Mix Examples - Constructions Questions in English

(A) To construct an angle using a ruler and a compass,the angle must be a multiple of $7.5^{\circ}$ (which is half of $15^{\circ}$,which in turn is half of $30^{\circ}$,which is half of $60^{\circ}$).
We can check the given options:
$A) 37.5^{\circ} = 7.5^{\circ} \times 5$. Since $37.5^{\circ}$ is a multiple of $7.5^{\circ}$,it can be constructed by bisecting $75^{\circ}$ (which is $60^{\circ} + 15^{\circ}$).
$B) 35^{\circ}$ is not a multiple of $7.5^{\circ}$.
$C) 40^{\circ}$ is not a multiple of $7.5^{\circ}$.
$D) 47.5^{\circ}$ is not a multiple of $7.5^{\circ}$.
Thus,$37.5^{\circ}$ is the only angle among the options that can be constructed.

(B) In the construction of a triangle $ABC$ where the base $AB$ and the base angle $\angle A$ are given,the construction of the triangle is possible if the difference between the other two sides ($BC - AC$ or $AC - BC$) is less than the length of the base $AB$.
According to the triangle inequality theorem,the difference of any two sides of a triangle must be less than the third side.
Here,the base $AB = 4 \, cm$.
Therefore,for the triangle to be constructible,the difference $|BC - AC|$ must be strictly less than $AB$.
$|BC - AC| < 4 \, cm$.
Checking the options:
$(A)$ $3.5 < 4$ (Possible)
$(B)$ $4.5 > 4$ (Not possible)
$(C)$ $3 < 4$ (Possible)
$(D)$ $2.5 < 4$ (Possible)
Thus,the construction is not possible when the difference is $4.5 \, cm$.

(C) Angles that can be constructed using a ruler and a compass are those that are multiples of $3.75^{\circ}$ or can be derived by bisecting standard angles like $60^{\circ}, 90^{\circ}, 45^{\circ}, 30^{\circ}$,etc.
$37.5^{\circ}$ is half of $75^{\circ}$,which is constructible.
$22.5^{\circ}$ is half of $45^{\circ}$,which is constructible.
$67.5^{\circ}$ is half of $135^{\circ}$,which is constructible.
However,$40^{\circ}$ cannot be constructed using only a ruler and a compass because it is not a multiple of $3.75^{\circ}$ and cannot be obtained by repeated bisection of standard constructible angles.

(D) In the construction of a triangle $ABC$ where the base $BC$ and base angle $\angle B$ are given,the construction is possible only if the difference between the other two sides ($AB$ and $AC$) is strictly less than the base $BC$.
Mathematically,for a triangle to be constructed,we must have $|AB - AC| < BC$.
Given $BC = 6 \, cm$,the condition for the triangle to be impossible is $|AB - AC| \geq 6 \, cm$.
Among the given options,$6.9 \, cm$ is the only value that is greater than or equal to $6 \, cm$.
Therefore,the construction is not possible when the difference is $6.9 \, cm$.

(A) In the construction of a triangle $ABC$ where the base $BC$,the base angle $\angle C$,and the difference between the other two sides $(AB - AC)$ are given,the construction is possible only if the difference between the two sides is strictly less than the length of the base $BC$.
Given:
$BC = 3 \, cm$
Difference $= |AB - AC|$
Condition for construction:
$|AB - AC| < BC$
$|AB - AC| < 3 \, cm$
Comparing this with the given options:
$A) 2.8 < 3$ (Possible)
$B) 3 < 3$ (Not possible)
$C) 3.1 < 3$ (Not possible)
$D) 3.2 < 3$ (Not possible)
Therefore,the only value that satisfies the condition is $2.8 \, cm$.

(A) True.
An angle of $67.5^{\circ}$ can be constructed because it is half of $135^{\circ}$.
Since $135^{\circ} = 90^{\circ} + 45^{\circ}$,we can construct $90^{\circ}$ and $45^{\circ}$ using a compass and ruler.
By bisecting the angle of $135^{\circ}$,we obtain $67.5^{\circ}$ $(67.5^{\circ} = \frac{135^{\circ}}{2})$.
Therefore,the statement is true.

(A) To determine if an angle of $52.5^{\circ}$ can be constructed,we check if it is a multiple of $7.5^{\circ}$,as angles that are multiples of $7.5^{\circ}$ can be constructed using a compass and ruler.
We can express $52.5^{\circ}$ as:
$52.5^{\circ} = \frac{105^{\circ}}{2} = \frac{1}{2} \times (60^{\circ} + 45^{\circ})$.
Since $60^{\circ}$ and $45^{\circ}$ are standard angles that can be constructed,their sum $105^{\circ}$ can be constructed. By bisecting the $105^{\circ}$ angle,we obtain $52.5^{\circ}$.
Alternatively,$52.5^{\circ} = \frac{210^{\circ}}{4} = \frac{180^{\circ} + 30^{\circ}}{4}$. Since $180^{\circ}$ and $30^{\circ}$ are constructible,$210^{\circ}$ is constructible,and bisecting it twice yields $52.5^{\circ}$.
Therefore,the statement is True.

(B) The statement is False.
To construct an angle using a ruler and compass,it must be possible to express the angle as a combination of bisections and additions/subtractions of angles that are constructible (like $60^{\circ}, 90^{\circ}, 45^{\circ}$,etc.).
An angle of $42.5^{\circ}$ is equal to $\frac{85^{\circ}}{2}$.
Since $85^{\circ}$ cannot be constructed using a ruler and compass (as it is not a multiple of $7.5^{\circ}$ or $3.75^{\circ}$ that can be derived from standard constructions),$42.5^{\circ}$ cannot be constructed.

(FALSE) The statement is False.
According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.
In this case,we are given $AB = 5 \, cm$ and $BC + AC = 5 \, cm$.
This implies $BC + AC = AB$,which violates the triangle inequality theorem.
Therefore,a triangle with these dimensions cannot be constructed.

(A) The statement is True.
In a triangle,the construction of a triangle $ABC$ is possible if the difference between two sides is less than the third side.
Here,the given difference is $AC - AB = 4 \, cm$ and the third side is $BC = 6 \, cm$.
Since $4 \, cm < 6 \, cm$,the condition $AC - AB < BC$ is satisfied.
Therefore,a triangle $ABC$ can be constructed with the given measurements.

(B) The given statement is False.
In any triangle $ABC$,the sum of all interior angles must be equal to $180^{\circ}$,i.e.,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given that $\angle B = 105^{\circ}$ and $\angle C = 90^{\circ}$.
Calculating the sum of the two given angles: $\angle B + \angle C = 105^{\circ} + 90^{\circ} = 195^{\circ}$.
Since $195^{\circ} > 180^{\circ}$,the sum of two angles alone exceeds the angle sum property of a triangle. Therefore,such a triangle cannot be constructed.

(TRUE) The given statement is true.
In any triangle $ABC$,the sum of the angles must be $180^{\circ}$. Here,$\angle B + \angle C = 60^{\circ} + 45^{\circ} = 105^{\circ}$. Since $105^{\circ} < 180^{\circ}$,the third angle $\angle A = 180^{\circ} - 105^{\circ} = 75^{\circ}$ can be determined.
According to the construction rules for a triangle given its perimeter and two base angles,a triangle can be constructed if the sum of the two base angles is less than $180^{\circ}$. Since $105^{\circ} < 180^{\circ}$,the construction is possible.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 7.5\, cm$.
$2$. At point $B$,construct an angle $\angle XBC = 45^{\circ}$.
$3$. Cut a line segment $BD = 4\, cm$ from the ray $BX$.
$4$. Join $DC$.
$5$. Construct the perpendicular bisector of $DC$,which intersects $BD$ (extended if necessary) at point $A$.
$6$. Join $AC$. Thus,$\triangle ABC$ is the required triangle.

(N/A) Given: An angle $\angle ABC = 110^{\circ}$.
Required: To draw the bisector of $\angle ABC$.
Steps of construction:
$1.$ With $B$ as centre and a convenient radius,draw an arc to intersect the rays $BA$ and $BC$ at $P$ and $Q$ respectively.
$2.$ With centre $P$ and a radius greater than half of $PQ$,draw an arc.
$3.$ With centre $Q$ and the same radius (as in step $2$),draw another arc to cut the previous arc at $R$.
$4.$ Draw ray $BR$. This ray $BR$ is the required bisector of $\angle ABC$.
Measurement: Each bisected angle,$\angle ABR$ and $\angle RBC$,will be $110^{\circ} / 2 = 55^{\circ}$.

(N/A) Given: $A$ line segment $AB$ of length $4 \,cm$.
Required: To draw perpendiculars to $AB$ through points $A$ and $B$,respectively.
Steps of construction:
$1.$ Draw a line segment $AB = 4 \,cm$.
$2.$ At point $A$,construct a $90^{\circ}$ angle using a compass and ruler to draw a perpendicular line $CD$.
$3.$ At point $B$,construct a $90^{\circ}$ angle using a compass and ruler to draw a perpendicular line $EF$.
$4.$ Since both lines $CD$ and $EF$ are perpendicular to the same line segment $AB$,they make interior angles of $90^{\circ}$ with $AB$.
$5.$ The sum of the interior angles on the same side of the transversal $AB$ is $90^{\circ} + 90^{\circ} = 180^{\circ}$.
$6.$ Therefore,the lines $CD$ and $EF$ are parallel to each other.

(N/A) Steps of construction:
$1.$ Draw a ray $OA$.
$2.$ With the help of a protractor,construct $\angle BOA = 80^{\circ}$.
$3.$ Taking $O$ as the center and any suitable radius,draw an arc to intersect rays $OA$ and $OB$ at points $P$ and $Q$ respectively.
$4.$ Bisect $\angle BOA$. Let ray $OC$ be the bisector of $\angle BOA$,then $\angle COA = \frac{1}{2} \angle BOA = \frac{1}{2} \times 80^{\circ} = 40^{\circ}$.
$5.$ With $Q$ as the center and radius equal to $PQ$,draw an arc to cut the extended arc $PQ$ at $R$. Join $OR$ and produce it to form ray $OD$,then $\angle DOA = 2 \angle BOA = 2 \times 80^{\circ} = 160^{\circ}$.
$6.$ Bisect $\angle DOB$. Let $OE$ be the bisector of $\angle DOB$. Then $\angle EOA = \angle EOB + \angle BOA = \frac{1}{2} \angle DOB + \angle BOA = \frac{1}{2}(80^{\circ}) + 80^{\circ} = 40^{\circ} + 80^{\circ} = 120^{\circ}$.

(N/A) Step $1$: Draw a line segment $AB = 4.8 \, cm$.
Step $2$: Taking $A$ as the center and radius $3.0 \, cm$,draw an arc. Taking $B$ as the center and radius $3.6 \, cm$,draw another arc that intersects the previous arc at point $C$.
Step $3$: Join $CA$ and $CB$ to obtain the required triangle $ABC$.
Step $4$: Measure all internal angles. The smallest angle is $\angle ABC$ (opposite to the shortest side $AC = 3.0 \, cm$).
Step $5$: To bisect $\angle ABC$,take any radius and with center $B$,draw an arc that intersects $AB$ at $P$ and $BC$ at $Q$.
Step $6$: With the same radius and centers $P$ and $Q$,draw two arcs that intersect each other at point $R$.
Step $7$: Join $BR$ and extend it to intersect $AC$ at point $D$. $BD$ is the angle bisector of $\angle ABC$.

(N/A) Given: In $\triangle ABC$,$BC = 5 \, cm$,$AC + AB = 7.5 \, cm$ and $\angle B = 60^{\circ}$.
Required: To construct $\triangle ABC$.
Steps of construction:
$1.$ Draw a ray $BX$ and cut off a line segment $BC = 5 \, cm$ from it.
$2.$ At $B$,construct $\angle XBY = 60^{\circ}$.
$3.$ With $B$ as centre and radius $= 7.5 \, cm$,draw an arc to meet $BY$ at $D$.
$4.$ Join $CD$.
$5.$ Draw the perpendicular bisector of $CD$,intersecting $BD$ at $A$.
$6.$ Join $AC$. Then,$ABC$ is the required triangle.

(N/A) Steps of construction:
$1.$ Draw a line segment $AB = 3 \, cm$.
$2.$ At point $A$,construct a perpendicular line $AY$ such that $\angle YAB = 90^{\circ}$.
$3.$ With $A$ as the centre and radius $= 3 \, cm$,draw an arc to intersect $AY$ at point $D$.
$4.$ With $B$ as the centre and radius $= 3 \, cm$,draw an arc above $B$.
$5.$ With $D$ as the centre and radius $= 3 \, cm$,draw another arc to intersect the previous arc at point $C$.
$6.$ Join $BC$ and $DC$. Thus,$ABCD$ is the required square of side $3 \, cm$.

(N/A) Steps of construction:
$1.$ Draw a line segment $AB = 5\,cm$.
$2.$ Construct an angle of $90^{\circ}$ at point $A$ and point $B$ using a compass or set square.
$3.$ With $A$ as the center and radius $= 3.5\,cm$,draw an arc to intersect the perpendicular line at $D$.
$4.$ With $B$ as the center and radius $= 3.5\,cm$,draw an arc to intersect the perpendicular line at $C$.
$5.$ Join $CD$. Thus,$ABCD$ is the required rectangle with adjacent sides $5\,cm$ and $3.5\,cm$.

(N/A) Steps of construction:
$1.$ Draw a line segment $AB = 3.4 \, cm$.
$2.$ At point $A$,construct an angle $\angle BAM = 45^{\circ}$.
$3.$ At point $B$,construct an angle $\angle TBN = 45^{\circ}$ (where $T$ is on the line extending from $AB$).
$4.$ From ray $AM$,cut off a segment $AD = 3.4 \, cm$.
$5.$ From ray $BN$,cut off a segment $BC = 3.4 \, cm$.
$6.$ Join $CD$. Thus,$ABCD$ is the required rhombus.

(N/A) Steps of construction:
$1$. Draw a line $XY$.
$2$. Take any point $D$ on the line $XY$.
$3$. Construct a perpendicular $PD$ to $XY$ at point $D$.
$4$. Cut a line segment $AD = 6\, cm$ from $PD$.
$5$. At point $A$,construct angles $\angle CAD = 30^{\circ}$ and $\angle BAD = 30^{\circ}$ such that $B$ and $C$ lie on the line $XY$.
$6$. Join $AB$ and $AC$. Thus,$\triangle ABC$ is the required equilateral triangle.
Justification:
In $\triangle ABC$,$AD \perp BC$. Since $\angle BAD = 30^{\circ}$ and $\angle CAD = 30^{\circ}$,then $\angle A = \angle BAD + \angle CAD = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Since $AD$ is the altitude,$\angle ADB = 90^{\circ}$. In $\triangle ABD$,$\angle B = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Similarly,in $\triangle ACD$,$\angle C = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Since all angles are $60^{\circ}$,$\triangle ABC$ is an equilateral triangle with altitude $AD = 6\, cm$.

(N/A) Steps of construction:
$1.$ Draw a line segment $XY = 10.4 \, cm$.
$2.$ Construct $\angle LXY = 45^{\circ}$ and $\angle MYX = 120^{\circ}$ at points $X$ and $Y$ respectively.
$3.$ Draw the angle bisectors of $\angle LXY$ and $\angle MYX$. Let these bisectors intersect at point $A$.
$4.$ Draw the perpendicular bisector of $AX$,which intersects $XY$ at point $B$.
$5.$ Draw the perpendicular bisector of $AY$,which intersects $XY$ at point $C$.
$6.$ Join $AB$ and $AC$. Thus,$\triangle ABC$ is the required triangle.
Justification:
Since $B$ lies on the perpendicular bisector of $AX$,$XB = AB$. Similarly,since $C$ lies on the perpendicular bisector of $AY$,$YC = AC$.
The perimeter of $\triangle ABC = AB + BC + AC = XB + BC + CY = XY = 10.4 \, cm$.
Also,$\angle XAB = \angle AXB$ (as $XB = AB$) and $\angle YAC = \angle AYC$ (as $YC = AC$).
By the exterior angle property,$\angle ABC = \angle XAB + \angle AXB = 2 \angle XAB = \angle LXY = 45^{\circ}$.
Similarly,$\angle ACB = \angle YAC + \angle AYC = 2 \angle YAC = \angle MYX = 120^{\circ}$.

(N/A) Steps of Construction:
$1.$ Draw a ray $QX$ and cut off a line segment $QR = 3 \, cm$.
$2.$ At $Q$,construct $\angle YQR = 45^{\circ}$.
$3.$ From the ray $QY$,cut off a line segment $QS = 2 \, cm$.
$4.$ Join $RS$.
$5.$ Draw the perpendicular bisector of $RS$. Let it intersect $QY$ at point $P$.
$6.$ Join $PR$. Then,$\triangle PQR$ is the required triangle.
Justification:
Since $P$ lies on the perpendicular bisector of $RS$,$PS = PR$.
Now,$QS = QP - PS = QP - PR$.
Since $QS = 2 \, cm$,we have $QP - PR = 2 \, cm$. This justifies the construction.

(N/A) In $\triangle ABC$,let the base $BC = 3.5 \, cm$,the sum of the other side and the hypotenuse be $AB + AC = 5.5 \, cm$,and $\angle ABC = 90^{\circ}$.
Steps of construction:
$1.$ Draw a ray $BX$ and cut off a line segment $BC = 3.5 \, cm$ from it.
$2.$ Construct $\angle XBY = 90^{\circ}$ at point $B$.
$3.$ From ray $BY$,cut off a line segment $BD = 5.5 \, cm$.
$4.$ Join $CD$.
$5.$ Draw the perpendicular bisector of $CD$,which intersects $BD$ at point $A$.
$6.$ Join $AC$. Then,$\triangle ABC$ is the required triangle.

(N/A) Steps of Construction:
$1.$ Draw a line $l$.
$2.$ Mark any point $D$ on the line $l$.
$3.$ At point $D$,draw a perpendicular line $\overline{DX} \perp l$ and mark a point $A$ on $\overline{DX}$ such that $DA = 3.2 \, cm$.
$4.$ At point $A$,construct rays $AB$ and $AC$ such that $\angle DAB = 30^{\circ}$ and $\angle DAC = 30^{\circ}$,meeting line $l$ at points $B$ and $C$ respectively.
$5.$ $\triangle ABC$ is the required equilateral triangle.
Justification:
In $\triangle ABD$,$\angle ADB = 90^{\circ}$ and $\angle DAB = 30^{\circ}$,so $\angle ABD = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Similarly,in $\triangle ACD$,$\angle ADC = 90^{\circ}$ and $\angle DAC = 30^{\circ}$,so $\angle ACD = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Also,$\angle BAC = \angle DAB + \angle DAC = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Since all angles are $60^{\circ}$,$\triangle ABC$ is an equilateral triangle.

(N/A) Steps of construction:
$1.$ Draw a line segment $AC = 6 \,cm$.
$2.$ Construct the perpendicular bisector of $AC$,which intersects $AC$ at point $M$.
$3.$ From point $M$,mark points $B$ and $D$ on the perpendicular bisector such that $MB = MD = 2 \,cm$ (since the total length of the second diagonal is $4 \,cm$,each half is $2 \,cm$).
$4.$ Join $AB$,$BC$,$CD$,and $DA$.
Thus,$ABCD$ is the required rhombus.
Justification:
In quadrilateral $ABCD$,the diagonals $AC$ and $BD$ bisect each other at right angles at point $M$. Since $AC = 6 \,cm$ and $BD = MB + MD = 2 \,cm + 2 \,cm = 4 \,cm$,and the diagonals bisect each other at $90^\circ$,the quadrilateral $ABCD$ is a rhombus.

(N/A) Steps of construction:
$(1)$ Let $AB$ be the given ray with initial point $A$. Extend $AB$ to the left to form a line $MAB$.
$(2)$ Taking $A$ as the center and any radius,draw an arc to intersect the line $MAB$ at points $X$ and $Y$.
$(3)$ Taking $X$ and $Y$ as centers and a radius greater than $\frac{1}{2} XY$,draw two arcs that intersect at point $P$ above the line $MAB$.
$(4)$ Draw ray $AC$ passing through $P$. Thus,$\angle CAB = 90^{\circ}$ is constructed.
$(5)$ Let the intersection of the arc (centered at $A$) and ray $AC$ be point $Z$.
$(6)$ Taking $Y$ and $Z$ as centers and a radius greater than $\frac{1}{2} YZ$,draw two arcs that intersect at point $Q$.
$(7)$ Draw ray $AQ$. Thus,$\angle QAB = 45^{\circ}$ is the required angle.
Justification:
Draw $PX$ and $PY$. In $\Delta PAX$ and $\Delta PAY$:
$AX = AY$ (Radii of the same arc)
$PX = PY$ (Radii of congruent arcs)
$PA = PA$ (Common side)
By $SSS$ congruence rule,$\Delta PAX \cong \Delta PAY$.
Therefore,$\angle PAX = \angle PAY$ $(CPCT)$.
Since $\angle PAX + \angle PAY = 180^{\circ}$ (Linear pair),$\angle PAY = \frac{180^{\circ}}{2} = 90^{\circ}$.
Thus,$\angle CAB = 90^{\circ}$.
Now,consider $\Delta AYQ$ and $\Delta AZQ$:
$AY = AZ$ (Radii of the same arc)
$YQ = ZQ$ (Radii of congruent arcs)
$AQ = AQ$ (Common side)
By $SSS$ congruence rule,$\Delta AYQ \cong \Delta AZQ$.
Therefore,$\angle QAY = \angle QAZ$ $(CPCT)$.
Since $\angle QAY + \angle QAZ = \angle ZAY = \angle CAB = 90^{\circ}$,
$\angle QAY = \frac{90^{\circ}}{2} = 45^{\circ}$.
Thus,$\angle QAB = 45^{\circ}$.

(N/A) Steps of construction:
$(1)$ Draw any ray $AB$. With centre $A$ and any radius,draw an arc to intersect $AB$ at $X$.
$(2)$ With centre $X$ and the same radius [as in step $(1)$],draw an arc to intersect the previous arc at $Y$. Draw ray $AY$. Then,$\angle YAB = 60^{\circ}$.
$(3)$ Draw ray $AT$,the bisector of $\angle YAB$.
Thus,$\angle TAB$ is the required angle of $30^{\circ}$.

(N/A) Steps of construction:
$(1)$ Draw any ray $AB$. Produce $AB$ on the side of $A$ to get line $CAB$.
$(2)$ Taking $A$ as centre and any radius,draw an arc of a circle to intersect line $CAB$ at $X$ and $Y$.
$(3)$ Taking $X$ and $Y$ as centres and radius more than $\frac{1}{2} XY$,draw arcs to intersect each other at $L$ on one side of line $CAB$. Draw ray $AL$. Then,$\angle LAB = 90^{\circ}$.
$(4)$ Draw ray $AM$,the bisector of $\angle LAB$. Then,$\angle MAB = 45^{\circ}$.
$(5)$ Draw ray $AN$,the bisector of $\angle MAB$. Then,$\angle NAB = 22 \frac{1}{2}^{\circ}$.
Thus,$\angle NAB$ is the required angle of $22 \frac{1}{2}^{\circ}$.

(N/A) To construct an angle of $45^{\circ}$ without first constructing $90^{\circ}$:
$1$. Draw a ray $OA$.
$2$. With $O$ as the center and any convenient radius,draw an arc cutting $OA$ at point $P$.
$3$. With $P$ as the center and the same radius,draw an arc cutting the previous arc at point $Q$. This forms a $60^{\circ}$ angle.
$4$. With $Q$ as the center and the same radius,draw another arc further along the circle to point $R$. This forms a $120^{\circ}$ angle.
$5$. Bisect the angle between $60^{\circ}$ $(Q)$ and $120^{\circ}$ $(R)$ to get $90^{\circ}$ (let this point be $S$).
$6$. Since we cannot use $90^{\circ}$,we use an alternative method: Construct a $60^{\circ}$ angle $(Q)$ and a $30^{\circ}$ angle (by bisecting $60^{\circ}$),then subtract or use the bisection of $60^{\circ}$ and $0^{\circ}$ to get $30^{\circ}$,then bisect the $30^{\circ}$ and $60^{\circ}$ range,or simply bisect the $90^{\circ}$ angle if allowed.
$7$. Correction: To construct $45^{\circ}$ without $90^{\circ}$ is geometrically impossible using standard compass and straightedge methods as $45^{\circ}$ is defined as half of $90^{\circ}$. However,one can construct $60^{\circ}$,then $30^{\circ}$ (bisect $60^{\circ}$),then $15^{\circ}$ (bisect $30^{\circ}$),and add $30^{\circ} + 15^{\circ} = 45^{\circ}$.

(N/A) Steps of construction:
$1$. Draw a line segment $PQ = 7.4\,cm$ using a ruler.
$2$. With $P$ as the center and a radius of more than half of $PQ$ (i.e.,$> 3.7\,cm$),draw two arcs,one above and one below the line segment $PQ$.
$3$. With $Q$ as the center and the same radius,draw two arcs that intersect the previous arcs at points $X$ and $Y$.
$4$. Join the points $X$ and $Y$ using a ruler. The line $XY$ is the required perpendicular bisector of $PQ$.

(N/A) An obtuse angle is an angle whose measure is greater than $90^{\circ}$ and less than $180^{\circ}$.
Steps of construction:
$1$. Draw a ray $OA$ using a ruler.
$2$. Place the center of the protractor at point $O$ and align the baseline with ray $OA$. Mark a point $B$ at an angle of $120^{\circ}$ (or any value between $90^{\circ}$ and $180^{\circ}$). Join $OB$.
$3$. With $O$ as the center and any convenient radius,draw an arc that intersects $OA$ at point $P$ and $OB$ at point $Q$.
$4$. With $P$ as the center and a radius more than half of $PQ$,draw an arc in the interior of the angle.
$5$. With $Q$ as the center and the same radius,draw another arc that intersects the previous arc at point $R$.
$6$. Join $OR$. The ray $OR$ is the angle bisector of $\angle AOB$.

(N/A) To construct an angle of $7 \frac{1}{2}^{\circ}$ (which is $7.5^{\circ}$),follow these steps:
$1$. Draw a ray $OA$.
$2$. Construct a $60^{\circ}$ angle using a compass and ruler.
$3$. Bisect the $60^{\circ}$ angle to get $30^{\circ}$.
$4$. Bisect the $30^{\circ}$ angle to get $15^{\circ}$.
$5$. Bisect the $15^{\circ}$ angle to get $7 \frac{1}{2}^{\circ}$.
Since $7 \frac{1}{2}^{\circ} = \frac{15^{\circ}}{2}$,this is the required angle.

(N/A) Steps of construction:
$1$. Draw a ray $OA$.
$2$. With $O$ as the center and any convenient radius,draw an arc cutting $OA$ at point $B$.
$3$. With $B$ as the center and the same radius,draw an arc cutting the previous arc at point $C$. This represents $60^{\circ}$.
$4$. With $C$ as the center and the same radius,draw an arc cutting the first arc at point $D$. This represents $120^{\circ}$.
$5$. With $D$ as the center and the same radius,draw an arc further along the circle to point $E$. This represents $180^{\circ}$.
$6$. Bisect the angle between $120^{\circ}$ (point $D$) and $180^{\circ}$ (point $E$) to get $150^{\circ}$. Let this point be $F$.
$7$. Now,bisect the angle between $120^{\circ}$ (point $D$) and $150^{\circ}$ (point $F$) to get $135^{\circ}$.
$8$. Draw a ray $OG$ passing through this point. The angle $\angle AOG = 135^{\circ}$.

(N/A) To construct an angle of $157.5^{\circ}$,we note that $157.5^{\circ} = 180^{\circ} - 22.5^{\circ}$.
$1$. Draw a ray $OA$.
$2$. Construct an angle of $180^{\circ}$ at point $O$ using a protractor or by extending the line.
$3$. Construct an angle of $45^{\circ}$ at point $O$ by bisecting the $90^{\circ}$ angle.
$4$. Bisect the $45^{\circ}$ angle to obtain $22.5^{\circ}$.
$5$. Subtract $22.5^{\circ}$ from $180^{\circ}$ to get $157.5^{\circ}$.

(N/A) Steps of construction for $75^{\circ}$:
$1$. Draw a ray $OA$ with initial point $O$.
$2$. With $O$ as the center and any radius,draw an arc that intersects $OA$ at a point $P$.
$3$. With $P$ as the center and the same radius,draw an arc that intersects the previous arc at point $Q$. This represents $60^{\circ}$.
$4$. With $Q$ as the center and the same radius,draw another arc that intersects the first arc at point $R$. This represents $120^{\circ}$.
$5$. With $Q$ and $R$ as centers and radius more than half of $QR$,draw two arcs that intersect each other at point $S$. Join $OS$. The angle $\angle SOA = 90^{\circ}$.
$6$. Now,the angle between $60^{\circ}$ (point $Q$) and $90^{\circ}$ (point $S$) needs to be bisected.
$7$. With $Q$ and $S$ as centers and radius more than half of $QS$,draw two arcs that intersect at point $T$.
$8$. Join $OT$. The angle $\angle TOA = 75^{\circ}$.

(D) $1$. Draw a ray $OA$.
$2$. Using a compass,construct an angle of $90^{\circ}$ at point $O$.
$3$. Construct an angle of $135^{\circ}$ at point $O$ (which is the angle bisector of $90^{\circ}$ and $180^{\circ}$).
$4$. Now,bisect the angle between $90^{\circ}$ and $135^{\circ}$.
$5$. The angle formed is $(90^{\circ} + 135^{\circ}) / 2 = 225^{\circ} / 2 = 112.5^{\circ}$.

(N/A) To construct an angle of $37.5^{\circ}$,we follow these steps:
$1$. Draw a ray $OA$.
$2$. Construct an angle of $75^{\circ}$ at point $O$ using a compass and ruler. Let this be $\angle AOB = 75^{\circ}$.
$3$. Bisect $\angle AOB$ to get an angle of $37.5^{\circ}$.
$4$. To bisect $\angle AOB$,place the compass at $O$ and draw an arc intersecting $OA$ and $OB$ at points $P$ and $Q$ respectively.
$5$. With $P$ and $Q$ as centers and radius more than half of $PQ$,draw two arcs intersecting each other at point $R$.
$6$. Join $OR$. The angle $\angle AOR$ is $37.5^{\circ}$.

(N/A) Steps of construction:
$1$. Draw a line segment $AB = 5 \, cm$ using a ruler.
$2$. With $A$ as the center and radius $5 \, cm$,draw an arc above the line segment $AB$.
$3$. With $B$ as the center and radius $5 \, cm$,draw another arc intersecting the previous arc at point $C$.
$4$. Join $AC$ and $BC$.
$5$. Thus,$\Delta ABC$ is the required equilateral triangle with side length $5 \, cm$.

(N/A) $1$. Calculate the side length of the equilateral triangle: Since the perimeter is $21\,cm$,each side length $= 21\,cm / 3 = 7\,cm$.
$2$. Draw a line segment $XY = 7\,cm$ using a ruler.
$3$. Set the compass width to $7\,cm$.
$4$. Place the compass point at $X$ and draw an arc above the line $XY$.
$5$. Place the compass point at $Y$ and draw another arc intersecting the previous arc at point $Z$.
$6$. Join $XZ$ and $YZ$ using a ruler.
$7$. The resulting $\Delta XYZ$ is the required equilateral triangle with each side $7\,cm$ and perimeter $21\,cm$.

(N/A) An equiangular triangle is an equilateral triangle where each angle is $60^{\circ}$.
Steps of construction:
$1$. Draw a line segment $PQ = 6\,cm$ using a ruler.
$2$. At point $P$,construct an angle of $60^{\circ}$ using a compass and ruler.
$3$. At point $Q$,construct an angle of $60^{\circ}$ using a compass and ruler.
$4$. Let the rays from $P$ and $Q$ intersect at point $R$.
$5$. $\Delta PQR$ is the required equiangular triangle.

(N/A) Steps of construction:
$(1)$ Draw any ray $BX$. With centre $B$ and radius $8 \, cm$,draw an arc to intersect $BX$ at $C$.
$(2)$ At $B$,construct $\angle YBC$ with measure $75^{\circ}$.
$(3)$ With centre $B$ and radius $15 \, cm$,draw an arc to intersect $BY$ at $M$.
$(4)$ Draw line segment $MC$. Draw the perpendicular bisector of $MC$ to intersect $BM$ at $A$.
$(5)$ Draw line segment $AC$.
Then,$\Delta ABC$ is the required triangle.

(N/A) Steps of construction:
$(1)$ Draw a ray $QX$ and cut off a line segment $QR = 9\, cm$ from it.
$(2)$ At point $Q$,construct a ray $QY$ such that $\angle YQR = 60^{\circ}$.
$(3)$ Produce the ray $QY$ backwards to form a ray $QZ$. Cut off a line segment $QS = 3\, cm$ from $QZ$.
$(4)$ Join $RS$. Draw the perpendicular bisector of the line segment $RS$.
$(5)$ Let the perpendicular bisector intersect the ray $QY$ at point $P$. Join $PR$.
$(6)$ Thus,$\Delta PQR$ is the required triangle.

(N/A) Steps of construction:
$(1)$ Draw a line segment $AB = 8 \, cm$.
$(2)$ Construct $\angle LAB = \frac{1}{2} \times 30^{\circ} = 15^{\circ}$ at point $A$.
$(3)$ Construct $\angle MBA = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$ at point $B$.
$(4)$ Let the rays $AL$ and $BM$ intersect at point $X$.
$(5)$ Draw the perpendicular bisector of $XA$ to intersect $AB$ at $Y$.
$(6)$ Draw the perpendicular bisector of $XB$ to intersect $AB$ at $Z$.
$(7)$ Join $XY$ and $XZ$. Then $\Delta XYZ$ is the required triangle.

(N/A) Steps of construction:
$(1)$ Draw any ray $BX$ and from that obtain the line segment $BC$ of length $6\,cm$.
$(2)$ Construct ray $BY$ such that $\angle YBC = 90^{\circ}$.
$(3)$ Taking $B$ as the centre and radius $9\,cm$,draw an arc to intersect $BY$ at $M$.
$(4)$ Draw line segment $CM$. Draw the perpendicular bisector of $CM$ to intersect $BM$ at $A$.
$(5)$ Draw line segment $AC$.
Then,$\Delta ABC$ is the required triangle in which $\angle B$ is a right angle,$BC = 6\,cm$ and $AB + AC = 9\,cm$.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 5\,cm$.
$2$. At point $B$,construct an angle $\angle XBC = 60^{\circ}$.
$3$. Cut a line segment $BD = 9\,cm$ from the ray $BX$.
$4$. Join $CD$.
$5$. Construct the perpendicular bisector of $CD$,which intersects $BD$ at point $A$.
$6$. Join $AC$. Thus,$\triangle ABC$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \, cm$.
$2$. At point $B$,construct an angle $\angle XBC = 60^{\circ}$.
$3$. Cut a line segment $BD = AB - AC = 1 \, cm$ from the ray $BX$.
$4$. Join $DC$.
$5$. Draw the perpendicular bisector of $DC$,let it intersect $BX$ at point $A$.
$6$. Join $AC$. Thus,$\Delta ABC$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 7 \, cm$.
$2$. At point $B$,construct an angle $\angle XBC = 45^{\circ}$.
$3$. Cut a line segment $BD = 2 \, cm$ from the ray $BX$ (since $AC - AB > 0$,$D$ is on the ray $BX$).
$4$. Join $DC$.
$5$. Draw the perpendicular bisector of $DC$,let it intersect $BX$ at point $A$.
$6$. Join $AC$. Thus,$\Delta ABC$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $PQ = 10 \, cm$ (where $PQ = XY + YZ + ZX$).
$2$. Construct an angle $\angle P = 45^{\circ}$ at point $P$ and $\angle Q = 60^{\circ}$ at point $Q$.
$3$. Bisect $\angle P$ and $\angle Q$. Let the bisectors intersect at point $X$.
$4$. Draw the perpendicular bisector of $PX$ and $QX$. Let these intersect $PQ$ at $Y$ and $Z$ respectively.
$5$. Join $XY$ and $XZ$. Thus,$\Delta XYZ$ is the required triangle.