Every shape is made of 6 squares, so counting the squares cannot tell them apart. Compare the shapes edge by edge instead.

Shapes A and D are both a long T-shape. Each fits inside a rectangle 3 squares by 4 squares. Every edge of one has an edge of the same number of squares on the other, and one of them can be turned through a half turn, then slid, to fit exactly on the other.

Of the other two shapes, one is a long Z-shape, which fits in the same rectangle but is a different shape, and the other is a C-shape. Neither matches any other shape, however it is turned or flipped over.

The congruent shapes are D and A.

Find the marked parts on the first triangle, then the same parts on the second. The second triangle is turned or reflected, so look at where the parts sit, not which way up the triangle is.

A right angle, the hypotenuse (the side opposite the right angle) and one other side are marked on each triangle, so this is RHS.

The reason is RHS.

Triangles A and C are congruent.

Triangle C shows 48° and 32°, so its third angle is 180° − 48° − 32° = 100°. It then has 48° and 100° at the ends of its 9 cm side, exactly as triangle A does, so they are congruent by ASA.

Triangle D: It has the same two angles, but the side between them is 11 cm, not 9 cm, so it is not congruent.

Triangle B: The 9 cm side is opposite the 48° angle here, not between the 48° and 100° angles, so the sides do not correspond.

Must be true. Each triangle has base angles equal to the parallelograms' acute angle, so between 36° and 40°, and an apex angle of 180° - 2 × base angle, so between 100° and 108°. The apex angle 180° - 2 × base angle is larger than the base angle only if the base angle is under 60°, and every base angle in the range is under 60°.

Must be false. Each triangle has base angles equal to the parallelograms' acute angle, so between 36° and 40°, and an apex angle of 180° - 2 × base angle, so between 100° and 108°. A triangle is equilateral only if all its angles are 60°, which needs a base angle of exactly 60°, and 60° is outside the range of base angles.

Must be false. Each triangle has base angles equal to the parallelograms' acute angle, so between 36° and 40°, and an apex angle of 180° - 2 × base angle, so between 100° and 108°. A triangle is right-angled only if its apex angle is 90°, which needs a base angle of exactly 45° (a base angle cannot be 90°), and 45° is outside the range of base angles.

Line the two triangles up. \(AB = AD\) is given. So is the second fact, that \(CB = CD\). And \(AC\) is a side of both triangles, so it is equal to itself.

That is three matching pieces of information, and they give SSS.

Once two triangles are known to be congruent, every pair of matching sides and matching angles is equal. Matching \(ABC\) with \(ADC\) pairs \(B\) with \(D\), so the answer is \(\angle DAC\).

Here two pairs of sides are equal and the angles between them are equal, so the condition is SAS.

\(P\) matches \(A\) and \(R\) matches \(C\), so \(PR\) matches \(AC\).

Congruent triangles are identical in every measurement, so \(PR = AC = 29\) cm.

Both triangles have a right angle at \(M\). Their hypotenuses \(AB\) and \(AC\) are equal, as the dashes show. And \(AM\) is a side of both, so it is equal to itself.

Right angle, hypotenuse, side: RHS.

It is not SAS. SAS needs the equal angle to lie between the two equal sides, and the right angle at \(M\) is not between \(AB\) and \(AM\).

The triangles are congruent, so \(BM = CM\). Together those two make up the whole of \(BC\).

\(BM = 12 \div 2 = 6\) cm

Two sides and an angle only force congruence when the angle is the one between the two sides. Check where the angle sits before deciding.

Here the angle given does not lie between the two sides given, and two different triangles can be drawn from that information.

So the answer is No.

The case to watch for is two sides and an angle that is not between them. Swing the third side round from the end of the first and it can meet the base line in two different places, giving one triangle with an acute angle and another with an obtuse one. Both fit the measurements, and they are not congruent.

In a parallelogram opposite sides are equal, so \(AB = CD\) and \(BC = DA\). The diagonal \(AC\) belongs to both triangles, so it is equal to itself. Three pairs of equal sides: SSS.

SAS is also a correct answer here, because opposite angles of a parallelogram are equal, so \(AB = CD\), \(\angle ABC = \angle CDA\) and \(BC = DA\) match a side, the angle between, and a side.

\(\angle CDA\) is the angle in triangle \(CDA\) that matches \(\angle ABC\) in triangle \(ABC\), and congruent triangles have equal matching angles.

\(\angle CDA = 78^\circ\)

Congruent means same shape and same size. Similar means same shape only — every length multiplied by the same number, with the angles unchanged.

Two angles are enough to fix the shape, because the third follows from the angle sum. So compare the angles first.

Triangle \(ABC\): \(48^\circ\), \(69^\circ\), and \(180 - 48 - 69 = 63^\circ\).

Triangle \(PQR\): \(48^\circ\), \(69^\circ\), and \(180 - 48 - 69 = 63^\circ\).

Then compare the one length each triangle gives: \(BC = 7\) cm and \(QR = 7\) cm.

Here two pairs of angles are equal and so are the sides between them, which is ASA.

So the two triangles are congruent.