Textbook - Constructions Questions in English

(N/A) Steps of construction:
$I.$ Draw a ray $OA$.
$II.$ Taking $O$ as the centre and a suitable radius,draw a semicircle,which cuts $OA$ at $B$.
$III.$ Keeping the radius the same,divide the semicircle into three equal parts such that $\widehat{BC} = \widehat{CD} = \widehat{DE}$.
$IV.$ Draw rays $\overrightarrow{OC}$ and $\overrightarrow{OD}$.
$V.$ Draw $\overrightarrow{OF}$,the bisector of $\angle COD$.
Thus,$\angle AOF = 90^{\circ}$.
Justification:
Since $O$ is the centre of the semicircle and it is divided into $3$ equal parts,
$\therefore \widehat{BC} = \widehat{CD} = \widehat{DE}$.
$\Rightarrow \angle BOC = \angle COD = \angle DOE$.
Since equal arcs subtend equal angles at the centre,and $\angle BOC + \angle COD + \angle DOE = 180^{\circ}$,we have $3 \angle BOC = 180^{\circ}$.
$\therefore \angle BOC = 60^{\circ}$.
Similarly,$\angle COD = 60^{\circ}$ and $\angle DOE = 60^{\circ}$.
Since $OF$ is the bisector of $\angle COD$,
$\therefore \angle COF = \frac{1}{2} \angle COD = \frac{1}{2}(60^{\circ}) = 30^{\circ}$.
Now,$\angle BOF = \angle BOC + \angle COF = 60^{\circ} + 30^{\circ} = 90^{\circ}$.
Thus,$\angle AOF = 90^{\circ}$.

(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Taking $O$ as the centre and with a suitable radius,draw a semicircle such that it intersects $\overrightarrow{OA}$ at $B$.
$III.$ Taking $B$ as the centre and keeping the same radius,cut the semicircle at $C$. Similarly,cut the semicircle at $D$ and $E$,such that $\widehat{BC} = \widehat{CD} = \widehat{DE}$.
$IV.$ Bisect $\angle BOC$ to get a ray $\overrightarrow{OF}$ such that $\angle BOF = 30^{\circ}$.
$V.$ Bisect $\angle BOF$ to get a ray $\overrightarrow{OG}$ such that $\angle BOG = 45^{\circ}$.
Justification:
Since $\widehat{BC} = \widehat{CD} = \widehat{DE}$,we have $\angle BOC = \angle COD = \angle DOE = 60^{\circ}$ (as they subtend equal arcs at the centre).
Since $\overrightarrow{OF}$ is the bisector of $\angle BOC$,$\angle BOF = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Since $\overrightarrow{OG}$ is the bisector of $\angle FOC$ (where $\angle FOC = 30^{\circ}$),$\angle FOG = \frac{1}{2} \times 30^{\circ} = 15^{\circ}$.
Therefore,$\angle BOG = \angle BOF + \angle FOG = 30^{\circ} + 15^{\circ} = 45^{\circ}$.

(N/A) $(i)$ Angle of $30^{\circ}$
Steps of construction:
$I.$ Draw a ray $OA$.
$II.$ With $O$ as the center and a suitable radius,draw an arc,cutting $\overrightarrow{OA}$ at $B$.
$III.$ With $B$ as the center and the same radius,draw an arc to cut the previous arc at $C$.
$IV.$ Join $\overrightarrow{OC}$. Now,$\angle BOC = 60^{\circ}$.
$V.$ Draw the bisector of $\angle BOC$ to get ray $\overrightarrow{OD}$.
Thus,$\angle BOD = \frac{1}{2} \angle BOC = \frac{1}{2}(60^{\circ}) = 30^{\circ}$.
Therefore,$\angle BOD = 30^{\circ}$.

(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Construct $\angle AOB = 90^{\circ}$.
$III.$ Draw $OC$,the angle bisector of $\angle AOB$,such that $\angle AOC = \frac{1}{2} \angle AOB = \frac{1}{2}(90^{\circ}) = 45^{\circ}$.
$IV.$ Now,draw $OD$,the angle bisector of $\angle AOC$,such that $\angle AOD = \frac{1}{2} \angle AOC = \frac{1}{2}(45^{\circ}) = 22 \frac{1}{2}^{\circ}$.
Thus,$\angle AOD = 22 \frac{1}{2}^{\circ}$ is the required angle.

(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Construct $\angle AOB = 60^{\circ}$.
$III.$ Draw $\overrightarrow{OC}$,the angle bisector of $\angle AOB$,such that $\angle AOC = \frac{1}{2} \angle AOB = \frac{1}{2}(60^{\circ}) = 30^{\circ}$.
$IV.$ Draw $\overrightarrow{OD}$,the angle bisector of $\angle AOC$,such that $\angle AOD = \frac{1}{2} \angle AOC = \frac{1}{2}(30^{\circ}) = 15^{\circ}$.
Thus,$\angle AOD = 15^{\circ}$ is the required angle.

(N/A) $75^{\circ} = 60^{\circ} + 15^{\circ}$
Steps of construction:
$I.$ Draw a ray $\overrightarrow{ OA }$.
$II.$ With $O$ as the center and a suitable radius,draw an arc that meets $\overrightarrow{ OA }$ at $B$.
$III.$ With center $B$ and the same radius,mark a point $C$ on the previous arc. Now,$\angle BOC = 60^{\circ}$.
$IV.$ With center $C$ and the same radius,mark another point $D$ on the arc. Now,$\angle COD = 60^{\circ}$.
$V.$ Draw $\overrightarrow{ OP }$,the bisector of $\angle COD$,such that $\angle COP = \frac{1}{2}(60^{\circ}) = 30^{\circ}$. Thus,$\angle BOP = 60^{\circ} + 30^{\circ} = 90^{\circ}$.
$VI.$ Draw $\overrightarrow{ OQ }$,the bisector of $\angle COP$,such that $\angle COQ = 15^{\circ}$.
Thus,$\angle BOQ = \angle BOC + \angle COQ = 60^{\circ} + 15^{\circ} = 75^{\circ}$.
Finally,verify the angle using a protractor.

(N/A) $105^{\circ} = 90^{\circ} + 15^{\circ}$
Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ With $O$ as the center and any suitable radius,draw an arc that meets $\overrightarrow{OA}$ at point $B$.
$III.$ With $B$ as the center and the same radius,draw an arc to cut the previous arc at point $C$. (This represents $60^{\circ}$).
$IV.$ With $C$ as the center and the same radius,draw another arc to cut the first arc at point $D$. (This represents $120^{\circ}$).
$V.$ Bisect the arc between $C$ and $D$ to get point $P$ such that $\angle AOP = 90^{\circ}$.
$VI.$ Bisect the arc between $P$ and $D$ to get point $Q$. The angle $\angle AOQ$ will be $105^{\circ}$ (since $90^{\circ} + 15^{\circ} = 105^{\circ}$).

(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{OP}$.
$II.$ With centre $O$ and a suitable radius,draw an arc to meet $\overrightarrow{OP}$ at $A$.
$III.$ Keeping the same radius and starting from $A$,mark points $Q, R,$ and $S$ on the arc such that $\angle AOQ = 60^{\circ}$,$\angle AOR = 120^{\circ}$,and $\angle AOS = 180^{\circ}$.
$IV.$ Draw $\overrightarrow{OL}$,the bisector of $\angle ROS$,which makes $\angle ROL = 30^{\circ}$. Thus,$\angle AOL = 120^{\circ} + 30^{\circ} = 150^{\circ}$.
$V.$ Draw $\overrightarrow{OM}$,the bisector of $\angle ROL$,which makes $\angle ROM = 15^{\circ}$. Thus,$\angle AOM = 120^{\circ} + 15^{\circ} = 135^{\circ}$.
Therefore,$\angle AOM = 135^{\circ}$.

(N/A) Let us construct an equilateral triangle,each of whose side $= PQ$.
Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Taking $O$ as centre and radius equal to $PQ$,draw an arc to cut $OA$ at $B$ such that $OB = PQ$.
$III.$ Taking $O$ as centre and radius $PQ$,draw an arc. Then,taking $B$ as centre and radius $PQ$,draw another arc to intersect the previous arc at $C$.
$IV.$ Join $OC$ and $BC$.
Thus,$\Delta OBC$ is the required equilateral triangle.
Justification:
By construction,$OB = PQ$ and $OC = PQ$ (radii of the same arc).
Also,$BC = PQ$ (radius of the arc with centre $B$).
Therefore,$OB = OC = BC = PQ$.
Since all three sides are equal,$\Delta OBC$ is an equilateral triangle.

(N/A) $1.$ Draw a line segment $PQ = 11 \, cm$ such that $PQ = AB + BC + CA$.
$2.$ At point $P$,construct an angle of $\frac{60^{\circ}}{2} = 30^{\circ}$ and at point $Q$,construct an angle of $\frac{45^{\circ}}{2} = 22.5^{\circ}$.
$3.$ Let these angle bisectors intersect at point $A$.
$4.$ Draw the perpendicular bisector of $AP$,which intersects $PQ$ at $B$.
$5.$ Draw the perpendicular bisector of $AQ$,which intersects $PQ$ at $C$.
$6.$ Join $AB$ and $AC$. Thus,$ABC$ is the required triangle.

(N/A) Steps of construction:
$I.$ Draw a ray $BX$.
$II.$ From ray $BX$,cut off a line segment $BC = 7\,cm$.
$III.$ At point $B$,construct an angle $\angle CBY = 75^\circ$.
$IV.$ From ray $BY$,cut off a line segment $BD = 13\,cm$.
$V.$ Join $D$ and $C$.
$VI.$ Construct the perpendicular bisector of the line segment $DC$. Let this bisector intersect $BD$ at point $A$.
$VII.$ Join $AC$.
Thus,$\Delta ABC$ is the required triangle.

(N/A) Steps of construction:
$I.$ Draw a ray $BX$.
$II.$ From $\overrightarrow{BX},$ cut off $\overline{BC} = 8\,cm$.
$III.$ Construct $\angle CBY = 45^{\circ}$.
$IV.$ From $\overrightarrow{BY},$ cut off $\overline{BD} = 3.5\,cm$.
$V.$ Join $D$ and $C$.
$VI.$ Draw the perpendicular bisector $PQ$ of $\overline{DC},$ which intersects $\overrightarrow{BY}$ at $A$.
$VII.$ Join $AC$.
Thus,$ABC$ is the required triangle.

(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{QX}$.
$II.$ From $\overrightarrow{QX}$,cut off $QR = 6 \, cm$.
$III.$ Construct a line $YQY'$ such that $\angle RQY = 60^{\circ}$.
$IV.$ Cut off $QS = 2 \, cm$ (from $\overrightarrow{QY'}$).
$V.$ Join $S$ and $R$.
$VI.$ Draw $MN$,the perpendicular bisector of $SR$,which intersects $QY$ at $P$.
$VII.$ Join $P$ and $R$.
Thus,$PQR$ is the required triangle.

(N/A) $I.$ Draw a line segment $AB = 11 \,cm = (XY + YZ + ZX)$.
$II.$ Construct $\angle BAP = 30^{\circ} = \angle Y$.
$III.$ Construct $\angle ABQ = 90^{\circ} = \angle Z$.
$IV.$ Draw $\overrightarrow{AR}$,the bisector of $\angle BAP$.
$V.$ Draw $\overrightarrow{BS}$,the bisector of $\angle ABQ$,such that $\overrightarrow{AR}$ and $\overrightarrow{BS}$ intersect each other at $X$.
$VI.$ Draw the perpendicular bisector of $AX$,which intersects $AB$ at $Y$.
$VII.$ Draw the perpendicular bisector of $XB$,which intersects $AB$ at $Z$.
$VIII.$ Join $XY$ and $XZ$.
Thus,$XYZ$ is the required triangle.

(N/A) $I.$ Draw a line segment $\overline{BC} = 12 \, cm$.
$II.$ Construct $\angle CBY = 90^{\circ}$ at point $B$.
$III.$ From the ray $\overrightarrow{BY}$,cut off a line segment $BX = 18 \, cm$.
$IV.$ Join $CX$.
$V.$ Construct the perpendicular bisector of $CX$,and let it intersect $BX$ at point $A$.
$VI.$ Join $AC$.
Thus,$\triangle ABC$ is the required right-angled triangle.