Congruent Figures

Congruent figures have exactly the same shape and size. Match their corners, name them in matching order with the sign $\equiv$ , and use a congruence to find unknown sides and angles.

Before you start

$\triangle ABC$ is the triangle with corners (vertices) $A$ , $B$ and $C$ . $AB$ is the side joining $A$ and $B$ , and $\angle ABC$ (or $\angle B$ ) is the angle at $B$ , between $BA$ and $BC$ .
The angles of a triangle add up to $180^{\circ}$ (∠ sum of △), and the angles of a quadrilateral add up to $360^{\circ}$ (∠ sum of quad.).
In a drawing, sides with the same number of small marks (ticks) are equal, and angles with the same number of arcs are equal.

What Congruent Means

Trace triangle A on a sheet of tracing paper. The tracing fits exactly on triangle B once it is turned, and exactly on triangle C once it is flipped over. The three triangles have the same shape and the same size: they are congruent.

Congruent figures

Two figures are congruent if they have exactly the same shape and the same size.

One figure can then be moved to fit exactly on the other: slid along (a translation), turned (a rotation) or flipped over (a reflection), or a mixture of these. None of these moves changes a length or an angle.

So a figure and its mirror image are congruent, and so are a figure and the same figure turned upside down. Where a figure is placed, and which way round it faces, makes no difference.

Example 1: Congruent shapes on a grid

Which of the shapes B, C and D are congruent to shape A? Each square of the grid is 1 cm by 1 cm.

Solution

Count the squares along each edge. Shape A is an L with arms 4 cm and 3 cm long, each 1 cm wide.

B has arms 4 cm and 3 cm, each 1 cm wide: it is A given a quarter turn. B fits exactly on A, so B is congruent to A.

C has arms 4 cm and 3 cm, each 1 cm wide: it is A flipped over. A flipped shape still fits exactly on A (turn the tracing paper over), so C is congruent to A.

D has the same kind of shape, but its long arm is 5 cm, not 4 cm. No slide, turn or flip changes a length, so D can never fit exactly on A. D is not congruent to A.

B and C are congruent to A; D is not.

Corresponding Parts and the Order of Letters

When one figure is placed exactly on a congruent figure, each corner lands on a corner, each side on a side and each angle on an angle. Parts that land on each other are called corresponding parts, and corresponding parts are equal.

In the drawing, $\triangle PQR$ is $\triangle ABC$ flipped over. $A$ lands on $P$ , $B$ on $Q$ and $C$ on $R$ . The ticks show the corresponding sides and the arcs show the corresponding angles.

Writing a congruence

$\triangle ABC \equiv \triangle PQR$ is read "triangle $ABC$ is congruent to triangle $PQR$ ". The sign $\equiv$ means "is congruent to".

The letters are written in matching order: the first letters ( $A$ and $P$ ) are corresponding vertices, so are the second ( $B$ and $Q$ ), and so are the third ( $C$ and $R$ ).

So $AB = PQ$ , $BC = QR$ and $CA = RP$ , and $\angle A = \angle P$ , $\angle B = \angle Q$ and $\angle C = \angle R$ .

Read from $\triangle ABC \equiv \triangle PQR$	Corresponding parts
Letters 1 and 2	$AB = PQ$
Letters 2 and 3	$BC = QR$
Letters 1 and 3	$AC = PR$
Letter 1, 2 or 3 in the middle	$\angle BAC = \angle QPR$ , $\angle ABC = \angle PQR$ , $\angle ACB = \angle PRQ$

The order is part of the statement

Read corresponding parts from the letters, never from where the figures sit on the page: a turned or flipped figure puts its corresponding side somewhere else.

For the drawing above, $\triangle ABC \equiv \triangle QPR$ would be wrong: it says $A$ matches $Q$ , so $\angle A = \angle Q$ , which is not true. The same pair can be written in other true ways, such as $\triangle BCA \equiv \triangle QRP$ , as long as each pair of corresponding letters stands in the same place.

Two quadrilaterals are written the same way, with the letters in order round each figure: $ABCD \equiv PQRS$ means $A$ , $B$ , $C$ , $D$ match $P$ , $Q$ , $R$ , $S$ in turn.

Example 2: Writing the congruence from the marks

The two triangles are congruent. Write the congruence statement, starting with $\triangle ABC$ , and list the three pairs of equal sides.

Solution

Match the corners by their arcs. One arc: $\angle A$ and $\angle Y$ , so $A$ corresponds to $Y$ . Two arcs: $B$ corresponds to $Z$ . Three arcs: $C$ corresponds to $X$ .

Write the corresponding letters in the same places: $A, B, C$ under $Y, Z, X$ .

Check with the sides. $AB$ has one tick and so has $YZ$ ; $BC$ has two and so has $ZX$ ; $CA$ has three and so has $XY$ . Each pair agrees with the order $Y, Z, X$ .

$\triangle ABC \equiv \triangle YZX$ , with $AB = YZ$ , $BC = ZX$ and $CA = XY$ .

Finding Unknown Sides and Angles

When two figures are given as congruent, every side and angle of one is known as soon as the corresponding part of the other is known. Find the corresponding part from the order of the letters, and give the reason.

Using a congruence

If two figures are congruent, then corresponding sides are equal and corresponding angles are equal.

Write the reason as (corr. sides of congruent △s) or (corr. ∠s of congruent △s); for other figures, (corr. sides of congruent figures).

Example 3: Sides and angles from a congruence

In the figure, $\triangle ABC \equiv \triangle PQR$ . $AB = 13$ cm, $BC = 12$ cm, $\angle ABC = 42^{\circ}$ , $\angle BAC = 63^{\circ}$ and $PR = 9$ cm. Find (a) $PQ$ , (b) $QR$ , (c) $AC$ , (d) $\angle PQR$ , (e) $\angle PRQ$ .

Solution

The order $ABC$ and $PQR$ matches $A$ with $P$ , $B$ with $Q$ and $C$ with $R$ .

(a) $PQ = AB = 13$ cm (corr. sides of congruent △s)

(b) $QR = BC = 12$ cm (corr. sides of congruent △s)

(d) $\angle PQR = \angle ABC = 42^{\circ}$ (corr. ∠s of congruent △s)

(e) First find $\angle ACB$ in $\triangle ABC$ : $\angle ACB = 180^{\circ} - 63^{\circ} - 42^{\circ} = 75^{\circ}$ (∠ sum of △).

Then $\angle PRQ = \angle ACB = 75^{\circ}$ (corr. ∠s of congruent △s).

$PQ = 13$ cm, $QR = 12$ cm, $AC = 9$ cm, $\angle PQR = 42^{\circ}$ and $\angle PRQ = 75^{\circ}$ .

Example 4: Unknowns in algebra

$\triangle ABC \equiv \triangle DEF$ . $AB = (2x + 1)$ cm, $DE = 11$ cm, $\angle ACB = (3y - 15)^{\circ}$ and $\angle DFE = 60^{\circ}$ . Find $x$ and $y$ .

Solution

$AB$ corresponds to $DE$ (letters 1 and 2 of each name), so $AB = DE$ (corr. sides of congruent △s):

$2x + 1 = 11$ , so $2x = 10$ and $x = 5$ .

$\angle ACB$ corresponds to $\angle DFE$ (letter 3 in the middle of each), so $\angle ACB = \angle DFE$ (corr. ∠s of congruent △s):

$3y - 15 = 60$ , so $3y = 75$ and $y = 25$ .

Check: $2(5) + 1 = 11$ and $3(25) - 15 = 60$ .

$x = 5$ and $y = 25$ .

Example 5: Congruent quadrilaterals

In the figure, $ABCD \equiv PQRS$ . $AB = 7$ cm, $BC = 3$ cm, $CD = 4$ cm, $\angle DAB = 45^{\circ}$ , and the angles at $B$ and $C$ are right angles. Find $PQ$ , $RS$ and $\angle PSR$ , marked $x$ .

Solution

The order matches $A$ with $P$ , $B$ with $Q$ , $C$ with $R$ and $D$ with $S$ .

$PQ = AB = 7$ cm (corr. sides of congruent figures)

$RS = CD = 4$ cm (corr. sides of congruent figures)

$\angle PSR$ corresponds to $\angle ADC$ . In $ABCD$ : $\angle ADC = 360^{\circ} - 45^{\circ} - 90^{\circ} - 90^{\circ} = 135^{\circ}$ (∠ sum of quad.).

So $x = \angle PSR = \angle ADC = 135^{\circ}$ (corr. ∠s of congruent figures).

$PQ = 7$ cm, $RS = 4$ cm and $x = 135^{\circ}$ .

Congruent or Not?

To show from measurements that two figures are congruent, match every side and every angle of one with an equal side or angle of the other, all in one matching order. A single side or angle that has no equal partner shows that they are not congruent.

Equal angles alone are not enough: a small triangle and a large one can have the same three angles. In Secondary 3 you will learn which few facts are enough to be sure two triangles are congruent; here every side and angle is compared.

Example 6: Comparing every side and angle

Every side and angle of three triangles is given below. (a) Show that $\triangle ABC$ and $\triangle LMN$ are congruent, and write the congruence in matching order. (b) Is $\triangle XYZ$ congruent to $\triangle ABC$ ?

$\triangle ABC$ : $AB = 9$ cm, $BC = 13$ cm, $CA = 12$ cm; $\angle A = 75^{\circ}$ , $\angle B = 63^{\circ}$ , $\angle C = 42^{\circ}$ .

$\triangle LMN$ : $LM = 12$ cm, $MN = 9$ cm, $NL = 13$ cm; $\angle L = 42^{\circ}$ , $\angle M = 75^{\circ}$ , $\angle N = 63^{\circ}$ .

$\triangle XYZ$ : $XY = 4.5$ cm, $YZ = 6.5$ cm, $ZX = 6$ cm; $\angle X = 75^{\circ}$ , $\angle Y = 63^{\circ}$ , $\angle Z = 42^{\circ}$ .

Solution

(a) Match the angles: $\angle A = \angle M = 75^{\circ}$ , $\angle B = \angle N = 63^{\circ}$ , $\angle C = \angle L = 42^{\circ}$ . So $A$ , $B$ , $C$ correspond to $M$ , $N$ , $L$ .

Check the sides in the same order: $AB = MN = 9$ cm, $BC = NL = 13$ cm, $CA = LM = 12$ cm. Every side and every angle has an equal partner.

So the triangles are congruent, and $\triangle ABC \equiv \triangle MNL$ .

(b) The angles of $\triangle XYZ$ are the same as those of $\triangle ABC$ , but its sides are each half as long: $XY = 4.5$ cm, while $AB = 9$ cm. $\triangle XYZ$ has the same shape but not the same size.

(a) $\triangle ABC \equiv \triangle MNL$ . (b) No: the angles are equal but the sides are not.

Congruent Figures Around Us

Things made to one design are congruent: two 50-cent coins, two EZ-Link cards, the floor tiles of a void deck. A pattern of shapes that covers a surface with no gaps and no overlaps is a tessellation. Many tessellations use one shape, so all their tiles are congruent.

A line of symmetry cuts a figure into two halves, and one half flipped over fits exactly on the other. So the two halves of a symmetric figure are congruent.

Example 7: A tiled floor

The floor of a void deck is tiled with congruent tiles, each the parallelogram $PQRS$ with $PQ = 30$ cm, $QR = 20$ cm and $\angle PQR = 70^{\circ}$ . (a) Find the other three angles of each tile. (b) Four tiles meet at a corner. Show that their angles there fit together with no gap. (c) Find the perimeter of one tile.

Solution

(a) Opposite sides of a parallelogram are parallel, so $PQ \parallel SR$ and $\angle PQR + \angle QRS = 180^{\circ}$ (int. ∠s, $PQ \parallel SR$ ). So $\angle QRS = 180^{\circ} - 70^{\circ} = 110^{\circ}\text{.}$

Opposite angles of a parallelogram are equal: $\angle RSP = \angle PQR = 70^{\circ}$ and $\angle SPQ = \angle QRS = 110^{\circ}\text{.}$

Every tile is congruent to $PQRS$ , so every tile has angles $70^{\circ}$ , $110^{\circ}$ , $70^{\circ}$ and $110^{\circ}$ (corr. ∠s of congruent figures).

(b) Where four tiles meet, two of the angles are $70^{\circ}$ and two are $110^{\circ}$ : $70^{\circ} + 110^{\circ} + 70^{\circ} + 110^{\circ} = 360^{\circ}\text{.}$ The angles at a point add up to $360^{\circ}$ (∠s at a pt.), so they fill the space round the corner exactly.

(a) $110^{\circ}$ , $70^{\circ}$ and $110^{\circ}$ ; (b) $70^{\circ} + 110^{\circ} + 70^{\circ} + 110^{\circ} = 360^{\circ}\text{;}$ (c) $100$ cm.

Example 8: A symmetric logo

The logo of a school CCA club is the kite $ABCD$ , symmetric about the line $AC$ . $AB = 4$ cm, $BC = 7$ cm and $\angle ABC = 100^{\circ}$ . Find $AD$ , $CD$ , $\angle ADC$ and the perimeter of the logo.

Solution

The line of symmetry $AC$ cuts the kite into $\triangle ABC$ and $\triangle ADC$ . Flipping $\triangle ABC$ over $AC$ puts $B$ on $D$ , while $A$ and $C$ stay where they are. So $\triangle ABC \equiv \triangle ADC$ .

$AD = AB = 4$ cm and $CD = CB = 7$ cm (corr. sides of congruent △s)

$\angle ADC = \angle ABC = 100^{\circ}$ (corr. ∠s of congruent △s)

Perimeter $= AB + BC + CD + DA = 4 + 7 + 7 + 4 = 22$ cm.

$AD = 4$ cm, $CD = 7$ cm, $\angle ADC = 100^{\circ}$ and the perimeter is $22$ cm.

Common Mistakes

Writing the letters in the wrong order. In $\triangle ABC \equiv \triangle PQR$ the order is the meaning: if $A$ matches $Q$ , the statement must put $Q$ first.
Matching sides by where they sit on the page. A turned or flipped figure has its corresponding side in a different place: follow the letters, ticks or arcs.
Thinking a flipped figure is not congruent. A mirror image has the same shape and size, so it is congruent.
Calling two figures congruent because their angles are equal, or because they have the same area. Every side and every angle must match.

Practice Questions

Try each question before you open its solution. Give a reason for each length or angle found from a congruence.

Question 1

State whether each pair of figures is congruent. (a) Two squares, each of side 5 cm. (b) A circle of radius 3 cm and a circle of diameter 6 cm. (c) A rectangle 4 cm by 9 cm and a square of side 6 cm. (d) A rectangle 8 cm by 5 cm and a rectangle 5 cm by 8 cm.

ShowHide solution

(a) Every angle of a square is $90^{\circ}$ and both have sides 5 cm: same shape and size. Congruent.

(b) A diameter of 6 cm is a radius of 3 cm. Two circles of the same radius are congruent.

(d) The second is the first turned a quarter turn: sides 8 cm and 5 cm, all angles $90^{\circ}$ . Congruent.

(a) Yes (b) Yes (c) No (d) Yes

Question 2

(a) $\triangle LMN \equiv \triangle XYZ$ . List the three pairs of equal sides and the three pairs of equal angles. (b) $PQRS \equiv WXYZ$ . Which side is equal to $XY$ , and which angle is equal to $\angle S$ ?

ShowHide solution

(a) $L$ , $M$ , $N$ correspond to $X$ , $Y$ , $Z$ . Sides: $LM = XY$ , $MN = YZ$ , $LN = XZ$ . Angles: $\angle L = \angle X$ , $\angle M = \angle Y$ , $\angle N = \angle Z$ .

(b) $P$ , $Q$ , $R$ , $S$ correspond to $W$ , $X$ , $Y$ , $Z$ . $XY$ is letters 2 and 3, so it equals $QR$ . $S$ is letter 4, so $\angle S = \angle Z$ .

(a) $LM = XY$ , $MN = YZ$ , $LN = XZ$ ; $\angle L = \angle X$ , $\angle M = \angle Y$ , $\angle N = \angle Z$ . (b) $QR = XY$ and $\angle S = \angle Z$ .

Question 3

The two triangles are congruent. Write the congruence statement, starting with $\triangle DEF$ .

ShowHide solution

One arc: $D$ and $T$ . Two arcs: $E$ and $U$ . Three arcs: $F$ and $S$ .

Check the sides: $DE$ and $TU$ (one tick), $EF$ and $US$ (two ticks), $FD$ and $ST$ (three ticks). Each agrees.

$\triangle DEF \equiv \triangle TUS$

Question 4

In the figure, $\triangle PQR \equiv \triangle XYZ$ . $PQ = 11.3$ cm, $QR = 8$ cm, $\angle QPR = 40^{\circ}$ , $\angle XYZ = 75^{\circ}$ and $XZ = 12$ cm. Find $XY$ , $YZ$ , $PR$ and $\angle XZY$ .

ShowHide solution

$P$ , $Q$ , $R$ correspond to $X$ , $Y$ , $Z$ .

$XY = PQ = 11.3$ cm and $YZ = QR = 8$ cm (corr. sides of congruent △s).

$PR = XZ = 12$ cm (corr. sides of congruent △s).

$\angle PQR = \angle XYZ = 75^{\circ}$ (corr. ∠s of congruent △s), so in $\triangle PQR$ : $\angle PRQ = 180^{\circ} - 40^{\circ} - 75^{\circ} = 65^{\circ}$ (∠ sum of △).

$\angle XZY = \angle PRQ = 65^{\circ}$ (corr. ∠s of congruent △s).

$XY = 11.3$ cm, $YZ = 8$ cm, $PR = 12$ cm and $\angle XZY = 65^{\circ}$ .

Question 5

In the figure, $ABCD \equiv EFGH$ . $AB = 6$ cm, $BC = 4$ cm, $\angle ABC = 105^{\circ}$ , $\angle BCD = 85^{\circ}$ , $GH = 5.4$ cm and $\angle FEH = 70^{\circ}$ . Find $EF$ , $FG$ , $CD$ and $\angle EHG$ , marked $x$ .

ShowHide solution

$A$ , $B$ , $C$ , $D$ correspond to $E$ , $F$ , $G$ , $H$ .

$EF = AB = 6$ cm and $FG = BC = 4$ cm (corr. sides of congruent figures).

$CD = GH = 5.4$ cm (corr. sides of congruent figures).

$\angle DAB = \angle FEH = 70^{\circ}$ (corr. ∠s of congruent figures). In $ABCD$ : $\angle ADC = 360^{\circ} - 70^{\circ} - 105^{\circ} - 85^{\circ} = 100^{\circ}$ (∠ sum of quad.).

$x = \angle EHG = \angle ADC = 100^{\circ}$ (corr. ∠s of congruent figures).

$EF = 6$ cm, $FG = 4$ cm, $CD = 5.4$ cm and $x = 100^{\circ}$ .

Question 6

$\triangle PQR \equiv \triangle KLM$ . $PQ = (3x - 2)$ cm, $KL = 13$ cm, $QR = 9$ cm, $LM = (x + 4)$ cm, $\angle PRQ = (2y + 10)^{\circ}$ and $\angle KML = 70^{\circ}$ . Find $x$ and $y$ , and check $x$ with the second pair of sides.

ShowHide solution

$PQ = KL$ (corr. sides of congruent △s): $3x - 2 = 13$ , so $3x = 15$ and $x = 5$ .

Check: $LM = x + 4 = 9$ cm, and $QR = 9$ cm. $QR$ and $LM$ correspond (letters 2 and 3), so they should be equal, and they are.

$\angle PRQ = \angle KML$ (corr. ∠s of congruent △s): $2y + 10 = 70$ , so $2y = 60$ and $y = 30$ .

$x = 5$ and $y = 30$ .

Question 7

Every side and angle of two triangles is given. Are they congruent? If so, write the congruence starting with $\triangle DEF$ . $\triangle DEF$ : $DE = 8.7$ cm, $EF = 7.7$ cm, $FD = 9.4$ cm, $\angle D = 50^{\circ}$ , $\angle E = 70^{\circ}$ , $\angle F = 60^{\circ}$ . $\triangle RST$ : $RS = 7.7$ cm, $ST = 9.4$ cm, $TR = 8.7$ cm, $\angle R = 70^{\circ}$ , $\angle S = 60^{\circ}$ , $\angle T = 50^{\circ}$ .

ShowHide solution

Match the angles: $\angle D = \angle T = 50^{\circ}$ , $\angle E = \angle R = 70^{\circ}$ , $\angle F = \angle S = 60^{\circ}$ . So $D$ , $E$ , $F$ correspond to $T$ , $R$ , $S$ .

Sides in the same order: $DE = TR = 8.7$ cm, $EF = RS = 7.7$ cm, $FD = ST = 9.4$ cm. All agree.

Yes: $\triangle DEF \equiv \triangle TRS$ .

Question 8

Two triangles each have angles $40^{\circ}$ , $60^{\circ}$ and $80^{\circ}$ . The shortest side of the first is 5 cm, and the shortest side of the second is 8 cm. Are the triangles congruent? Explain.

ShowHide solution

If the triangles were congruent, every side of one would equal a side of the other, so their shortest sides would be equal. They are 5 cm and 8 cm, which are not equal.

Corresponding sides of congruent figures are equal, so the triangles cannot be congruent, even though all their angles are equal.

No. They have the same shape, but their shortest sides, 5 cm and 8 cm, are not equal.

Question 9

$\triangle ABC$ and $\triangle PQR$ are congruent, with $AB = QR$ , $BC = RP$ and $CA = PQ$ . Siti writes $\triangle ABC \equiv \triangle PQR$ . Explain her mistake and write the congruence correctly.

ShowHide solution

$AB = QR$ and $BC = RP$ share the letter $B$ on one side and $R$ on the other, so $B$ corresponds to $R$ .

Then $AB = QR$ gives $A$ to $Q$ , and $BC = RP$ gives $C$ to $P$ . Check with $CA = PQ$ : $C$ to $P$ and $A$ to $Q$ . It agrees.

Siti put the letters in alphabetical order instead of matching order: her statement says $AB = PQ$ , which is not given.

$\triangle ABC \equiv \triangle QRP$ .

Question 10

$ABCD$ is a rectangle with $AB = 8$ cm and $BC = 6$ cm. The diagonal $AC$ cuts it into two triangles, and $\triangle ABC \equiv \triangle CDA$ . $\angle BAC = 37^{\circ}$ . Find $CD$ , $DA$ , $\angle DCA$ , $\angle ACB$ and $\angle DAC$ .

ShowHide solution

$A$ , $B$ , $C$ correspond to $C$ , $D$ , $A$ .

$CD = AB = 8$ cm and $DA = BC = 6$ cm (corr. sides of congruent △s).

$\angle DCA = \angle BAC = 37^{\circ}$ (corr. ∠s of congruent △s).

In $\triangle ABC$ : $\angle ACB = 180^{\circ} - 90^{\circ} - 37^{\circ} = 53^{\circ}$ (∠ sum of △).

$\angle DAC = \angle BCA = 53^{\circ}$ (corr. ∠s of congruent △s).

$CD = 8$ cm, $DA = 6$ cm, $\angle DCA = 37^{\circ}$ , $\angle ACB = 53^{\circ}$ and $\angle DAC = 53^{\circ}$ .

Question 11

A banner for Racial Harmony Day is the pentagon $ABCDE$ , symmetric about the dashed line $AM$ , so $ABCM \equiv AEDM$ . $AB = 5$ cm, $BC = 4$ cm, $CM = 3$ cm and $\angle ABC = 143^{\circ}$ ; the angles at $C$ and $M$ are right angles. Find (a) $DE$ and $AE$ , (b) $\angle AED$ , (c) $\angle BAE$ , (d) the perimeter of the banner.

ShowHide solution

$A$ , $B$ , $C$ , $M$ correspond to $A$ , $E$ , $D$ , $M$ .

(a) $DE = CB = 4$ cm and $AE = AB = 5$ cm (corr. sides of congruent figures).

(b) $\angle AED = \angle ABC = 143^{\circ}$ (corr. ∠s of congruent figures).

(c) In $ABCM$ : $\angle BAM = 360^{\circ} - 143^{\circ} - 90^{\circ} - 90^{\circ} = 37^{\circ}$ (∠ sum of quad.). $\angle EAM = \angle BAM = 37^{\circ}$ (corr. ∠s of congruent figures), so $\angle BAE = 37^{\circ} + 37^{\circ} = 74^{\circ}\text{.}$

(d) $MD = MC = 3$ cm (corr. sides of congruent figures), so $CD = 3 + 3 = 6$ cm. Perimeter $= AB + BC + CD + DE + EA = 5 + 4 + 6 + 4 + 5 = 24$ cm.

(a) $DE = 4$ cm, $AE = 5$ cm (b) $143^{\circ}$ (c) $74^{\circ}$ (d) $24$ cm

Question 12

Mr Tan tiles a wall with congruent triangular tiles. Each tile has angles $50^{\circ}$ , $60^{\circ}$ and a third angle. At one point six tiles meet, and each of the three angles of the tile appears there twice. Find the third angle, and show that the six angles at the point add up to $360^{\circ}$ .

ShowHide solution

Third angle $= 180^{\circ} - 50^{\circ} - 60^{\circ} = 70^{\circ}$ (∠ sum of △). Every tile has these angles, since the tiles are congruent.

At the point: $2 \times 50^{\circ} + 2 \times 60^{\circ} + 2 \times 70^{\circ} = 100^{\circ} + 120^{\circ} + 140^{\circ} = 360^{\circ}\text{.}$

This is $2 \times 180^{\circ}$ : twice the angle sum of one tile. The angles fill the space round the point with no gap (∠s at a pt.).

The third angle is $70^{\circ}$ , and $2(50^{\circ} + 60^{\circ} + 70^{\circ}) = 360^{\circ}\text{.}$

Key Points

Congruent figures have exactly the same shape and the same size.
A slide, a turn or a flip does not change a length or an angle, so a figure and its moved or mirror image are congruent.
$\triangle ABC \equiv \triangle PQR$ lists corresponding vertices in matching order: $A$ with $P$ , $B$ with $Q$ , $C$ with $R$ .
Corresponding sides of congruent figures are equal, and corresponding angles are equal: give this as the reason.
Read corresponding parts from the letters, ticks and arcs, never from where the figures sit on the page.
Equal angles alone, or equal areas, do not make two figures congruent: every side and angle must match.
The two halves of a symmetric figure are congruent, and so are the tiles of a tessellation made from one shape.