ Y
- C ↔ Z

Therefore:
- AC corresponds to XZ → x corresponds to 6
- BC corresponds to YZ → ? corresponds to 8 (but we don’t have BC labeled — wait, actually in diagram, AB = 25, XY = 10)

Actually, let’s match based on positions:

In ΔABC:
AC = x (vertical leg)
BC = ? (horizontal leg — not given)
AB = 25 (hypotenuse)

In ΔXYZ:
XZ = 6 (vertical leg)
YZ = 8 (horizontal leg)
XY = 10 (hypotenuse)

Since they’re similar, ratio of corresponding sides should be equal.

Let’s use hypotenuses first:
AB / XY = 25 / 10 = 2.5

That means every side in ΔABC is 2.5 times the corresponding side in ΔXYZ.

So AC (which is x) corresponds to XZ = 6 → x = 6 × 2.5 = 15

Check with other side: BC should correspond to YZ = 8 → BC = 8 × 2.5 = 20. But we don’t need it since question asks for x.

✔ So x = 15

---

Problem 2: ΔABC ~ ΔXYZ

Right angles:
- ΔABC: right angle at C → AC and BC are legs, AB = 50 is hypotenuse
- ΔXYZ: right angle at Z → XZ and YZ are legs, XY = x is hypotenuse? Wait, look at labels:

In ΔXYZ:
Y to Z = 20 (horizontal)
Z to X = ? (vertical)
X to Y = x (hypotenuse)

But in similarity statement: ΔABC ~ ΔXYZ → so A↔X, B↔Y, C↔Z

So:
- AB corresponds to XY → 50 corresponds to x
- BC corresponds to YZ → 40 corresponds to 20
- AC corresponds to XZ → ? corresponds to ?

Wait — BC = 40, YZ = 20 → so ratio is 40/20 = 2 → meaning ΔABC is twice as big as ΔXYZ.

Then AB = 50 corresponds to XY = x → so 50 / x = 2 → x = 50 / 2 = 25

✔ So x = 25

---

Problem 3: ΔABC ~ ΔXYZ

ΔABC: right angle at C → AC = 16, BC = 30, AB = 34
ΔXYZ: right angle at Z → YZ = y, XZ = ?, XY = x, and YZ is vertical? Wait, diagram shows:

In ΔXYZ:
Y to Z = 15 (horizontal)
Z to X = y (vertical)
X to Y = x (hypotenuse)

Similarity: ΔABC ~ ΔXYZ → A↔X, B↔Y, C↔Z

So:
- AC (16) corresponds to XZ (y) → because both are vertical legs from right angle
- BC (30) corresponds to YZ (15) → horizontal legs
- AB (34) corresponds to XY (x) → hypotenuses

Use BC and YZ: 30 / 15 = 2 → scale factor is 2 (ABC is bigger)

So:
- AC = 16 → corresponds to XZ = y → 16 / y = 2 → y = 8
- AB = 34 → corresponds to XY = x → 34 / x = 2 → x = 17

✔ So x = 17, y = 8

---

Problem 4: ΔABC ~ ΔXYZ

No right angles shown, but similarity given: ΔABC ~ ΔXYZ → A↔X, B↔Y, C↔Z

Sides:
In ΔABC:
AB = 14.4
BC = y
AC = x

In ΔXYZ:
XY = 12
YZ = 13
XZ = 5

Corresponding sides:
- AB ↔ XY → 14.4 ↔ 12
- BC ↔ YZ → y ↔ 13
- AC ↔ XZ → x ↔ 5

Find ratio using AB and XY: 14.4 / 12 = 1.2

So scale factor is 1.2 (ABC is 1.2 times XYZ)

Then:
- y = 13 × 1.2 = 15.6
- x = 5 × 1.2 = 6

✔ So x = 6, y = 15.6

---

Problem 5: ΔJKL ~ ΔGKH

This is a bit different — smaller triangle inside larger one.

Given: ΔJKL ~ ΔGKH

Points: G is on JK, H is on JL, GH connects them.

So:
- J ↔ G
- K ↔ K (common vertex?)
Wait — similarity statement: ΔJKL ~ ΔGKH

So order matters: J↔G, K↔K, L↔H

Wait that doesn't make sense unless K is shared.

Actually, looking at diagram:

Triangle JKL has points J, K, L.

Inside it, triangle GKH: G on JK, H on JL, and GH parallel to KL? Probably.

Given lengths:
JK = 6 + 4 = 10? Wait — J to G is 4, G to K is 6 → so JK = 10

JL = 15 (base)

GH = x (middle line)

Since ΔJKL ~ ΔGKH, and assuming GH || KL, then yes, similar.

Correspondence: J↔G, K↔K, L↔H? That seems off.

Better: since GH is drawn connecting G on JK and H on JL, and if GH || KL, then ΔJGH ~ ΔJKL.

But problem says ΔJKL ~ ΔGKH — which might mean GKH is the small triangle.

Looking at vertices:

If ΔJKL ~ ΔGKH, then:

J ↔ G
K ↔ K
L ↔ H

But K is common — so perhaps it's rotated.

Alternatively, maybe it's ΔJKL ~ ΔGKH meaning:

Side JK corresponds to GK
KL corresponds to KH
LJ corresponds to HG

But let’s use proportions.

From diagram:

In large triangle JKL:
JK = JG + GK = 4 + 6 = 10
JL = 15
KL = ? (not needed)

Small triangle GKH:
GK = 6
KH = ?
GH = x

Since similar, ratio of corresponding sides.

Assume correspondence: J↔G, K↔K, L↔H → then side JK corresponds to GK → 10 corresponds to 6 → ratio 10/6 = 5/3

Then side JL corresponds to GH → 15 corresponds to x → so 15 / x = 5/3 → x = 15 * 3 / 5 = 9

Check: also, KL would correspond to KH — but we don’t have those.

Alternatively, if ΔGKH ~ ΔJKL, then GK/JK = GH/JL → 6/10 = x/15 → x = (6/10)*15 = 9

Same answer.

✔ So x = 9

---

Problem 6: ΔABC ~ ΔADE

Diagram: Point A at bottom left, D above A, E to the right.

B is on AD, C is on AE.

So ΔABC is inside ΔADE, sharing angle A.

Given: ΔABC ~ ΔADE → so A↔A, B↔D, C↔E

Sides:
In ΔABC:
AB = 3
AC = 4
BC = 5

In ΔADE:
AD = y (whole side)
AE = x (whole base)
DE = 12.5

Corresponding sides:
- AB ↔ AD → 3 ↔ y
- AC ↔ AE → 4 ↔ x
- BC ↔ DE → 5 ↔ 12.5

Use BC and DE to find scale factor: 12.5 / 5 = 2.5 → so ΔADE is 2.5 times bigger than ΔABC.

Then:
- AB = 3 → AD = y = 3 × 2.5 = 7.5
- AC = 4 → AE = x = 4 × 2.5 = 10

✔ So x = 10, y = 7.5

---

Final Answers:

1. x = 15
2. x = 25
3. x = 17, y = 8
4. x = 6, y = 15.6
5. x = 9
6. x = 10, y = 7.5

──────────────────────────────────────

Final Answer:
1. x = 15
2. x = 25
3. x = 17, y = 8
4. x = 6, y = 15.6
5. x = 9
6. x = 10, y = 7.5