Sec 4 E Maths: Congruence and Similarity, practice questions & worked solutions

$AW=BX$ (Given)
$\angle WAZ=\angle XBW=90^\circ$ (int angle of a square)
$AD-DZ=AB-AW$
($AW=DZ$ given, $AD=AB$ sides of square)
$AZ=BW$ By SAS, triangle $AWZ$ is congruent to triangle $BXW$

Answer: $AW=BX$, $\angle WAZ=\angle XBW=90^\circ$ and $AZ=BW$, so $\triangle AWZ\equiv\triangle BXW$ (SAS)

(a)(i)

$\dfrac{V_K}{V_J}=\left(\dfrac95\right)^3=729:125$

(a)(ii)

$\dfrac{S_J}{S_K}=\left(\dfrac59\right)^2=25:81$

(b)

$$\begin{aligned}K&=\frac{729}{125}\times250\\&=1.458\text{kg}\end{aligned}$$

Answer: (a)(i) $729:125$ (a)(ii) $25:81$ (b) $1.458$kg

(a)

$$\begin{aligned}\angle CAB&=\angle BAD\ (\text{common angle})\\\angle ABC&=\angle ADB\ (\text{given})\end{aligned}$$

Since 2 pairs of corresponding angles are equal, triangle $ABC$ and $ADB$ are similar.
OR by AA Similarity Test

(b)

$$\begin{aligned}\frac{AC}{AB}&=\frac{AB}{AD}\\\frac{AC}8&=\frac85\\AC&=\frac85\times8\\AC&=12.8\text{ m or }12\frac45\text{ m o.e.}\end{aligned}$$

Answer: (b) $12.8$ m or $12\frac45$ m

$PA=PB$ (given)
$AD=BC$ (opp. sides of rectangle)
$\angle PAB=\angle PBA$ (base angles of isosceles triangle)

$$\begin{aligned}\angle PAD&=\angle PAB+90^\circ\\&=\angle PBA+90^\circ\\&=\angle PBC\end{aligned}$$

Therefore, Triangle $PAD\equiv$ Triangle $PBC$ (SAS)
$\therefore PD=PC$,
triangle is an isosceles triangle.

Answer: $\triangle PAD\equiv\triangle PBC$ (SAS), so $PD=PC$ and $\triangle PDC$ is isosceles

(a)

In $\triangle ABE$ and $\triangle CDE$,

$$\begin{aligned}\angle EAB&=\angle ECD\text{ (alt. }\angle\text{s)}\\\angle AEB&=\angle CED\text{ (vert. opp. }\angle\text{s)}\end{aligned}$$

$\therefore\triangle ABE$ is similar to $\triangle CDE$. (AA similarity test)

(b)

(i)  $\dfrac{\text{area of }\triangle ABE}{\text{area of }\triangle CDE}=\left(\dfrac54\right)^2=\dfrac{25}{16}$

(ii)  $\dfrac{\text{area of }\triangle ABE}{\text{area of }\triangle ABC}=\dfrac59$

Answer: (b)(i) $\dfrac{25}{16}$  (ii) $\dfrac59$

Method 1

Since $\angle EDC=\angle ECD=55^\circ$, $\triangle EDC$ is an isosceles triangle, $ED=EC$.

$$\begin{aligned}\angle ADE&=\angle BCE=90^\circ-55^\circ=35^\circ\\AD&=BC\quad\text{(sides of sq)}\\\triangle ADE&\equiv\triangle BCE\quad(\mathrm{SAS})\end{aligned}$$

$\therefore AE=BE$, $\triangle AEB$ is an isosceles triangle.

Method 2

Let $P$ be the foot of the perpendicular from $E$ to $DC$.

Since $\angle EDC=\angle ECD=55^\circ$,

$\Rightarrow\triangle EDC$ is an isosceles triangle,

$\Rightarrow EP\perp DC$,

$\Rightarrow DP=PC$.

Let $EP$ extended to cut $AB$ at $Q$.

Since $ABCD$ is a square and $EP$ extended to $Q$ on $AB$,

$$\left.\begin{aligned}\angle AQE&=\angle BQE=90^\circ\\AQ&=BQ\end{aligned}\right\}$$

$EQ$ is a shared side for $\triangle QAE$ and $\triangle QBE$,

$$\triangle QAE\equiv\triangle QBE\quad(\mathrm{SAS})$$

$$\therefore\angle QAE=\angle QBE$$

$\Rightarrow\triangle AEB$ is an isosceles triangle.

Method 3

Since $\angle EDC=\angle ECD=55^\circ$,

$\Rightarrow\triangle EDC$ is an isosceles triangle

$\Rightarrow E$ lies on the perpendicular bisector of $DC$.

Since $ABCD$ is a square,

$\Rightarrow E$ lies on the perpendicular bisector of $AB$.

Let $Q=$ foot of the perpendicular bisector of $AB$ from $E$.

$\Rightarrow AQ=BQ$ and $\angle AQE=\angle BQE$

$EQ$ is a shared side for $\triangle QAE$ and $\triangle QBE$,

$$\triangle QAE\equiv\triangle QBE\quad(\mathrm{SAS})$$

$$\therefore\angle QAE=\angle QBE$$

$\Rightarrow\triangle AEB$ is an isosceles triangle.

Answer: $ED=EC$, $\angle ADE=\angle BCE=35^\circ$ and $AD=BC$, so $\triangle ADE\equiv\triangle BCE$ (SAS); hence $AE=BE$ and $\triangle AEB$ is isosceles

(a)

$\angle BDC=\angle FED$ (corresponding $\angle$s, $BD\parallel AE$),
$\dfrac{BD}{FE}=\dfrac{15}{10}=\dfrac32$ and $\dfrac{CD}{DE}=\dfrac{12}8=\dfrac32$

$\therefore\triangle CDB$ and $\triangle DEF$ are similar. (Ratio of corr sides and included angle are equal / SAS similarity test)

(b)

$(12:8)^2=9:4$

(c)

$\angle CBD=\angle DFE$ (from part (a) $\triangle CDB$ and $\triangle DEF$ are similar)

$\angle BAF=\angle CBD$ (corresponding $\angle$s, $BD\parallel AE$) and so $\angle BAF=\angle DFE$

$\therefore AB\parallel FD$ and hence $ABDF$ is a parallelogram.

Answer: (b) $9:4$

(a)

$$\begin{aligned}\angle DCE&=\angle DBF\text{ (corresponding angles)}\\\angle CDE&=\angle BDE\text{ (common angle)}\end{aligned}$$

$\therefore$ Triangle $DCE$ is similar to triangle $DBF$

(b)

$$\begin{aligned}\frac{\text{Area of }\triangle DCE}{\text{Area of }\triangle DBF}&=\left(\frac12\right)^2=\frac14\\\frac{\text{Area of }\triangle DCE}{\text{Area of }\triangle DAG}&=\left(\frac13\right)^2=\frac19\end{aligned}$$

$$\begin{aligned}\frac{\text{area of trapezium }AGFB}{\text{area of trapezium }BFEC}&=\frac{9-4}{4-1}=\frac53\\\frac{\text{area of trapezium }AGFB}{15}&=\frac53\\\text{area of trapezium }AGFB&=25\text{ cm}^2\end{aligned}$$

Answer: (b) $25$ cm$^2$