Important Questions for CBSE Class 9 Maths Chapter 11 - Constructions

Short Answer Questions (2 Marks)

1. Construct a triangle $\text{ABC}$ in which $\text{BC}=8~\text{cm},\angle \text{B}={{45}^{\prime }}$ and $\text{AB}-\text{AC}=3.5~\text{cm}$.

Ans: Given that the  Base $\text{BC}=8~\text{cm}$, One Base angle $\angle \text{B}=45{}^\circ $ and $\text{AB}-\text{AC}=3.5~\text{cm}$

To construct the  $A$ triangle $\text{ABC}$

Steps of construction are 

Draw a ray $BX$ and cut off a line segment $\text{BC}=8~\text{cm}$ from it.

Cut $\angle \text{YBC}={{45}^{{}^\circ }}$.

Cut off a line segment $\text{BD}=3.5~\text{cm}$$(\because AB-AC=3.5~\text{cm})$ from $BY$

Join$\text{CD}$.

Draw perpendicular bisector $PQ$of $CD$ intersecting $BY$ at a point$\text{A}$.

Join $AC$.

Then $\text{ABC}$ is the required triangle.

Justification: $A$ lies on the perpendicular bisector of $\text{CD}$.

$\therefore AD=AC$

Now $\text{BD}=\text{AB}-\text{AD}$

$\Rightarrow \text{BD}=\text{AB}-\text{AC}=3.5~\text{cm}$.

2. Construct a triangle $PQR$ in which $QR=6~\text{cm},\angle Q={{60}^{\prime }}$ and $\text{PR}-\text{PQ}=2~\text{cm}$.

Ans: Given that the Base $QR=6~\text{cm}$, one base angle $\angle Q={{60}^{{}^\circ }}$ and $\text{PR}-\text{PQ}=2~\text{cm}$.

To construct the  A triangle \[PQR\].

Steps of construction are

Draw a ray $QX$ and cut off a line segment $\text{QR}=6~\text{cm}$ from it.

Construct a ray $QY$ with the  angle of ${{60}^{{}^\circ }}$ at $QR$ and yield  $YQ$ to form the line $YQY'$.

Cut off a line segment $QO=2~\text{cm}(\because \text{PR}-\text{PQ}=2~\text{cm})$ from $Q{{Y}^{\prime }}$.

Join $OR$.

Draw perpendicular bisector $\text{MN}$ of $\text{OR}$.

Join $PR$.

Then $PQR$ is the required triangle.

Justification:P lies on perpendicular bisector of $OR$. $\therefore \text{PO}=\text{PR}$

$\Rightarrow PQ+QO=PR$

$\Rightarrow QO=PR-PQ=2~\text{cm}$

Short Answer Questions (3 Marks)

1. Construct an angle of ${{90}^{{}^\circ }}$ at the initial point of a given ray and justify the construction.

Ans: Steps of construction are

Draw a ray $OA$.

With $\text{O}$ as centre and convenient radius, draw an arc $LM$ cutting $\text{OA}$ at $\text{L}$.

Now with $L$ as centre and radius $OL$, draw an arc cutting the arc $LM$ at $P$.

Then taking $P$ as centre and radius , draw an arc cutting arc $PM$ at the point $Q$.

Join ${OP}$ to draw the ray $OB$. Also join $O$ and $Q$ to draw the $OC$. 

$\Rightarrow \text{  }$We observe that $\angle \text{AOB}=\angle \text{BOC}={{60}^{{}^\circ }}$

Now, With $\text{P}$ as centre and radius greater than$\frac{1}{2}\text{PQ}$ to bisect .  

draw an arc.

Now with $Q$ as centre and the same radius as in step $6$ , draw another arc cutting the arc drawn in step $6$ at $\text{R}$.

Join $O$ and $R$ and draw ray $OD$.

Then $\angle \text{AOD}$ is the required angle of ${{90}^{{}^\circ }}$

Justification: Join $PL$, then $\text{OL}=\text{OP}=\text{PL}$ (by construction)

Therefore $\Delta $ \[OQP\]is an equilateral triangle and $\angle $ \[POL\] which is same as $\angle BOA$ is equal to ${{60}^{{}^\circ }}$

Now join $\text{QP}$, then $\text{OP}=\text{OQ}=\text{PQ}$ (by construction)

Therefore $\Delta $ \[OQP\] is an equilateral triangle.

$\therefore \angle $ POQ which is same as $\angle BOC$ is equal to ${{60}^{{}^\circ }}$.

By construction $OD$ is bisector of $\angle BOC$.

$\therefore \angle \text{DOC}=\angle \text{DOB}=\frac{1}{2}\angle \text{BOC}=\frac{1}{2}\times {{60}^{{}^\circ }}={{30}^{{}^\circ }}$

Now, $\angle \text{DOA}=\angle \text{BOA}+\angle \text{DOB}\Rightarrow \angle \text{DOA}={{60}^{{}^\circ }}+{{30}^{{}^\circ }}$

$\Rightarrow \angle \text{DOA}={{90}^{{}^\circ }}$

2. Construct an angle of $45{}^\circ $ at the initial point of a given ray and justify the construction.

Ans: Steps of construction are 

Draw a ray $OA$.

 Draw an arc $LM$ with $O$ as centre and radius cutting $OA$at $L$.

Now with $OL$as  a radius and $L$ as centre , draw an arc cutting the arc $LM$ at $P$.

Then taking $\text{P}$ as centre and radius $OL$, draw an arc cutting arc $\text{PM}$ at the point $\text{Q}$.

Join $OP$ to draw the ray $OB$. 

Also, join $O$ and $Q$ to draw the \[OC.\]We observe that:$\angle \text{AOB}=\angle \text{BOC}={{60}^{{}^\circ }}$

Now, With $\text{P}$ as centre and radius greater than$\frac{1}{2}\text{PQ}$ to bisect $\angle \text{BOC}$. 

Draw an arc. 

Now, With $\text{P}$ as centre and radius greater than$\frac{1}{2}\text{PQ}$ to bisect $\angle \text{BOC}$.  

Now with $Q$ as centre and the same radius as in step $6$ , draw another arc cutting the arc drawn in step $6$ at $\text{R}$.

Join $O$ and $R$ and draw ray $OD$.Then $\angle $$AOD$ is the required angle of ${{90}^{{}^\circ }}$.

With $\text{L}$ as centre and radius greater than $\frac{1}{~\text{L}}~\text{L}$, draw an arc.

Now with $\text{S}$ as centre and the same radius as in step $2$, draw another arc cutting the arc

draw in step $2$ at $\text{T}$.

Join $O$ and $\text{T}$ and draw ray $\text{OE}$. Thus $OE$ bisects $\angle AOD$ and therefore $\angle AOE=\angle DOE={{45}^{\prime }}$

Justification:

Join \[LS\] 

Then

$\Delta $ \[OLS\] is isosceles right angle  triangle(right angled at $O$) .

$\therefore O~\text{L}=OS$

Therefore, $\text{O}$ is a perpendicular bisector of $SL$.

$\therefore SF=FL$

And $\angle $ $OFS$ $=\angle $ $OFL$ $\left[ {} \right.$ Each $\left. {{90}^{{}^\circ }} \right]$

Now in\[~\Delta ~OFS~\] and \[\Delta ~OFL\],

$OF$$=OF$ (Common)

$OS$  $= OL$ (By construction)

$SF=FL$ (Proved)

$\therefore {{\Delta }^{\text{OFS}}}\simeq {{\Delta }^{\text{OFL }}}[\text{By}$ SSS rule]

$\Rightarrow \angle \text{SOF}=\angle \text{LOF}[\text{By  CPCT}]$

Now $\angle \text{SOF}+\angle \text{LOF}=\angle \text{SOL}$

$\Rightarrow \angle \text{SOF}+\angle \text{LOF}={{90}^{{}^\circ }}$

$\Rightarrow 2\angle \text{LOF}={{90}^{{}^\circ }}$

$\Rightarrow \angle $ $LOF$ $=\frac{1}{2}\times {{90}^{{}^\circ }}={{45}^{{}^\circ }}$

And $\angle \text{AOE}={{45}^{{}^\circ }}$

3. Construct an equilateral triangle, given its side and justify the construction.

Ans: Steps of construction are

Draw a line segment $\text{BC}$ with the  length  of $6~\text{cm}$.

At B draw $\angle \text{XBC}={{60}^{{}^\circ }}$.

Draw perpendicular bisector PQ of the line segment $BC$.

Let $\text{A}$ and $\text{D}$ be the points where $PQ$ intersects the ray $BX$ and side \[BC\]respectively.

Join $AC$.

Thus $\text{ABC}$ is the required equilateral triangle.

Justification:

In right triangle $ADB$ and right triangle $ADC$,

$\text{AD}=\text{AD}$ (Common)

$\angle \text{ADB}=\angle \text{ADC}={{90}^{{}^\circ }}$ (By construction)

$\text{BD}=\text{CD}$ (By construction)

\[\therefore \Delta \text{ADB}\simeq \Delta \text{ADC}\] (By SAS congruency)

$\therefore \angle \text{B}=\angle \text{C}={{60}^{{}^\circ }}[\text{ByCPCT}]$

$\angle \text{A}={{180}^{{}^\circ }}-(\angle \text{B}+\angle \text{C})$

$={{180}^{{}^\circ }}-\left( {{60}^{{}^\circ }}+{{60}^{{}^\circ }} \right)={{180}^{{}^\circ }}-{{120}^{{}^\circ }}={{60}^{{}^\circ }}$

$\because \angle \text{A}=\angle \text{B}=\angle \text{C}={{60}^{{}^\circ }}$

. $\vartriangle ABC$ is an equilateral triangle.

4. Construct a triangle $\text{ABC}$ in which $\text{BC}=7~\text{cm},\angle \text{B}={{75}^{{}^\circ }}$ and $\text{AB}+\text{AC}=13~\text{cm}$.

Ans: 

Given  that the Base $BC=7~\text{cm},\angle B={{75}^{{}^\circ }}$ and sum of two sides $AB+AC=13~\text{cm}$.

To construct a triangle $\text{ABC}$.

Steps of construction are 

Draw a ray $BX$ and cut off a line segment $\text{BC}=7~\text{cm}$ from it.

At $B$, construct $\angle \text{YBX}={{75}^{{}^\circ }}$.

With $B$ as centre and radius = $13~\text{cm}(\because \text{AB}+\text{AC}=13~\text{cm})$ draw an arc to meet $\text{BY}$ at $\text{D}$.

Join $CD$.

Draw perpendicular bisector $PQ$ of $CD$ intersecting $BD$ at $A$.

Join $AC$.

Then $ABC$ is the required triangle.

Justification: $A$ lies on the perpendicular bisector of $\text{CD}$.

$\therefore AC=AD$

And $\text{AB}=\text{BD}-\text{AD}$

$\Rightarrow \text{AB}=\text{BD}-\text{AC}$

$\Rightarrow \text{AB}+\text{AC}=\text{BD}=13~\text{cm}$

5. Construct a triangle $\text{XYZ}$ in which $\angle \text{Y}={{30}^{{}^\circ }},\angle \text{Z}={{90}^{\prime }}$ and $\text{XY}+\text{YZ}+\text{ZX}=11~\text{cm}$.

Ans: Given: Base angles $\angle \text{Y}={{30}^{{}^\circ }}$ and $\angle \text{Z}={{90}^{{}^\circ }}$ and $\text{XY}+\text{YZ}+\text{ZX}=11~\text{cm}$.

To construct: $\Delta \text{XYZ}$

Steps of construction:

(a) Draw a line segment $\text{PQ}=11~\text{cm}$.

(b) Draw $\angle \text{KPQ}={{30}^{{}^\circ }}$ and $\angle \text{LQP}={{90}^{{}^\circ }}$

(c) Bisect the $\angle \text{KPQ}$ and $\angle $ $LQP$. Let these intersect at a point $\text{X}$.

(d) Draw perpendicular bisectors, MN of $\text{PX}$ and $\text{RS}$ of $\text{XQ}$.

(e) Let MN intersects $PQ$ at $Y$ and $RS$ intersects $PQ$ at $Z$.

(f) Join $\text{XY}$ and $\text{XZ}$.

Then\[~XYZ\]is the required triangle.

Justification:

$Y$ lies on the perpendicular bisector $MN$ of $PX$.

$\therefore \text{PY}=\text{XY}$ and similarly $Q\text{Z}=\text{XZ}$

This gives $\text{XY}+\text{YZ}+\text{XZ}=\text{PY}+\text{Y}Z+\text{QZ}=\text{PQ}=11~\text{cm}$

Again $\angle \text{YXP}=\angle \text{XPY}$ [Since $\text{XY}=\text{PY}]$

$\Rightarrow \angle \text{XYZ}=\angle \text{YXP}+\angle \text{XPY}=2\angle \text{XPY}=\angle \text{KPQ}$

$\Rightarrow \angle \text{XYZ}={{30}^{{}^\circ }}$

Similarly, $\angle \text{XZY}=\angle \text{LQP}$

$\Rightarrow \angle \text{XZY}={{90}^{{}^\circ }}$

6. Construct a right triangle whose base is $12~\text{cm}$ and the sum of its hypotenuse and other side is $18~\text{cm}$.

Ans: Given that the Base $\text{BC}=12~\text{cm}$ and $\text{AB}+\text{AC}=18~\text{cm}$.

Steps of construction are

Draw a ray $BX$ and cut off a line segment $\text{BC}=12~\text{cm}$ from it.

Draw an angle $\text{XBY}=90{}^\circ $.

From the ray $BY$, cut off a line segment with $\text{BD}=18~\text{cm}$.

Join $\text{CD}$.

Draw the perpendicular bisector of $\text{CD}$ intersecting $\text{BD}$ at $\text{A}$.

Join $\text{AC}$.

Then $\text{ABC}$ is the required right- angled triangle.

Justification:

A lies on the perpendicular bisector of $\text{CD}$.

$\therefore AC=AD$

And then $\text{AB}=\text{BD}-\text{AD}$

$\Rightarrow \text{AB}=\text{BD}-\text{AC}$

$\Rightarrow AB+AC=BD=18~\text{cm}$

7. Construct the angle of the measurement ${{90}^{{}^\circ }}$

Ans: Steps of construction are 

Draw any ray $\text{OA}$

 Draw an arc intersecting $OA$ at $P$ with $O$ as centre and any convenient radius.

With $P$ as a centre and the same radius as above draw an arc intersecting the previous radius at $Q$

Again taking $Q$ as a centre and arc of the same radius draw another arc intersecting the previous arc at $\text{R}$.

Again with $\text{R}$ and $Q$ as a centre and radius more than $\frac{1}{2}$ $QR$ draw two arcs intersecting each other at $\underset{\scriptscriptstyle-}{S}$

Draw Ray $OX$.

Then $\angle \text{YOA}={{90}^{{}^\circ }}$

8. Construct an equilateral triangle whose side is $4~\text{cm}$

Ans: The Steps of Construction are 

Draw a ray $O\text{X}$

Taking $O$ as a centre draw an arc of radius $4~\text{cm}$ which cut $O\text{X}$ at $\text{A}$.

Now taking $O$ and $\text{A}$ as a centre now draw two arcs with a radius of $4~\text{cm}$ which intersect each other at $\text{B}$

Join $OB$ and$AB$

(5) $\Delta ABC$ is required triangle

9. Construct the Perpendicular bisector of a line segment of length $12.5~\text{cm}$

Ans: Steps of construction:

Draw a line segment $AB$ with the  length of $12.5~\text{cm}$

Taking $A$ as a centre and arc of radius more than $\frac{1}{2}AB$ draw both sides of $\text{AB}$

Again taking $B$ as a centre and arc of the previous radius draw both sides of $\text{AB}$ which intersect previous arcs at $C$ and $D$

Join $CD$, Which intersect $\text{AB}$ at $\text{O}$.

Point $O$ bisects $\text{AB}$.

10. Construct an angle of $22{{\frac{1}{2}}^{{}^\circ }}$

Ans: Steps of construction are 

Draw ray $OX$

Draw $\angle XOY={{90}^{{}^\circ }}$

Bisect $\angle YOY$

$\angle ROP={{45}^{{}^\circ }}$

Now bisect angle $\angle ROP$

$\angle TOP$ is required angle $=22\frac{1}{2}=\frac{{{45}^{{}^\circ }}}{2}$

11. Construct an equilateral triangle of each sides $5.6~\text{cm}$

Ans: Steps of construction are 

Draw $\text{AB}=5.6~\text{cm}$

Draw $\angle KAB={{60}^{{}^\circ }}$

Taking $A$ as centre draw an arc of radius $5.6~\text{cm}$ which intersect $\text{AX}$ at point $\text{C}$

Join $BC$

$\vartriangle ABC$ is required equilateral $\Delta $

12. Construct the perpendicular bisector of a line segment of side $6.5~\text{cm}$

Ans: Step of construction:

Draw line segment $\text{AB}=6.5~\text{cm}$

Taking $A$ as a centre and draw two arcs of radius more than $\frac{1}{2}AB$ on both sides of $\text{AB}$.

Again taking $B$ as a centre and draw arcs of same radius which intersect previous arcs at point $C$ and $D$.

Join $CD$, $CD$ is perpendicular bisector of $AB$.

13. Construct an angle of ${{105}^{{}^\circ }}$

Ans:  Steps of construction are

Draw ray $OA$

Taking $\text{X}$ as a centre draw an arc of any radius which intersects $\text{OA}$ at the point $\text{X}$

Taking $\text{X}$ as a centre draw two arcs of same radius which intersect previous arcs at point $\text{R}$ and $\text{Q}$

Bisect $QR$

(5) $\angle BOA={{90}^{{}^\circ }}$

Now, bisect $\text{PQ}$ and Join $\text{OC}$

$\angle COA  =  105{}^\circ $

14. Construct an angle of ${{45}^{{}^\circ }}$ at initial Point of the given ray and justify the construction

Ans: Steps of construction are

Draw ray $OA$of any length

Draw $\angle QOA={{90}^{{}^\circ }}$

Bisect angle $\angle QOA$

 $\angle ROA$ is required angle and $\angle ROA={{45}^{{}^\circ }}$

15. Construct the angle of ${{15}^{{}^\circ }}$

Ans: Steps of construction are

Draw a ray $\text{OA}$

Taking $\text{P}$ as centre draw an arc of any radius which intersect $\text{OA}$ at point $\text{P}$

Now take $\text{P}$  as a centre draw an arc of the same radius which intersect previous arc at point $Q$

Now taking $\text{P}$ and $Q$ as a centre draw arcs of the same radius which intersects each other at $\text{R}$.

Now bisect $\angle ROP$

 $\angle POS={{15}^{{}^\circ }}$

16. Construct an equilateral triangle whose each side is $4.5~\text{cm}$

Ans: Steps of construction are

Draw a ray $\text{OX}$

Taking $\text{O}$ as centre  and draw an arc of radius $4.5~\text{cm}$ which cut $\text{OX}$ at $\text{A}$

Now taking $O$and $\text{A}$ as a centre draw two arcs which intersect each other at $\text{B}$.

Join $\text{OB}$ and $\text{AB}$

(5) $\vartriangle ABC$ is required triangle

17. Construct an angle  ${{30}^{{}^\circ }}$ at the initial point of a ray and Justify your construction

Ans: Steps of construction are 

Draw ray $OX$

Taking $O$ as a centre, draw an arc of any radius which intersect $O\text{X}$ at point $\text{P}$

Taking $\text{P}$ as a centre draw an arc of same radius which intersect previous radius at $\text{Q}$

Now taking $\text{P}$ and $Q$ as a centre draw arcs which intersect each other at point $R$

Join $OR$

(6) $\angle ROX={{30}^{{}^\circ }}$

18. Construct a line segment of length $5.5~\text{cm}$ and bisect it.

Ans: Steps of construction are

Draw line segment $AB=5.5~\text{cm}$

Taking $A$ as a centre draw arc of radius more than $\frac{1}{2}AB$ on both sides of $\text{AB}$

Taking $B$as a centre draw arc equal to previous radius which intersects previous arcs at point $\text{C}$ and $\text{D}$

Join $CD$ which intersects $\text{AB}$ at point $O$ .

Point $O$ is bisector of $AB$

19. Construct an equilateral triangle whose each side is $4.9~\text{cm}$

Ans: Steps of construction are

Draw line segment $\text{AB}=4.9~\text{cm}$

Taking $A$ as a centre draw an arc of radius $4.9~\text{cm}$

Taking $\text{B}$ as a centre draw an arc of radius $4.9~\text{cm}$ which intersect previous arc at point $\text{C}$

Join $AC$ and $\text{BC}$

 $\Delta ABC$ is required triangle

20. Construct an angle of ${{135}^{{}^\circ }}$

Ans: Steps of construction are

Draw ray $OA$

Taking $O$ as a centre draw an arc of any radius which intersect ray $\text{OA}$ at point $\text{P}$.

Now taking $\text{P}$ as a centre draw arcs of same radius which intersect previous arc at point

$Q$, $R$ and $S$ respectively

Now taking $R$ and $\text{S}$ as a centre draw arcs of same radius which intersect each other at point $\text{T}$.

Join $OT$

(6) $\angle TOA={{135}^{{}^\circ }}$

21. Construct perpendicular bisector of line segment $8~\text{cm}$

Ans: Steps of construction are

Draw a line segment $\text{AB}=8~\text{cm}$

Taking $A$ as a centre draw arcs of radius more $\frac{1}{2}AB$ on both side of $\text{AB}$.

Taking $B$ as a centre draw arcs of same radius on both sides of AB which intersect previous arcs at point $\text{C}$ and $\text{D}$.

Join $CD$ which intersect $\text{AB}$ at point $\text{O}$

$\text{OA}=\text{OB}=4~\text{cm}$

22. Construct an angle of ${{60}^{{}^\circ }}$ at the initial point of a given ray and bisect it.

Ans: Steps of construction are

Draw ray $\text{OA}$

Taking $O$ as a centre draw an arc of any radius which intersects $OA$ at point $P$

Now taking $\text{P}$ as a centre draw an arc of same radius which intersect previous are at point $Q$

$\angle QOA={{60}^{{}^\circ }}$

Taking $\text{P}$ and $Q$ as a centre draw arcs of same radius which intersect each other at point $\text{R}$

Join $OR$, OR bisects $\angle QOA$

$\angle ROA=\angle ROQ={{30}^{{}^\circ }}$

23. Construct the angle of the measurement $7{{\frac{1}{2}}^{{}^\circ }}$

Ans: Steps of construction are

Draw ray $\text{AB}$

Taking $A$ as a centre draw an arc of any radius which intersects $\text{AB}$ at point $\text{P}$

Taking $\text{P}$ as a centre draw an arc of same radius which intersect previous arc at point $Q$

Now taking $\text{P}$ and $Q$ as a centre draw two arcs of same radius which intersects each other at point $\text{R}$.

$\angle RAP={{30}^{{}^\circ }}$ ,Bisect $\angle RAP$

$\angle SAB$ is bisector of $\angle RAP$ then $\angle SAP={{\frac{15}{2}}^{{}^\circ }}=7\frac{1}{2}{}^\circ $

24. Construct the angle of the measurement $37\frac{1}{2}$

Ans: Steps of construction are

Draw ray $OA$

Taking $O$ as a centre draw an arc of any radius which intersects $OA$ at point $B$

Taking $B$ as a centre draw two arcs of same radius which intersect previous arc at point $\text{C}$ and D

Now taking $C$ and D as a centre draw two arcs of same radius which intersect previous arc at point $E$

Bisect $\angle EOC$

(6) $\angle FOB={{60}^{{}^\circ }}+{{15}^{{}^\circ }}={{75}^{{}^\circ }}$

25. Construct an equilateral triangle of side $5~\text{cm}$

Ans: Steps of construction are

Draw a line segment $\text{AB}=5~\text{cm}$

Taking $A$ as a centre draw an arc of radius $5~\text{cm}$

Taking $\text{B}$ as a centre draw an arc of same radius which intersect previous arc at point $C$

Join $BC$

Triangle $\text{ABC}$ is required triangle

26. Draw a line segment of length $4.5~\text{cm}$ and bisect it

Ans: Steps of construction are

Draw a line segment $\text{AB}=4.5~\text{cm}$

Taking $A$ as a centre draw an arc of radius more then $\frac{1}{2}AB$

Taking $B$ as a centre draw another arc of same radius. Which intersect previous arc at point $\text{C}$ and $\text{D}$

Join $CD$

$CD$ bisect $\text{AB}$ at point $\text{O}$

27. Construct a $ \Delta $ whose all angles are ${{60}^{{}^\circ }}$ each

Ans: Steps of construction are

Take a line segment $\text{AB}$ of any length

Taking $\text{A}$ and $\text{B}$ as a centre draw an arc of any radius. Which intersect $\text{AB}$ at point $\text{P}$ and $Q$.

Taking $\text{P}$ and $Q$ as a centre draw an arc of same radius which intersect previous arc at point $\text{R}$ and $S$

Join $AR$ and $BS$ ,which intersect at the point $C$

Triangle $\text{ABC}$ is required triangle

28. Draw a line segment of length $12.6~\text{cm}$ bisect it and measure each part

Ans: Steps of construction are

Draw a line segment $\text{AB}=12.6~\text{cm}$

Taking $\text{A}$ as a centre draw an arc of radius more than $\frac{1}{2}AB$

Taking $\text{B}$ as a centre draw an arc of  radius more than $\frac{1}{2}AB$ ,which intersect previous arc at point $C$and $D$

Join $CD$

$\text{CD}$ bisect $\text{AB}$ at point $\text{O}$

$\text{AO}=\text{OB}=6.3~\text{cm}$

29. Construct an angle of ${{60}^{{}^\circ }}$ and bisect it and measure each angle

Ans: Steps of construction are

Take a ray $\text{OA}$

Taking $\text{O}$ as a centre draw an arc of any radius which intersects $\text{OA}$ at point $\text{P}$.

Taking $\text{P}$ as a centre draw arc of same radius which intersect previous arc at point $\text{Q}$

Join $0Q$ and expand to $Y$

Bisect $\angle YOA$

Ray OX bisect $\angle YOA$

$\angle YOX=\angle XOA={{30}^{{}^\circ }}$

30. Construct an angle of ${{30}^{{}^\circ }}$ whose initial point is given ray.

Ans: Steps of construction are

Draw a ray $O~\text{A}$

Taking $O$ as a centre draw an arc of any radius which intersects $OA$ at point $P$.

Taking $\text{P}$ as a centre draw an arc of same radius which intersect previous arc at point $\text{Q}$

Bisect $\angle POQ$

 $\angle AOR$ is bisector of $\angle AOQ$

 $\angle AOR={{30}^{{}^\circ }}$

Long Answer Questions (4 Marks)

1. Construct the angles of the following measurements:

${{30}^{{}^\circ }}$

Ans:  Steps of construction: ${{30}^{{}^\circ }}$

Draw a ray $OA$.

With $O$ as centre and a suitable radius, draw an arc $LM$ that cuts $OA$ at $L$.

With $L$ as centre and radius $OL$, draw an arc to cut $LM$ at $\text{N}$.

Join $O$ and $\text{N}$ draw ray $\text{OB}$. Then $\angle \text{AOB}={{60}^{{}^\circ }}$.

With $L$ as centre and radius greater than $\frac{1}{2}{{L}_{N}}$, draw an arc.

With $\text{N}$ as centre and same radius as in step $5$, draw another arc cutting the arc drawn in step $5$ at $P$.

Join $O$ and $P$ and draw ray $OC$

Thus $OC$bisects $\angle \text{AOB}$ and therefore $\angle \text{AOC}=\angle \text{BOC}={{30}^{{}^\circ }}$

 $22\frac{1}{2}{}^\circ $

Ans: Steps of construction of $22\frac{1}{2}{}^\circ $ 

Draw a ray $\text{OA}$.

With $O$ as centre and convenient radius, draw an arc $LM$ cutting $O~\text{A}$ at $\text{L}$.

Now with $L$ as centre and radius OL, draw an arc cutting the arc $LM$ at $P$.

Then taking $P$ as centre and radius OL, draw an arc cutting arc PM at the point $Q$.

Join $OP$ to draw the ray $\text{OB}$. Also, join $O$ and $\text{Q}$ to draw the $\text{OC}$. We observe that:$\angle \text{AOB}=\angle \text{BOC}={{60}^{2}}$

Now, With $\text{P}$ as centre and radius greater than$\frac{1}{2}\text{PQ}$ to bisect $\angle \text{BOC}$.  

Now with $Q$ as centre and the same radius as in step 6, draw another arc cutting the arc drawn in step 6 at$\text{R}$.

Join $O$ and $\text{R}$ and draw ray$\text{OD}$. Then $\angle \text{AOD}$ is the required angle of${{90}^{{}^\circ }}$.

With $\text{L}$ as centre and radius greater than $\frac{1}{2}$ $LS$, draw an arc.

Now with $\text{S}$ as centre and the same radius as in step 2, draw another arc cutting the arc draw in step 2 at$\text{T}$.

Join $O$ and $\text{T}$ and draw ray$\text{OE}$. Thus ${OE}$ bisects $\angle \text{AOD}$ and therefore $\angle \text{AOE}=\angle \text{DOE}=$ ${{45}^{\prime }}$.

Let ray $OE$ intersect the arc of circle at$\text{N}$.

Now with $L$ as centre and radius greater than$\frac{1}{2}\text{LN}$, draw an arc.

With $\text{N}$ as centre and same radius as in above step and draw another arc-cutting arc drawn in above step at$\text{I}$.

Join $O$ and $I$ and draw ray$OF$. Thus OF bisects $\angle AOE$ and therefore $\angle AOF=\angle $ EOF= $22{{\frac{1}{2}}^{{}^\circ }}$

${{15}^{{}^\circ }}$

Ans: (iii) Steps of construction of ${{15}^{{}^\circ }}$

Draw a ray$OA$.

With $O$ as centre and a suitable radius, draw an arc $L$ that cuts $OA$ at$L$.

With $\text{L}$ as centre and radius OL, draw an arc to cut $LM$ at$\text{N}$.

Join $O$ and $\text{N}$ draw ray$\text{OB}$. Then$\angle \text{AOB}={{60}^{{}^\circ }}$.

With $L$as centre and radius greater than$\frac{1}{3}\text{LN}$, draw an arc.

Now with $\text{N}$ as centre and same radius as in step $5$ , draw another arc cutting the arc drawn in step $5$ at$\underset{\scriptscriptstyle-}{P}$.

Join $O$ and $\text{P}$ and draw ray$0\text{c}$. Thus $0\text{c}$ bisects $\angle \text{AOB}$ and therefore $\angle \text{AOC}=\angle \text{BOC}=$ ${{30}^{{}^\circ }}.$

Let ray OC intersects the arc of circle at point$Q$.

Now with $\text{L}$ as centre and radius greater than$\frac{1}{2}\text{LQ}$, draw an arc.

With $Q$ as centre and same radius as in above step, draw another arc cutting the arc shown in above step at$\text{R}$.

Join $O$ and $\text{R}$ and draw ray $OS$.

Thus $OS$ bisects $\angle AOC$ and therefore $\angle \operatorname{COS}=\angle AOS={{15}^{{}^\circ }}$

2. .Construct the following angles and verify by measuring them by a protector:

${{75}^{{}^\circ }}$

Ans:  Step of construction are

Draw $\angle \text{ABE}={{60}^{\prime }}$ and $\angle \text{ABF}={{90}^{{}^\circ }}$ (Follow the same steps as done in Question 1 and Question 3 (I))

Let ray $BF$ intersects the arc of circle at$G$.

Now with $\text{M}$ as centre and radius greater than $\frac{1}{2}\text{MG}$ draw an arc.

With $\text{G}$ as centre and with same radius as in step (c), draw an arc, which intersects the previous arc at point$\text{H}$.

Draw a ray $BC$ passing through $\text{H}$ which bisects$\angle \text{EBF}$.

Thus $\angle ABC={{75}^{{}^\circ }}$ is the required angle.

Justification:

$\angle \text{EBF}=\angle \text{ABF}-\angle \text{ABE}$

$={{90}^{{}^\circ }}-{{60}^{{}^\circ }}={{30}^{{}^\circ }}$

Now $\angle \text{EBF}=\angle \text{CBF}=\frac{1}{2}\angle \text{EBF}$

$=\frac{1}{2}\times {{30}^{{}^\circ }}={{15}^{{}^\circ }}\text{I}\because \text{BC}$ Is the bisector of? $\left. \angle \text{EBF} \right]$

$\therefore \angle \text{ABC}=\angle \text{ABE}+\angle \text{EBC}$

$={{60}^{{}^\circ }}+{{15}^{{}^\circ }}={{75}^{{}^\circ }}$

${{105}^{{}^\circ }}$

Ans: Steps of construction of ${{105}^{{}^\circ }}$

Draw $\angle \text{ABE}={{90}^{{}^\circ }}$ and$\angle \text{ABF}={{120}^{{}^\circ }}$.

Let ray $BE$ intersects the arc of circle at $\text{M}$ and ray BF intersects the arc of circle

$\text{N}$.

With point $\text{M}$ as centre and radius greater than$\frac{1}{2}\text{MN}$, draw an arc.

With $\text{N}$ as centre and with same radius as in step (c), draw another arc, which intersects the previous arc at$\text{P}$.

Draw a ray $BC$ passing through $\text{P}$ which bisects $\angle \text{EBF}$.

Thus $\angle \text{ABC}={{105}^{{}^\circ }}$ is the required angle.

Justification:

$\angle EBF=\angle ABF-\angle ABE$

$={{120}^{{}^\circ }}-{{90}^{{}^\circ }}={{30}^{{}^\circ }}$

Now $\angle EBC=\angle CBF$

$=\frac{1}{2}\angle EBF=\frac{1}{2}\times {{30}^{{}^\circ }}={{15}^{{}^\circ }}[\because BC$ Is the bisector of? $\angle \text{EBF}]$

$\therefore \angle \text{ABC}=\angle \text{ABE}+\angle \text{EBC}$

$={{90}^{{}^\circ }}+{{15}^{{}^\circ }}={{105}^{{}^\circ }}$

${{135}^{{}^\circ }}$

Ans: Steps of construction of ${{135}^{{}^\circ }}$

Draw a ray $OA$.

With $O$ as centre and convenient radius, draw an arc $LM$ (having length more than the semicircle) cutting $OA$ at $L$.

Now with $\text{L}$ as centre and radius = OL; draw an arc cutting the arc $LM$ at $\text{P}$.

Then taking $\text{P}$ as centre and radius $\text{OL}$, draw an arc cutting arc $\text{PM}$ at $Q$.

Now bisect $\angle $ $POQ$ by ray $O~\text{B}$, we get $\angle \text{AOB}={{90}^{{}^\circ }}$.

Now taking $Q$ as centre and radius $OL$, draw an arc cutting $QM$at $N$.

Join $\text{O}$ and $\text{N}$ to draw the ray $\text{OC}$

Thus we get $\angle \text{AOC}={{180}^{{}^\circ }}$.

$\Rightarrow \angle \text{BOC}=\angle \text{AOB}={{90}^{{}^\circ }}$

Now bisect $\angle \text{BOC}$ by ray $OD$.

Then $\angle AOD$ is the required angle of ${{135}^{{}^\circ }}$.

$\angle \text{AOD}=\angle \text{AOB}+\angle \text{BOD}={{90}^{{}^\circ }}+{{45}^{{}^\circ }}={{135}^{{}^\circ }}$

3. Construct a triangle $\text{ABC}$ in which $\text{BC}=7~\text{cm}    \angle B={{75}^{{}^\circ }}$ and $\text{AB}+\text{AC}=9~\text{cm}$.

Ans: Steps of construction are

Draw $\text{BC}=7~\text{cm}$

Draw $\angle DBC={{75}^{{}^\circ }}$

Cut a line segment $\text{BD}=9~\text{cm}$

Join DC and make $\angle DCY=\angle BDC$

Let $CY$ intersect $\text{BX}$ at $\text{A}$

Triangle $\text{ABC}$ is required triangle.

4. Construct a triangle $\text{XYZ}$ in which $\angle y={{30}^{{}^\circ }}\angle Z={{90}^{{}^\circ }}$ and $XY+\mathbb{Z}+ZX=11~\text{cm}$

Ans: Steps of construction

Draw line segment $\text{PQ}=11~\text{cm}$

At P construct an angle ${{30}^{{}^\circ }}$ and at $\text{Q}$ an angle ${{90}^{{}^\circ }}$

Bisect these angles. Let the bisectors of these angles intersect each other at point $\text{X}$.

Draw perpendicular bisector $DE$ of$PX$ and$FG$ of$XQ$ intersect $PQ$ at point $Y$ and $Z$ respectively.

Join$XY$ and$XZ$

$XYZ$ is required triangle.

5. Construct a triangle of $\text{ABC}$ in which $\text{BC}=8~\text{cm}\angle B={{45}^{{}^\circ }}$ and $AB-AC=3.5~\text{cm}$.

Ans: Steps of construction are

(1) Draw line segment $\text{AB}=8~\text{cm}$

(2) Construct $\angle YBC={{45}^{{}^\circ }}$

(3) Taking $\text{B}$ as a centre draw an arc of radius $3.5~\text{cm}$ which intersect at point $\text{D}$

(4) Join $DC$

(5) Draw perpendicular bisector of$DC$ which intersect$BY$ at point $A$

(6) Join $AC$

(7) $\vartriangle ABC$ is required triangle.

6. Construct a right triangle whose base is $12~\text{cm}$ and sum of its hypotenuse and other side is $18~\text{cm}$.

Ans: Steps of constructions are

(1) Draw line segment $\text{AB}=12~\text{cm}$

(2) Construct $\angle A={{90}^{{}^\circ }}$

(3) Draw $\text{AD}=18~\text{cm}$

(1) Construct $\angle DBY=\angle ADB$ by intersecting $\text{AD}$ at point $\text{C}$

(5) $\Delta ABC$ is required right triangle.

7. Construct a triangle $PQR$ in which $QR=6~\text{cm}\angle Q={{60}^{{}^\circ }}$ and $PR-PQ=2~\text{cm}$

Ans: Steps of construction are

(1) Draw line segment $QR$$=$$6cm$

(2) Cut line segment $\text{QD}=\text{PR}-\text{PQ}=2~\text{cm}$ from line $\text{x}$ extended on opposite side of line segment $QR$

(3) Join $DR$ and draw the perpendicular bisector say $MN$ of $DR$

(4) Let $MN$ bisect$DX$ at point $P$. join $PR$

(5) $PQR$ is required triangle.

8. Construct a triangle $\text{ABC}$, in which $\angle B={{60}^{{}^\circ }},\angle C={{45}^{{}^\circ }}$ and $AB+BC+CA=11~\text{cm}$

Ans: Steps of construction are

(1) Draw a line segment $PQ=11~\text{cm}(=AB+BC+CA)$

(2) At $P$ construct an angle of ${{60}^{{}^\circ }}$ and at $Q$ an angle of ${{45}^{{}^\circ }}$

(3) Bisect these angle at the point $\text{A}$and intersect at the point $\text{A}$.

(4) Draw perpendicular bisectors $DE$ of$AP$ to intersect$PQ$ at$B$ and$FG$ of $AQ$ to intersect $PQ$ at $\text{C}$.

(5) Join $\text{AB}$ and $\text{AC}$. Then $\text{ABC}$ is required triangle.

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