Geometric triangle problems with variables and angles for solving.

Looking at the triangles, we need to find which ones have the same shape and size — meaning they are congruent. Congruent triangles have exactly the same side lengths and angles.

Let’s go one by one:

Triangle 1: Right triangle with legs x and 12cm, angle 40° between them? Wait — actually, looking again: it’s a right triangle (implied by the corner), with one leg = x, hypotenuse = 12cm? No — wait, in triangle 1, the side labeled “12cm” is opposite the right angle? Actually, no — let’s look carefully.

Actually, all these triangles appear to be right triangles (based on the drawing style — one corner looks like a right angle). But only some show the 40° angle explicitly.

Wait — perhaps we should use the Law of Sines or just compare corresponding parts.

But here’s a better approach: For two triangles to be congruent, their corresponding sides and angles must match. Since many have a 40° angle and a 12cm side, we can check if the 12cm side is in the same position relative to the 40° angle.

Let me label each triangle’s known parts:

Triangle 1:
- Right triangle (assumed from shape)
- One acute angle = 40°
- Side adjacent to 40° = x
- Hypotenuse = 12cm? Or is 12cm the side opposite?

Actually, in triangle 1: the side labeled “12cm” is the hypotenuse (longest side, opposite right angle). The side labeled “x” is one leg. The 40° angle is at the bottom right — so it’s between the base (unknown) and the hypotenuse.

This is getting messy. Let’s try a different method.

Notice that in several triangles, we have:
- A 40° angle
- A side of 12cm
- A side labeled x

We can use the Law of Sines: In any triangle, a/sin(A) = b/sin(B) = c/sin(C)

But since most are right triangles, we can use trig ratios.

Assume all are right triangles (as drawn). Then:

In a right triangle with an acute angle θ:
- sin(θ) = opposite / hypotenuse
- cos(θ) = adjacent / hypotenuse
- tan(θ) = opposite / adjacent

Let’s analyze each triangle assuming it’s right-angled (the right angle is not marked but implied by the shape — especially in triangles 1,6,7,8,9).

Triangle 1:
- Angle = 40°
- Side adjacent to 40° = x
- Hypotenuse = 12cm
→ So cos(40°) = x / 12 → x = 12 * cos(40°)

Triangle 2:
- Angle = 40°
- Side opposite to 40° = x? Wait — let's see: the 40° is at the bottom left. The side labeled x is the left side — which would be opposite the 40° angle if the right angle is at the top? This is ambiguous.

Perhaps I should look for triangles where the 12cm side and the 40° angle are in the same configuration.

Let me list each triangle with its known elements:

1. Right triangle: 40° angle, side adjacent to 40° = x, hypotenuse = 12cm → x = 12*cos(40°)
2. Not obviously right-angled? Wait, all seem to have a right angle except maybe 2,3,4,5? Actually, looking again, triangles 2,3,4,5 do NOT have a right angle marked — they are general triangles with one 40° angle and two sides given.

Ah! That’s key. Triangles 1,6,7,8,9 appear to be right triangles (with right angle at the corner where the two perpendicular sides meet). Triangles 2,3,4,5 are not necessarily right triangles — they are just triangles with one 40° angle and two sides labeled.

So let’s separate them:

Group A: Right triangles (1,6,7,8,9)
Group B: General triangles (2,3,4,5)

For Group A (right triangles):

Triangle 1:
- 40° angle
- Adjacent side to 40° = x
- Hypotenuse = 12cm
→ x = 12 * cos(40°)

Triangle 6:
- 40° angle
- Adjacent side to 40° = x
- Hypotenuse = 12cm
→ Same as triangle 1 → x = 12 * cos(40°)
So triangle 1 and 6 are identical.

Triangle 7:
- 40° angle
- Opposite side to 40° = x? Let's see: the 40° is at the bottom left. The side labeled x is the top side — which is opposite the 40° angle? And the 12cm is the hypotenuse?
Actually, in triangle 7: the side labeled 12cm is the hypotenuse (diagonal). The side labeled x is the top horizontal side. The 40° angle is at the bottom left — so it’s between the bottom side and the hypotenuse. Therefore, the side opposite to 40° is the top side (x). So sin(40°) = x / 12 → x = 12 * sin(40°)

Triangle 8:
- 40° angle at bottom right
- Side labeled x is the vertical side — which is opposite the 40° angle?
The 12cm is the top horizontal side — which is adjacent to the 40° angle?
Actually, if 40° is at bottom right, and the triangle is right-angled at top right, then:
- The side opposite 40° is the vertical side (x)
- The side adjacent to 40° is the bottom side (unknown)
- The hypotenuse is the diagonal (not labeled)
But the 12cm is labeled on the top side — which is adjacent to the right angle, not directly related to 40°.

This is confusing. Let me redraw mentally:

Triangle 8:
- Right angle at top right corner.
- 40° angle at bottom right corner.
- So the third angle (top left) is 50°.
- Side labeled 12cm is the top side — which is opposite the 40° angle? No — in a right triangle, the side opposite an angle is across from it.

If right angle is at C (top right), 40° at B (bottom right), then:
- Side opposite 40° (angle at B) is AC (left side)
- Side adjacent to 40° is BC (bottom side)
- Hypotenuse is AB (diagonal)

But in the diagram, the 12cm is labeled on the top side — which is AC? If AC is the left side, but in the diagram it's drawn as the top side? I think I'm misinterpreting the orientation.

Perhaps it's better to notice that in triangle 8, the 12cm side is adjacent to the 40° angle, and x is the side opposite the 40° angle. Because the 40° is at the bottom right, and x is the vertical side on the right — which would be opposite the 40° angle if the right angle is at the top right.

Yes: if right angle is at top right, 40° at bottom right, then the side opposite 40° is the left side (which is not labeled), and the side adjacent is the bottom side. But x is labeled on the right side — which is actually the side adjacent to the 40° angle? No.

Let's define:

In triangle 8:
- Vertices: let's say A (top left), B (top right, right angle), C (bottom right, 40° angle)
- Then side AB = top side = 12cm
- Side BC = bottom side = ?
- Side AC = diagonal = ?
- Angle at C = 40°
- Side x is labeled on side BC? No, in the diagram, x is on the right side — which is