Samacheer Kalvi 10th Maths Exercise 4.1 Solutions

$AB = AD + DB = 3 + 5 = 8$

$AC = AE + EC = 2 + 3\dfrac{1}{2} = \dfrac{11}{2}$

In $\Delta ADE$ and $\Delta ABC$, $\angle A$ is common.

For similarity the sides about $\angle A$ must be proportional.

$\dfrac{AD}{AB} = \dfrac{3}{8}$ and $\dfrac{AE}{AC} = \dfrac{2}{11/2} = \dfrac{4}{11}$

$\dfrac{3}{8} \neq \dfrac{4}{11}$, so the sides are not proportional.

Hence $\Delta ADE$ is not similar to $\Delta ABC$ (so no value of $x$ arises from this figure; $x$ is marked in figure (ii)).

$\angle AQB = 110^\circ$, and $B,Q,C$ lie on a line.

So $\angle PQC = 180^\circ - 110^\circ = 70^\circ$.

In $\Delta CQP$ and $\Delta CBA$:

$\angle PQC = \angle ABC = 70^\circ$

$\angle C$ is common.

By AA similarity, $\Delta CQP \sim \Delta CBA$.

So $\dfrac{CQ}{CB} = \dfrac{QP}{BA}$

$BC = BQ + QC = 3 + 3 = 6$

$\dfrac{3}{6} = \dfrac{x}{5}$ gives $x = \dfrac{5}{2} = 2.5$

Let $AB$ be the lamp post of height $h$ m, $C$ the mirror, $DE$ the girl.

$BC = 6.6$ m, $CD = 2.5$ m, $DE = 1.25$ m.

In $\Delta ABC$ and $\Delta EDC$:

$\angle ABC = \angle EDC = 90^\circ$ (post and girl stand vertically)

$\angle ACB = \angle ECD$ (angle of incidence = angle of reflection)

By AA similarity, $\Delta ABC \sim \Delta EDC$.

$\dfrac{AB}{ED} = \dfrac{BC}{DC}$

$\dfrac{h}{1.25} = \dfrac{6.6}{2.5}$

$h = \dfrac{1.25 \times 6.6}{2.5} = 3.3$

$PR$ and $SQ$ meet at $T$.

In $\Delta PTQ$ and $\Delta STR$:

$\angle TPQ = \angle TSR = 90^\circ$ (given right angles at $P$ and $S$)

$\angle PTQ = \angle STR$ (vertically opposite angles)

By AA similarity, $\Delta PTQ \sim \Delta STR$.

Hence $\dfrac{PT}{ST} = \dfrac{TQ}{TR}$

Cross multiplying, $PT \times TR = ST \times TQ$.

In $\Delta ABC$ and $\Delta ADE$:

$\angle A$ is common

$\angle ACB = \angle AED = 90^\circ$ (given $\angle C = 90^\circ$, $DE \perp AB$)

By AA similarity, $\Delta ABC \sim \Delta ADE$.

$AC = AD + DC = 3 + 2 = 5$

In right $\Delta ABC$, $AB^2 = AC^2 + BC^2 = 5^2 + 12^2 = 169$, so $AB = 13$.

From similarity, $\dfrac{AB}{AD} = \dfrac{AC}{AE} = \dfrac{BC}{DE}$

$\dfrac{13}{3} = \dfrac{5}{AE}$ gives $AE = \dfrac{15}{13}$

$\dfrac{13}{3} = \dfrac{12}{DE}$ gives $DE = \dfrac{36}{13}$

$OPRQ$ is a square, so $\angle POQ = 90^\circ$ and $OQ \perp MN$.

In $\Delta LOP$ and $\Delta QMO$:

$\angle OLP = 90^\circ$ (given $\angle MLN = 90^\circ$) and $\angle OQM = 90^\circ$, so $\angle OLP = \angle OQM$.

Since $L, O, M$ are collinear, $\angle LOP + \angle POQ + \angle QOM = 180^\circ$

As $\angle POQ = 90^\circ$, we get $\angle LOP + \angle QOM = 90^\circ$ … (1)

In $\Delta QMO$, $\angle OQM = 90^\circ$, so $\angle QOM + \angle QMO = 90^\circ$ … (2)

From (1) and (2), $\angle LOP = \angle QMO$.

By AA similarity, $\Delta LOP \sim \Delta QMO$.

$OPRQ$ is a square, so $\angle OPR = 90^\circ$ and $PR \perp MN$.

In $\Delta LOP$ and $\Delta RPN$:

$\angle OLP = 90^\circ = \angle PRN$

Since $L, P, N$ are collinear, $\angle LPO + \angle OPR + \angle RPN = 180^\circ$

As $\angle OPR = 90^\circ$, $\angle LPO + \angle RPN = 90^\circ$ … (1)

In $\Delta RPN$, $\angle PRN = 90^\circ$, so $\angle RPN + \angle PNR = 90^\circ$ … (2)

From (1) and (2), $\angle LPO = \angle PNR$.

By AA similarity, $\Delta LOP \sim \Delta RPN$.

Let $AP = 6$ m, $CR = 3$ m and $QB = y$, with $AB = x_1$, $BC = x_2$, $AC = x_1 + x_2$.

In $\Delta ABQ$ and $\Delta ACR$: $\angle A$ common, $\angle ABQ = \angle ACR = 90^\circ$, so they are similar.

$\dfrac{QB}{RC} = \dfrac{AB}{AC} \Rightarrow \dfrac{y}{3} = \dfrac{x_1}{x_1+x_2}$

In $\Delta CBQ$ and $\Delta CAP$: $\angle C$ common, $\angle CBQ = \angle CAP = 90^\circ$, so they are similar.

$\dfrac{QB}{PA} = \dfrac{CB}{CA} \Rightarrow \dfrac{y}{6} = \dfrac{x_2}{x_1+x_2}$

Adding: $\dfrac{y}{3} + \dfrac{y}{6} = \dfrac{x_1+x_2}{x_1+x_2} = 1$

$y\left(\dfrac{1}{3} + \dfrac{1}{6}\right) = 1 \Rightarrow y \times \dfrac{1}{2} = 1$

$y = 2$

Scale factor $\dfrac{2}{3} < 1$, so the new triangle is smaller.

Step 1: Construct $\Delta PQR$ with any measurement.

Step 2: Draw a ray $QX$ making an acute angle with $QR$ on the side opposite to the vertex $P$.

Step 3: Locate 3 points (greater of 2 and 3) $Q_1, Q_2, Q_3$ on $QX$ so that $QQ_1 = Q_1Q_2 = Q_2Q_3$.

Step 4: Join $Q_3R$. Draw a line through $Q_2$ (2 being smaller of 2 and 3) parallel to $Q_3R$ meeting $QR$ at $R'$.

Step 5: Draw a line through $R'$ parallel to $RP$ meeting $QP$ at $P'$.

$\Delta P'QR'$ is the required triangle.

Scale factor $\dfrac{4}{5} < 1$, so the new triangle is smaller.

Step 1: Construct $\Delta LMN$ with any measurement.

Step 2: Draw a ray $MX$ making an acute angle with $MN$ on the side opposite to the vertex $L$.

Step 3: Locate 5 points $M_1, M_2, M_3, M_4, M_5$ on $MX$ so that $MM_1 = M_1M_2 = M_2M_3 = M_3M_4 = M_4M_5$.

Step 4: Join $M_5N$. Draw a line through $M_4$ parallel to $M_5N$ meeting $MN$ at $N'$.

Step 5: Draw a line through $N'$ parallel to $NL$ meeting $ML$ at $L'$.

$\Delta L'MN'$ is the required triangle.

Scale factor $\dfrac{6}{5} > 1$, so the new triangle is larger.

Step 1: Construct $\Delta ABC$ with any measurement.

Step 2: Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.

Step 3: Locate 6 points (greater of 6 and 5) $B_1, B_2, \ldots, B_6$ on $BX$ with $BB_1 = B_1B_2 = \cdots = B_5B_6$.

Step 4: Join $B_5$ (5 being smaller of 6 and 5) to $C$. Draw a line through $B_6$ parallel to $B_5C$ meeting the extended $BC$ at $C'$.

Step 5: Draw a line through $C'$ parallel to $CA$ meeting the extended $BA$ at $A'$.

$\Delta A'BC'$ is the required triangle.

Scale factor $\dfrac{7}{3} > 1$, so the new triangle is larger.

Step 1: Construct $\Delta PQR$ with any measurement.

Step 2: Draw a ray $QX$ making an acute angle with $QR$ on the side opposite to the vertex $P$.

Step 3: Locate 7 points $Q_1, Q_2, \ldots, Q_7$ on $QX$ with $QQ_1 = Q_1Q_2 = \cdots = Q_6Q_7$.

Step 4: Join $Q_3R$ (3 being smaller of 7 and 3). Draw a line through $Q_7$ parallel to $Q_3R$ meeting the extended $QR$ at $R'$.

Step 5: Draw a line through $R'$ parallel to $RP$ meeting the extended $QP$ at $P'$.

$\Delta P'QR'$ is the required triangle.