Geometric Twins: questions with answers

All four symbols have the same arm lengths. Symbols with the same angle between their arms are congruent; symbols with different angles cannot fit exactly over each other. The different angles are $30^\circ$, $50^\circ$ and $70^\circ$, giving 3 different shapes.

Compare the two radii. Equal radii mean the circles are congruent; different radii mean they are not. Measuring diameters and circumferences as well adds no needed information. The centres may be in different positions. Tracing can check congruence, but measurements can check it too.

They are congruent. A figure may be turned over before it is laid on the other, and a mirror image is exactly the turned-over copy. Turning a figure over changes no length and no angle, so the two fit each other exactly. This does not depend on the figure having a line of symmetry: a figure and its mirror image are congruent whatever the shape is.

A congruent rectangle must have the same length and breadth: 5 cm and 3 cm. Doubling both measurements keeps the shape but changes the size. Each remaining choice changes one measurement.

Both triangles share the base and have the same two other side lengths, so they are congruent by SSS. They are mirror images across the base.

The line through XY is a line of symmetry for the construction. Flipping across this line takes one triangle exactly onto the other, so they are congruent. This works whether the two arcs have equal or unequal radii. No angle measurements are needed.

The pairing lays $MN$ on $ON$. These are adjacent sides of the rectangle, and they are unequal, so one cannot cover the other. $OP$ is the side equal to $MN$, and the diagonal does match itself. The correct correspondence is $\triangle MNP$ $\cong$ $\triangle OPN$.

The triangle name on the left has 3 choices for its first vertex, then 2 for its second, then 1 for its last. This gives $3 \times 2 \times 1 = 6$ orders. For each order, the fixed vertex pairs give exactly one matching name on the right. Thus there are six statements, including the given one.

Both triangles have right angles at Q, common leg QR, and hypotenuses of 5 cm. RHS establishes their congruence. The perpendicular intersections are reflections across QR, so one triangle fits the other after a flip.

$AC$ and $XZ$ are opposite the right angles, so they are the hypotenuses. Only the hypotenuses are given equal. Without another corresponding side equality, the information is insufficient to establish congruence. Right angles alone do not establish congruence.

Both triangles share PQ, have QR = QS = 7 cm, and have the given angle at P. Since R and S are distinct points on the same ray, PR and PS differ. The triangles therefore have different third side lengths and are not congruent. These SSA data admit two solutions; SAS would determine a triangle up to congruence.

For AB and AC, angle A is the included angle because it lies between these sides. SAS uses this included angle and guarantees congruence. SSA uses a non-included angle, and some such measurements fit two triangles that are not congruent. The difference is the angle's position, not the number of sides given.

The third angles are both $80^\circ$ by the triangle angle sum. The equal side BC and YZ lies between the equal angles at B and C, and Y and Z. ASA therefore establishes congruence. This also explains the AAS condition; angles alone do not fix size.

Equal sides have equal opposite angles, so all three interior angles are equal. The angles of a triangle sum to $180^\circ$, so each angle is $180^\circ \div 3 = 60^\circ$.

Since A is the midpoint of both segments, the named triangles have two pairs of equal sides. The angles between these sides are vertically opposite angles at A, so they are equal. Therefore, $\triangle DAE \cong \triangle GAF$ by SAS.