Exercise 4.2 - Chapter 4 Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.2$
Question 1.
In $\triangle \mathrm{ABC}, \mathrm{D}$ and $\mathrm{E}$ are points on the sides $\mathrm{AB}$ and $\mathrm{AC}$ respectively such that $\mathrm{DE} \| \mathrm{BC}$
(i) If $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{4}$ and $\mathrm{AC}=15 \mathrm{~cm}$ find $\mathrm{AE}$.
(ii) If $\mathrm{AD}=8 \mathrm{x}-7, \mathrm{DB}=5 \mathrm{x}-3, \mathrm{AE}=4 \mathrm{x}-3$ and $\mathrm{EC}=3 \mathrm{x}-1$, find the value of $\mathrm{x}$.
Solution:
(i) If $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{4}, \mathrm{AC}=15 \mathrm{~cm}, \mathrm{DE} \| \mathrm{BC}$, then by basic proportionality theorem.
$\begin{aligned}
\frac{\mathrm{AD}}{\mathrm{AB}} &=\frac{\mathrm{AE}}{\mathrm{AC}} \\
\frac{3}{7} &=\frac{\mathrm{AE}}{15} \\
7 \mathrm{AE} &=3 \times 15 \\
\mathrm{AE} &=\frac{45}{7}=6.43 \mathrm{~cm} .
\end{aligned}$
(ii) By basic proportionality theorem.
$\begin{aligned}
\frac{\mathrm{AD}}{\mathrm{DB}} &=\frac{\mathrm{AE}}{\mathrm{EC}} \\
\frac{8 x-7}{5 x-3} &=\frac{4 x-3}{3 x-1} \\
(8 x-7)(3 x-1) &=(5 x-3)(4 x-3) \\
24 x^{2}-21 x-8 x+7 &=20 x^{2}-12 x-15 x+9 \\
24 x^{2}-29 x+7-20 x^{2}+27 x-9=0
\end{aligned}$

$\begin{gathered}
24 x^{2}-29 x+7-20 x^{2}+27 x-9=0 \\
4 x^{2}-2 x-2=0 \\
2 x^{2}-x-1=0 \\
(2 x+1)(x-1)=0 \\
\mathrm{x}=1, \frac{-1}{2} \Rightarrow \mathrm{x}=1
\end{gathered}$

Question 2.
$\mathrm{ABCD}$ is a trapezium in which $\mathrm{AB} \| \mathrm{DC}$ and $\mathrm{P}, \mathrm{Q}$ are points on $\mathrm{AD}$ and $\mathrm{BC}$ respectively, such that $\mathrm{PQ} \|$ $\mathrm{DC}$ if $\mathrm{PD}=18 \mathrm{~cm}, \mathrm{BQ}=35 \mathrm{~cm}$ and $\mathrm{QC}=15 \mathrm{~cm}$, find $\mathrm{AD}$.
Solution:
Any line parallel to the parallel sides of a trapezium dives the non-parallel sides proportionally.

$\therefore$ By thales theorem, In $\triangle \mathrm{ACD}$, we have
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{AG}}{\mathrm{GC}} \Rightarrow \frac{x}{18}=\frac{\mathrm{AG}}{\mathrm{GC}}$
In $\mathrm{DABC}$, we have
$\frac{\mathrm{AG}}{\mathrm{GC}}=\frac{\mathrm{BQ}}{\mathrm{QC}} \Rightarrow \frac{\mathrm{AG}}{\mathrm{GC}}=\frac{35}{15}$
From (1) and $(2)$, we have

$\frac{x}{18}=\frac{35}{15} \Rightarrow 3 x=126$
$x=42$
$\mathrm{AD}=x+18=(42+18)=60 \mathrm{~cm}$

Question $3 .$
In $\triangle \mathrm{ABC}, \mathrm{D}$ and $\mathrm{E}$ are points on the sides $\mathrm{AB}$ and $\mathrm{AC}$ respectively. For each of the following cases show that $\mathrm{DE} \| \mathrm{BC}$
(i) $\mathrm{AB}=12 \mathrm{~cm}, \mathrm{AD}=8 \mathrm{~cm}, \mathrm{AE}=12 \mathrm{~cm}$ and $\mathrm{AC}=18 \mathrm{~cm}$.
(ii) $\mathrm{AB}=5.6 \mathrm{~cm}, \mathrm{AD}=1.4 \mathrm{~cm}, \mathrm{AC}=7.2 \mathrm{~cm}$ and $\mathrm{AE}=1.8 \mathrm{~cm}$.
Solution:

(i) In $\triangle \mathrm{ABC}, \mathrm{AB}=12 \mathrm{~cm}$,
$\mathrm{AD}=8 \mathrm{~cm}$
$\mathrm{AE}=12 \mathrm{~cm}$,
$\mathrm{AC}=18 \mathrm{~cm}$.
If $\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}} \Rightarrow \frac{12}{8}=\frac{18}{12}$
$\Rightarrow \quad \frac{3}{2}=\frac{3}{2}$
$\therefore$ It is satisfied
$\therefore \mathrm{DE} \| \mathrm{BC}$
(ii) $\mathrm{AB}=5.6 \mathrm{~cm}$,
$\mathrm{AD}=1.4 \mathrm{~cm}$
$\mathrm{AC}=7.2 \mathrm{~cm}$
$\mathrm{AE}=1.8 \mathrm{~cm}$.
If $\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}$ is satisfied then $\mathrm{BC} \| \mathrm{DE}$
$\frac{5.6}{1.4}=\frac{7.2}{1.8}$
$5.6 \times 1.8=1.4 \times 7.2$
$10.08=10.08$
L.H.S $=$ R.H.S
$\therefore$ It is satisfied
$\therefore \mathrm{DE} \| \mathrm{BC}$
 

Question $4 .$
In fig. if $P Q \| B C$ and $P R \| C D$ prove that
(i) $\frac{\mathbf{A R}}{\mathbf{A D}}=\frac{\mathbf{A Q}}{\mathbf{A B}}=$ (ii) $\frac{\mathbf{Q B}}{\mathbf{A Q}}=\frac{\mathbf{D R}}{\mathbf{A R}}$.

Solution:

In the figure $\mathrm{PQ}\|\mathrm{BC}, \mathrm{PR}\| \mathrm{CD}$.
(i) In $\triangle \mathrm{ADC}$, by $\mathrm{BPT} \frac{\mathrm{AR}}{\mathrm{AD}}=\frac{\mathrm{AP}}{\mathrm{AC}}$
In $\Delta \mathrm{ACB}$, by $\mathrm{BPT} \frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{AQ}}{\mathrm{AB}}$
From (1) and (2) we get
$\begin{aligned}
\frac{\mathrm{AR}}{\mathrm{AD}} &=\frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{AQ}}{\mathrm{AB}} \\
\Rightarrow \quad \frac{\mathrm{AR}}{\mathrm{AD}} &=\frac{\mathrm{AQ}}{\mathrm{AB}}
\end{aligned}$
It is proved.
(ii) In $\triangle \mathrm{ABC}, \frac{\mathrm{QB}}{\mathrm{AQ}}=\frac{\mathrm{PC}}{\mathrm{AP}}$ by $\mathrm{BPT}$
In $\Delta \mathrm{ACD}, \frac{\mathrm{PC}}{\mathrm{AP}}=\frac{\mathrm{DR}}{\mathrm{AR}}$ by $\mathrm{BPT}$.
From (1) \& (2)
$\begin{aligned}
\frac{\mathrm{QB}}{\mathrm{AQ}} &=\frac{\mathrm{PC}}{\mathrm{AP}}=\frac{\mathrm{DR}}{\mathrm{AR}} \\
\therefore \frac{\mathrm{QB}}{\mathrm{AQ}} &=\frac{\mathrm{DR}}{\mathrm{AR}}
\end{aligned}$
It is proved.

Question $5 .$
Rhombus $\mathrm{PQRB}$ is inscribed in $\triangle \mathrm{ABC}$ such that $\angle \mathrm{B}$ is one of its angle. $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ lie on $\mathrm{AB}, \mathrm{AC}$ and $B C$ respectively. If $A B=12 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$, find the sides $P Q, R B$ of the rhombus.
Solution:
In $\triangle \mathrm{CRQ}$ and $\triangle \mathrm{CBA}$
$\angle \mathrm{CRQ}=\angle \mathrm{CBA}$ (as $\mathrm{RQ} \| \mathrm{AB}$ )

From (1) and (2) we get
$\frac{\mathrm{AR}}{\mathrm{AD}}=\frac{\mathrm{AP}}{\mathrm{AC}}=\frac{\mathrm{AQ}}{\mathrm{AB}}$
$\Rightarrow \quad \frac{\mathrm{AR}}{\mathrm{AD}}=\frac{\mathrm{AQ}}{\mathrm{AB}}$
It is proved.
(ii) In $\triangle \mathrm{ABC}, \frac{\mathrm{QB}}{\mathrm{AQ}}=\frac{\mathrm{PC}}{\mathrm{AP}}$ by $\mathrm{BPT}$
In $\triangle \mathrm{ACD}, \frac{\mathrm{PC}}{\mathrm{AP}}=\frac{\mathrm{DR}}{\mathrm{AR}}$ by $\mathrm{BPT}$.
From (1) \& (2)
$\begin{aligned}
\frac{\mathrm{QB}}{\mathrm{AQ}} &=\frac{\mathrm{PC}}{\mathrm{AP}}=\frac{\mathrm{DR}}{\mathrm{AR}} \\
\therefore \frac{\mathrm{QB}}{\mathrm{AQ}} &=\frac{\mathrm{DR}}{\mathrm{AR}}
\end{aligned}$
It is proved.

Question $5 .$
Rhombus PQRB is inscribed in $\triangle \mathrm{ABC}$ such that $\angle \mathrm{B}$ is one of its angle. $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ lie on $\mathrm{AB}, \mathrm{AC}$ and $\mathrm{BC}$ respectively. If $\mathrm{AB}=12 \mathrm{~cm}$ and $\mathrm{BC}=6 \mathrm{~cm}$, find the sides $\mathrm{PQ}, \mathrm{RB}$ of the rhombus.
Solution:
In $\triangle \mathrm{CRQ}$ and $\triangle \mathrm{CBA}$
$\angle \mathrm{CRQ}=\angle \mathrm{CBA}$ (as $\mathrm{RQ} \| \mathrm{AB}$ )
$\angle \mathrm{CQR}=\angle \mathrm{CAB}($ as $\mathrm{RQ} \| \mathrm{AB})$

$
\therefore \triangle \mathrm{CRQ} \cong \triangle \mathrm{CBA}
$
$
\therefore \frac{\mathrm{CR}}{\mathrm{CB}}=\frac{\mathrm{RQ}}{\mathrm{BA}}
$
Let side of Rhombus be ' $a$ '.
$\begin{aligned}
&\therefore \frac{6-a}{6}=\frac{a}{12} \\
&\Rightarrow 72-12 a=6 a \\
&\Rightarrow 18 a=72 \\
&a=4
\end{aligned}$

Side of rhombus $P Q, R B=4 \mathrm{~cm}, 4 \mathrm{~cm}$.

Question 6.
In trapezium $\mathrm{ABCD}, \mathrm{AB} \| \mathrm{DC}, \mathrm{E}$ and $\mathrm{F}$ are points on non-parallel sides $\mathrm{AD}$ and $\mathrm{BC}$ respectively, such that $\mathrm{EF} \| \mathrm{AB}$. Show that $\frac{A E}{E D}=\frac{B F}{F C}$
Solution:

Question 7.
In figure $\mathrm{DE} \| \mathrm{BC}$ and $\mathrm{CD} \| \mathrm{EF}$. Prove that $\mathrm{AD}^{2}=\mathrm{AB} \times \mathrm{AF}$.

Solution:
$\mathrm{TPT} \Rightarrow \quad \mathrm{AD}^{2}=\mathrm{AB} \times \mathrm{AF}$
$\triangle \mathrm{AFE} \cong \triangle \mathrm{ADC}$
$\frac{\mathrm{AF}}{\mathrm{AD}}=\frac{\mathrm{AE}}{\mathrm{AC}}$
$\triangle \mathrm{ADE} \cong \triangle \mathrm{ABC}$
$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$
Equating RHS of (1) and (2)
$\frac{\mathrm{AF}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{AB}}$
$\Rightarrow \quad \quad \mathrm{AD}^{2}=\mathrm{AF} \times \mathrm{AB}$
It is proved.

Question $8 .$
In a $\triangle \mathrm{ABC}, \mathrm{AD}$ is the bisector of $\angle \mathrm{A}$ meeting side $\mathrm{BC}$ at $\mathrm{D}$, if $\mathrm{AB}=10 \mathrm{~cm}, \mathrm{AC}=14 \mathrm{~cm}$ and $\mathrm{BC}=6$ $\mathrm{cm}$, find $\mathrm{BD}$ and $\mathrm{DC}$.
Solution:

$\begin{aligned}
&\text { Let }\lfloor\mathrm{BAD}=\lfloor\mathrm{CAD}=\theta\\
&\text { Assume } \mathrm{BD}=y\\
&\mathrm{BC}-\mathrm{CD}=y\\
&6-\mathrm{CD}=y\\
&C D=6-y\\
&\text { Assume } \triangle \mathrm{ADB}=\propto\\
&\overline{\mathrm{ADC}}=180-\infty\\
&\text { In } \Delta \mathrm{ABD}, \frac{\mathrm{BD}}{\sin \theta}=\frac{\mathrm{AB}}{\sin \propto} \Rightarrow \frac{y}{\sin \theta}=\frac{10}{\sin \alpha}\\
&\Rightarrow \sin \propto=\frac{10}{y} \sin \theta\\
&\text { In } \Delta \mathrm{ACD}, \frac{\mathrm{CD}}{\sin \theta}=\frac{\mathrm{AC}}{\sin (180-\propto)}\\
&\Rightarrow \quad \frac{6-y}{\sin \theta}=\frac{14}{\sin \propto}\\
&\frac{6-y}{\sin \theta}=\frac{14}{\frac{10}{y} \sin \theta} \Rightarrow 6-y=\frac{14 y}{10}\\
&\Rightarrow y\left(1+\frac{14}{10}\right)=6 \Rightarrow y\left(\frac{24}{10}\right)=6\\
&\Rightarrow y \frac{60}{24} \Rightarrow y=2.5
\end{aligned}$

$\therefore B D=2.5 \mathrm{~cm} \text { and } C D=3.5 \mathrm{~cm}$

Question $9 .$
Check whether $\mathrm{AD}$ is bisector of $\angle \mathrm{A}$ of $\triangle \mathrm{ABC}$ in each of the following
(i) $\mathrm{AB}=5 \mathrm{~cm}, \mathrm{AC}=10 \mathrm{~cm}, \mathrm{BD}=1.5 \mathrm{~cm}$ and $\mathrm{CD}=3.5 \mathrm{~cm}$.
(ii) $\mathrm{AB}=4 \mathrm{~cm}, \mathrm{AC}=6 \mathrm{~cm}, \mathrm{BD}=1.6 \mathrm{~cm}$ and $\mathrm{CD}=2.4 \mathrm{~cm}$.
Solution:
$\begin{aligned}
&\mathrm{AB}=5 \mathrm{~cm} \\
&\mathrm{AC}=10 \mathrm{~cm} \\
&\mathrm{BD}=1.5 \mathrm{~cm} \\
&\mathrm{CD}=3.5 \mathrm{~cm}
\end{aligned}$
By ABT, check whether $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$
$\begin{aligned}
\frac{1.5 \times 10}{3.5 \times 10} &=\frac{5}{10} \\
\frac{15^{3}}{35_{7}} & \neq \frac{\not 5}{10_{2}} \\
\frac{3}{7} & \neq \frac{1}{2}
\end{aligned}$

$\therefore \mathrm{AD}$ is not the bisector of $\angle \mathrm{BAC}$.
(ii)
$\begin{aligned}
&\mathrm{AB}=4 \mathrm{~cm} \\
&\mathrm{AC}=6 \mathrm{~cm} \\
&\mathrm{BD}=1.6 \mathrm{~cm} \\
&\mathrm{CD}=2.4 \mathrm{~cm}
\end{aligned}$
By ABT, check
$\begin{aligned}
&\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{1.6 \times 10}{2.4 \times 10}=\frac{16^{2}}{24_{3}}=\frac{2}{3} \\
&\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4^{2}}{6_{3}}=\frac{2}{3}
\end{aligned}$
$\therefore A D$ is the bisector of $\triangle \mathrm{ABC}$.
 

Question $10 .$
In figure $\angle \mathrm{QPC}=90^{\circ}, \mathrm{PS}$ is its bisector. If $\mathrm{ST} \perp \mathrm{PR}$, prove that $\mathrm{ST} \times(\mathrm{PQ}+\mathrm{PR})=\mathrm{PQ} \times \mathrm{PR}$.

Solution:

In $\triangle \mathrm{PQR}$, since PS is angle bisector \& applying angle.
bisector theorem $\frac{\mathrm{PR}}{\mathrm{PQ}}=\frac{\mathrm{SR}}{\mathrm{SQ}}$
$\Delta \mathrm{RTS} \approx \Delta \mathrm{RPQ}$ (similarity)
$\Rightarrow \quad \frac{\mathrm{SR}}{\mathrm{SQ}}=\frac{\mathrm{TR}}{\mathrm{TP}}$
Given $\angle \mathrm{PTS}=90^{\circ}$
$\therefore$ In $\Delta \mathrm{PTS}$, since $\angle \mathrm{TPS}=45^{\circ}$ (PS - angle
bisector)
$\angle \mathrm{PST}$ also $=45^{\circ}$
$\therefore \angle \mathrm{PTS}$ is an isosceles $\Delta$
$\Rightarrow \mathrm{PT}=\mathrm{ST}$
Using (2) in (1), we get $\frac{\mathrm{SR}}{\mathrm{SQ}}=\frac{\mathrm{TR}}{\mathrm{ST}}$
$T R=P R-P T$
$=P R-S T$
From (A) \& (3), we get $\frac{\mathrm{PR}}{\mathrm{PQ}}=\frac{\mathrm{SR}}{\mathrm{SQ}}=\frac{\mathrm{TR}}{\mathrm{ST}}$
$\therefore \quad \mathrm{PR} \times \mathrm{ST}=\mathrm{TR} \times \mathrm{PQ}$
$=(\mathrm{PR}-\mathrm{ST}) \times \mathrm{PQ}$
$=\mathrm{PR} \times \mathrm{PQ}-\mathrm{ST} \times \mathrm{PQ}$
$\therefore \mathrm{PR} \times \mathrm{ST}+\mathrm{ST} \times \mathrm{PQ}=\mathrm{PR} \times \mathrm{PQ}$
$\Rightarrow \mathrm{ST}(\mathrm{PR}+\mathrm{PQ})=\mathrm{PR} \times \mathrm{PQ}$
Hence proved.

Question $11 .$
$\mathrm{ABCD}$ is a quadrilateral in which $\mathrm{AB}=\mathrm{AD}$, the bisector of $\angle \mathrm{BAC}$ and $\angle \mathrm{CAD}$ intersect the sides $\mathrm{BC}$ and $\mathrm{CD}$ at the points $\mathrm{E}$ and $\mathrm{F}$ respectively. Prove that $\mathrm{EF} \| \mathrm{BD}$.
Solution:
By angle bisector theorem in $\triangle \mathrm{ABC}$,
$\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$
By angle bisector theorem in $\triangle \mathrm{ADC}$,
$\begin{aligned} \frac{\mathrm{AD}}{\mathrm{AC}} &=\frac{\mathrm{DF}}{\mathrm{FC}} \\ \text { Since } \mathrm{AB} &=\mathrm{AD} \text {, equating (1) \& (2) } \\ \frac{\mathrm{BE}}{\mathrm{EC}} &=\frac{\mathrm{DF}}{\mathrm{FC}} \\ \text { In } \Delta \mathrm{BDC}, \text { as } \mathrm{EF} \text { is such that, } \\ \frac{\mathrm{DF}}{\mathrm{FC}} &=\frac{\mathrm{BE}}{\mathrm{EC}} \end{aligned}$
$\therefore \mathrm{EF} \| \mathrm{BD}$
 

Question $12 .$
Construct a $\triangle \mathrm{PQR}$ which the base $\mathrm{PQ}=4.5 \mathrm{~cm}, \angle \mathrm{R}=35^{\circ}$ and the median from $\mathrm{R}$ to $\mathrm{RG}$ is $6 \mathrm{~cm}$.
Solution:
Construction:
Step (1) Draw a line segment $P Q=4.5 \mathrm{~cm}$
Step (2) At $\mathrm{P}$, draw PE such that $\angle \mathrm{QPE}=35^{\circ}$.
Step (3) At P, draw PF such that $\angle E P F=90^{\circ}$.
Step (4) Draw $\perp^{\mathrm{r}}$ bisector to $\mathrm{PQ}$ which intersects $\mathrm{PF}$ at $\mathrm{O}$.
Step (5) With $\mathrm{O}$ centre $\mathrm{OP}$ as raidus draw a circle.
Step (6) From G mark arcs of $6 \mathrm{~cm}$ on the circle.
Mark them as $\mathrm{R}$ and $\mathrm{S}$.
Step (7) Join PR and RQ.

Step (8) PQR is the required triangle.

Question $13 .$
Construct a $\triangle \mathrm{PQR}$ in which $\mathrm{QR}=5 \mathrm{~cm}, \mathrm{P}=40^{\circ}$ and the median $\mathrm{PG}$ from $\mathrm{P}$ to $\mathrm{QR}$ is $4.4 \mathrm{~cm}$. Find the length of the altitude from $\mathrm{P}$ to $\mathrm{QR}$.
Solution:
Construction:
Step (1) Draw a line segment $Q R=5 \mathrm{~cm}$.
Step (2) At Q, draw QE such that $\angle R Q E=40^{\circ}$.
Step (3) At Q, draw QF such that $\angle \mathrm{EQF}=90^{\circ}$.
Step (4) Draw perpendicular bisector to $\mathrm{QR}$, which intersects $\mathrm{QF}$ at $\mathrm{O}$.
Step (5) With $\mathrm{O}$ as centre and $\mathrm{OQ}$ as raidus, draw a circle.
Step (6) From G mark arcs of radius $4.4 \mathrm{~cm}$ on the circle. Mark them as P and P'.
Step (7) Join PQ and PR.
Step (8) PQR is the required triangle.
Step(9) From P draw a line PN which is $\perp^{1}$ to LR. LR meets $\mathrm{PN}$ at $M$.
Step (10) The length of the altitude is $P M=2.2 \mathrm{~cm}$.

Question $14 .$
Construct a $\triangle \mathrm{PQR}$ such that $\mathrm{QR}=6.5 \mathrm{~cm}, \angle \mathrm{P}=60^{\circ}$ and the altitude from $\mathrm{P}$ to $\mathrm{QR}$ is of length $4.5 \mathrm{~cm}$.
Solution:

Construction:
Steps (1) Draw $Q R=6.5 \mathrm{~cm}$.
Steps (2) Draw $\angle R Q E=60^{\circ}$.
Steps (3) Draw $\angle \mathrm{FQE}=90^{\circ}$.
Steps (4) Draw $\perp^{\mathrm{r}}$ bisector to $\mathrm{QR}$.
Steps (5) The $\perp^{\mathrm{r}}$ bisector meets $\mathrm{QF}$ at $\mathrm{O}$.
Steps (6) Draw a circle with $\mathrm{O}$ as centre and $\mathrm{OQ}$ as raidus.
Steps (7) Mark an arc of $4.5 \mathrm{~cm}$ from $\mathrm{G}$ on the $\perp^{\mathrm{r}}$ bisector. Such that it meets $\mathrm{LM}$ at $\mathrm{N}$.
Steps (8) Draw PP' $\|$ QR through N.
Steps (9) It meets the circle at $P, P^{\prime}$.
Steps (10) Join $P Q$ and PR.
Steps (11) $\triangle \mathrm{PQR}$ is the required triangle.

Question $15 .$
Construct a $\triangle \mathrm{ABC}$ such that $\mathrm{AB}=5.5 \mathrm{~cm}, \mathrm{C}=25^{\circ}$ and the altitude from $\mathrm{C}$ to $\mathrm{AB}$ is $4 \mathrm{~cm}$. Solution:

Construction:
Step (1) Draw $\overline{\mathrm{AB}}=5.5 \mathrm{~cm}$
Step (2) Draw $\angle \mathrm{BAE}=25^{\circ}$
Step (3) Draw $\angle \mathrm{FAE}=90^{\circ}$
Step (4) Draw $\perp^{\mathrm{r}}$ bisector to $\mathrm{AB}$.
Step (5) The $\perp^{\mathrm{r}}$ bisector meets AF at $\mathrm{O}$.
Step (6) Draw a circle with $\mathrm{O}$ as centre and $\mathrm{OA}$ as radius.
Step (7) Mark an arc of length $4 \mathrm{~cm}$ from $\mathrm{G}$ on the $\perp^{\mathrm{r}}$ bisector and name as $\mathrm{N}$.
Step (8) Draw $\mathrm{CC}^{1} \| \mathrm{AB}$ through $\mathrm{N}$.
Step (9) Join $\mathrm{AC}$ \& $\mathrm{BC}$.
Step (10) $\triangle \mathrm{ABC}$ is the required triangle.

Question $16 .$
Draw a triangle $\mathrm{ABC}$ of base $\mathrm{BC}=5.6 \mathrm{~cm}, \angle \mathrm{A}=40^{\circ}$ and the bisector of $\angle \mathrm{A}$ meets $\mathrm{BC}$ at $\mathrm{D}$ such that $\mathrm{CD}$ $=4 \mathrm{~cm}$.
Solution:

Construction:
Steps (1) Draw a line segment $B C=5.6 \mathrm{~cm}$.
Steps (2) At B, draw BE such that $\angle \mathrm{CBE}=60^{\circ}$.
Steps (3) At B draw BF such that $\angle \mathrm{EBF}=90^{\circ}$.

Steps (4) Draw $\perp^{\mathrm{r}}$ bisector to $\mathrm{BC}$, which intersects $\mathrm{BF}$ at 0 .
Steps (5) With $\mathrm{O}$ as centre and $\mathrm{OB}$ as radius draw a circle.
Steps (6) From C, mark an arc of $4 \mathrm{~cm}$ on BC at D.
Steps (7) The $\perp^{\mathrm{r}}$ bisector intersects the circle at I. Join ID.
Steps (8) ID produced meets the circle at $A$.
Now join $\mathrm{AB}$ and $\mathrm{AC} \cdot \triangle \mathrm{ABC}$ is the required triangle.

Question $17 .$
Draw $\triangle \mathrm{PQR}$ such that $\mathrm{PQ}=6.8 \mathrm{~cm}$, vertical angle is $50^{\circ}$ and the bisector of the vertical angle meets the base at $\mathrm{D}$ where $\mathrm{PD}=5.2 \mathrm{~cm}$.
Solution:

Steps (1) Draw a line segment $\mathrm{PQ}=6.8 \mathrm{~cm}$
Steps (2) At P, draw PE such that $\angle \mathrm{QPE}=50^{\circ}$.
Steps (3) At P, draw PF such that $\angle \mathrm{FPE}=90^{\circ}$.
Step (4) Draw $\perp^{\mathrm{r}}$ bisector to $\mathrm{PQ}$, which intersects $\mathrm{PF}$ at 0 .
Step (5) With $\mathrm{O}$ as centre and $\mathrm{OP}$ as radius draw a circle.
Step (6) From P mark an arc of $5.2 \mathrm{~cm}$ on PQ at D.
Step (7) The $\perp^{\mathrm{r}}$ bisector intersects the circle at I. Join ID.
Step (8) ID produced meets the circle at R. Now join $\mathrm{PR}$ \& $\mathrm{QR} . \triangle \mathrm{PQR}$ is the required triangle.