Sec 2 Maths: Congruent & Similar Triangles practice questions with worked solutions

From the order of the letters, $A \leftrightarrow V$, $B \leftrightarrow W$, $C \leftrightarrow X$, $D \leftrightarrow Y$ and $E \leftrightarrow Z$.

(a) Length of $CD$

$$CD = XY = 2.2 \text{ cm} \quad(\text{corresponding sides of congruent figures})$$

(b) Angle $XYZ$

$$\angle XYZ = \angle CDE = 108^\circ \quad(\text{corresponding angles of congruent figures})$$

From the order of the letters, $A \leftrightarrow D$, $B \leftrightarrow E$ and $C \leftrightarrow C$.

(a) $\angle BAC$

$$\angle BAC = \angle EDC = 46^\circ \quad(\text{corresponding angles of congruent triangles})$$

(b) $AB$ and $ED$

With transversal $AD$, $\angle BAC = \angle CDE = 46^\circ$. These alternate angles are equal, so $AB$ is parallel to $ED$.

(a) Length $CD$

$$CD = CB = 8 \text{ cm} \quad(\text{corresponding sides of congruent triangles})$$

(b) Angle $BAD$

The diagram marks $AD = AC$, so triangle $ADC$ is isosceles with $\angle ACD = \angle ADC = 72^\circ$.

$$\begin{aligned}\angle DAC &= 180^\circ – 72^\circ – 72^\circ = 36^\circ \quad(\text{angle sum of triangle})\\ \angle BAC &= \angle DAC = 36^\circ \quad(\text{congruent triangles})\\ \angle BAD &= 36^\circ + 36^\circ = 72^\circ\end{aligned}$$

(a) Congruent quadrilateral

Match the sides by length: $AB = 7.8$ must match the unlabelled $DE$, $FC$ is shared, and $CD = 5.2$ matches $AF$. So $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$, $F \leftrightarrow C$:

$$ABCF \equiv DEFC$$

(b) Length of $AF$

$$AF = DC = 5.2 \text{ cm} \quad(\text{corresponding sides})$$

(c) Obtuse angle $BCD$

Corresponding angles: $\angle ABC = \angle DEF = 80^\circ$ and $\angle AFC = \angle DCF = 110^\circ$. Angle sum of quadrilateral $ABCF$:

$$\begin{aligned}\angle BCF &= 360^\circ – 63^\circ – 80^\circ – 110^\circ = 107^\circ\end{aligned}$$

Angles at a point $C$:

$$\begin{aligned}\angle BCD &= 360^\circ – 107^\circ – 110^\circ = 143^\circ\end{aligned}$$

From the order of the letters, $P \leftrightarrow S$, $Q \leftrightarrow P$ and $R \leftrightarrow M$.

(a) Length of $TM$

$$\begin{aligned}SM = PR &= 12.2 + 11 = 23.2 \text{ cm} \quad(\text{corresponding sides})\\ TM &= SM – ST = 23.2 – 18.4 = 4.8 \text{ cm}\end{aligned}$$

(b) Angle $PRQ$

$\angle QPR = \angle PSM = 29^\circ$ (corresponding angles). Angle sum of triangle $PQR$:

$$\angle PRQ = 180^\circ – 75^\circ – 29^\circ = 76^\circ$$

(c) Angle $QTS$

$\angle PMS = \angle PRQ = 76^\circ$ (corresponding angles). In triangle $TPM$, $\angle QTM$ is an exterior angle:

$$\begin{aligned}\angle QTM &= \angle TPM + \angle TMP = 29^\circ + 76^\circ = 105^\circ \quad(\text{ext. angle of triangle})\\ \angle QTS &= 180^\circ – 105^\circ = 75^\circ \quad(\text{angles on a straight line } STM)\end{aligned}$$

(d) Is $QR$ parallel to $SM$?

Yes. On the line $RMP$, $\angle QRP = 76^\circ$ and $\angle TMP = 76^\circ$. These corresponding angles are equal, so $QR$ is parallel to $SM$.

From the order of the letters, $A \leftrightarrow D$, $B \leftrightarrow C$ and $C \leftrightarrow E$.

(a) Angle $DCE$

$$\angle DCE = \angle ABC = 90^\circ \quad(\text{corresponding angles of similar triangles})$$

(b) Length of $DC$

$$\begin{aligned}\frac{DC}{AB} &= \frac{CE}{BC}\\ \frac{DC}{9} &= \frac{8}{12}\\ DC &= 6 \text{ cm}\end{aligned}$$

(c) Length of $DE$

Triangle $DCE$ is right-angled at $C$, so by Pythagoras’ theorem:

$$\begin{aligned}DE^2 &= 6^2 + 8^2 = 100\\ DE &= 10 \text{ cm}\end{aligned}$$

From the order of the letters, $A \leftrightarrow A$, $B \leftrightarrow D$ and $E \leftrightarrow C$, so $BE$ matches $DC$, $AE$ matches $AC$ and $AB$ matches $AD$.

(a) Scale factor

$$\frac{BE}{DC} = \frac{5}{7.5} = \frac23$$

(b) Length of $AC$

$$\begin{aligned}\frac{AE}{AC} &= \frac23\\ \frac{8}{AC} &= \frac23\\ AC &= 12 \text{ cm}\end{aligned}$$

(c) Length of $AB$

Use the perimeter of triangle $ADC$ to find $AD$ first.

$$\begin{aligned}AD &= 28.5 – 12 – 7.5 = 9 \text{ cm}\\ AB &= \tfrac23 \times 9 = 6 \text{ cm}\end{aligned}$$

(a) Value of $y$

$B$ corresponds to $Y$, so $\angle ABC = \angle XYZ$.

$$\begin{aligned}2y – 26 &= y\\ y &= 26\end{aligned}$$

(b) Length of $YC$

Let $YC = t$ cm. Then $BC = 12 + t$ and $YZ = t + 6$. Corresponding sides are in the same ratio:

$$\begin{aligned}\frac{YZ}{BC} &= \frac{XZ}{AC}\\ \frac{t + 6}{t + 12} &= \frac{7}{10}\\ 10t + 60 &= 7t + 84\\ 3t &= 24\\ t &= 8\end{aligned}$$

So $YC = 8$ cm.

From the order of the letters, $A \leftrightarrow A$, $B \leftrightarrow D$ and $C \leftrightarrow B$. So $AB$ matches $AD$, $BC$ matches $DB$ and $AC$ matches $AB$, with scale factor $\dfrac{BC}{DB} = \dfrac{15}{6} = 2.5$.

(a) $AD$

$$\begin{aligned}\frac{AD}{AB} &= \frac{DB}{BC}\\ \frac{AD}{12} &= \frac{6}{15}\\ AD &= 4.8 \text{ cm}\end{aligned}$$

(b) $CD$

$$\begin{aligned}\frac{AC}{AB} &= \frac{BC}{DB}\\ \frac{AC}{12} &= \frac{15}{6}\\ AC &= 30 \text{ cm}\\ CD &= AC – AD = 30 – 4.8 = 25.2 \text{ cm}\end{aligned}$$

(a) Forming the equation

$MNQP$ is a trapezium with parallel sides $MN = x + 3$ and $PQ = 7x – 5$, and height $NQ = 2x + 2$.

$$\begin{aligned}\tfrac12\big[(x+3) + (7x-5)\big](2x+2) &= 228\\ (8x – 2)(x + 1) &= 228\\ 8x^2 + 6x – 2 – 228 &= 0\\ 8x^2 + 6x – 230 &= 0\\ 4x^2 + 3x – 115 &= 0 \text{ (shown)}\end{aligned}$$

(b) Solving

$$\begin{aligned}(4x + 23)(x – 5) &= 0\\ x = -\tfrac{23}{4} \quad&\text{or}\quad x = 5\end{aligned}$$

$x = -\tfrac{23}{4}$ is rejected because it makes the lengths negative (for example $MN = x + 3 = -2.75$ cm).

(c) Exact length of $RN$

With $x = 5$: $MN = 8$ cm, $PQ = 30$ cm and $NQ = 12$ cm, so $RQ = RN + 12$. From the similar triangles $RMN$ and $RPQ$:

$$\begin{aligned}\frac{RN}{RQ} &= \frac{MN}{PQ}\\ \frac{RN}{RN + 12} &= \frac{8}{30}\\ 30\,RN &= 8\,RN + 96\\ 22\,RN &= 96\\ RN &= \frac{48}{11} = 4\tfrac{4}{11} \text{ cm}\end{aligned}$$