Congruence & similarity

Fluency

SSS. All three pairs of sides match ( $3, 3, 3$ ).
RHS. Both have a right angle, equal hypotenuses ( $5$ ) and a matching leg ( $3$ ).
SSS. All three pairs of sides match ( $4, 5, 6$ ).
No valid test (SSA). The sides $AB$ and $BC$ meet at vertex $B$ , but the given $30^\circ$ angle is at vertex $A$ - it is not between the two named sides. SSA is not a valid congruence test (the “ambiguous case”).
RHS. Both are right-angled with equal hypotenuses ( $10$ ) and one matching leg ( $6$ ).
ASA. Two angles ( $40^\circ$ at $A$ , $60^\circ$ at $B$ ) and the included side $AB = DE = 5$ match.
No valid test (SSA). Sides $AB$ and $BC$ meet at $B$ , but the given $40^\circ$ angle is at $C$ , not between the two named sides. SSA is not a valid congruence test.

Fluency

Yes, scale factor $3$ . ( $3 : 9 = 4 : 12 = 5 : 15$ .)
Yes, by AA. The third angle in each must be $70^\circ$ ; sorry - $\angle B = 70^\circ$ for the first, $\angle R = 70^\circ$ for the second; both have angles $40, 70, 70$ . Similar by AA.
Yes, scale factor $2$ .
No. Ratios: $10/5 = 2$ , $12/6 = 2$ , $15/8 = 1.875$ - not equal.

Reasoning

Yes - by SAS (the right angle is included between the two legs). The hypotenuse is then forced to be $5$ , matching.
Not correct. Counter-example: two sides $5$ and $3$ with a $30^\circ$ non-included angle gives two possible triangles (the “ambiguous case”).
Scale factor $1$ means every corresponding side is the same length. Same sides and same angles ⇒ congruent.
In rhombus $ABCD$ with centre $O$ , $\triangle AOB$ and $\triangle COD$ have $AB = CD$ (given), $\angle OAB = \angle OCD$ (alternate angles, parallel sides), $\angle OBA = \angle ODC$ (similarly). By ASA, $\triangle AOB \cong \triangle COD$ , so $AO = OC$ and $BO = OD$ - the diagonals bisect each other.

Problem solving

$6$ m. Method: scale factor $= 3/0.5 = 6$ ; ramp height $= 1 \times 6 = 6$ m.
Linear scale factor $3$ . Area ratio $= 3^2 = 9$ .
They are similar (same angle sum, same side ratio). To be congruent, you also need actual side lengths to match.
$9$ m. Method: $\dfrac{1.8}{1.2} = \dfrac{h}{6}$ ; $h = 9$ .

Reasoning

The “equal pair of sides” could be the legs in one triangle and the base in the other. Without specifying which sides, SAS is not established.
In $ABCD$ with $AC$ a diagonal: $\triangle ABC \cong \triangle CDA$ by ASA (alternate angles, shared side $AC$ ), so $AB = CD$ and $BC = AD$ .
$9$ and $12$ . Scale factor $15/10 = 1.5$ ; $6 \times 1.5 = 9$ , $8 \times 1.5 = 12$ .
All squares are similar (all angles $90^\circ$ ; all sides equal). Not all rectangles are similar - e.g. $2 \times 3$ and $2 \times 5$ rectangles have different side ratios.

Year 8 core - answers