Solution

By using the help of a ruler and a compass, that is possible construct the angels, $90^{\circ}, 60^{\circ}$, $45^{\circ}, 22.5^{\circ}, 30^{\circ}$ etc., and its bisector of an angle. So, it is not possible to construct an angle of $40^{\circ}$.

Hence, the correct option is (B).

Solution

Given: $BC=6 ~cm$ and $\angle B=45^{\circ}$

As, we know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle.

So, the difference between other two sides $AB$ and $AC$ should be equal to or greater then $BC$. Hence, the correct option is (A).

Solution

Given, $BC=3 ~cm$ and $\angle C=60^{\circ}$

As, we know that, the construction of a triangle is possible, if sum of two sides is greater than the third side of the triangle.

So, the difference between other two sides $AB$ and $AC$ should not be equal to or greater than $BC$.

Hence, the correct option is (D).

Solution

As, $42.5^{\circ}=\dfrac{1}{4} \times 210^{\circ}$ and an angle of $210^{\circ}=180^{\circ}+30^{\circ}$ cannot be constructed with the help of ruler and compass.

Hence, the given statement is true.

Solution

As, $42.5^{\circ}=\dfrac{1}{2} \times 85^{\circ}$ and an angle of $85^{\circ}$ cannot be constructed with the help of ruler and compass.

Hence, the given statement is false.

Solution

As, a triangle can be constructed, if sum of its two sides is greater than third side.

Here, $BC+AC=AB=5 ~cm$

So, triangle $ABC$ cannot be constructed.

Hence, the given statement is false.

Solution

As, a triangle can be constructed if sum of its two sides is greater than third side.

So, in triangle $ABC, AB+BC>AC$

$BC>AC-AB$

$6>4$, which is true, so triangle $ABC$ with given conditions can be constructed.

Hence, the given statement is true.

Solution

Given: $\angle B=105^{\circ}, \angle C=90^{\circ}$

We know that, sum of angles of a triangle is $180^{\circ}$.

$\angle A+\angle B+\angle C=180^{\circ}$

Here, $\angle B+\angle C=105^{\circ}+90^{\circ}$

$195^{\circ}>180^{\circ}$ which is not true.

Therefore, a triangle $ABC$ with given conditions cannot be constructed.

Hence, the given statement is false.

Solution

We know that, sum of angles of a triangle is $180^{\circ}$.

$\angle A+\angle B+\angle C=180^{\circ}$

Here, $\angle B+\angle C=60^{\circ}+45^{\circ}=105^{\circ}<180^{\circ}$,

Therefore, a triangle $ABC$ with given conditions can be constructed.

Hence, the given statement is true.

Solution

Given: An angle $AXB=110^{\circ}$.

To construct: the bisector of angle AXB as follows:

Steps of construction:

1. With $X$ as center and any radius draw an arc to intersect the rays $XA$ and $XB$, say at $E$ and $D$, respectively. 2. With $D$ and $E$ as center’s and with the radius more than $\dfrac{1}{2} DE$, draw arcs to intersect each other, say at $F$. 3. Draw the ray XF.

So, ray XF is the required bisector of the angle BXA. On measuring each angle, get:

$\angle BXC=\angle AXC=55^{\circ} .[As, \angle BXC=\angle AXC=\dfrac{1}{2} \angle BXA=\dfrac{1}{2} \times 110^{\circ}=55^{\circ}]$

Solution

Given: A line segment $AB$ of length $4 ~cm$.

To construct: To draw a line perpendicular to $AB$ through $A$ and $B$, respectively.

Use the following steps of construction.

1. Draw $AB=4 ~cm$. 2. With 4 as centre and radius more than $\dfrac{1}{2} AB$, draw an arc that is intersect $AB$ at $E$. 3. With $E$ as centre and with same radius as above draw an arc which intersect previous arc at $F$. 4. Again, taking $F$ as centre and with same radius as above draw an arc which intersect previous arc (obtained in step ii) at $G$. 5. With $G$ and $F$ are centres, draw arcs which intersect each other at $H$. 6. Join AH. So, AX is perpendicular to AB at A. Similarly, draw BY $\perp AB$ at B.

Now, we know that if two lines are parallel, then the angle between them will be $0^{\circ}$ or $180^{\circ}$. So,

$\angle XAB=90^{\circ}[XA \perp AB]$

and $\angle YBA=90^{\circ}[YB \perp AB]$

$\angle XAB+\angle YBA=90^{\circ}+90^{\circ}=180^{\circ}$

Hence, the lines XA and YS are parallel. [It sum of interior angle on same side of transversal is $180^{\circ}$, then the two lines are parallel]

Solution

Steps of construction:

Given: Draw an angle of $80^{\circ}$ say $\angle QOA=80^{\circ}$ with the help of protractor.

Use the following steps to construct angles of:

1. With $O$ as centre and any radius draw an arc which intersect $OA$ at $E$ and $OO$ at $F$. 2. With $E$ and $F$ as centres and radius more than $\dfrac{1}{2} EF$ draw arcs which intersect each other at $P$. 3. Join OP Since, $\angle POA=40^{\circ}[40^{\circ}=\dfrac{1}{2} \times 80^{\circ}]$ 4. With $F$ as centre and radius equal to $EF$ draw an arc which intersect previous arc obtained in step 2 at $S$. 5. Join OS. Thus, $\angle SOA=160^{\circ}[160^{\circ}=2 \times 80^{\circ}]$ 6. With $S$ and $F$ as centre and radius more than $\dfrac{1}{2} SF$ draw arcs which intersect each other at $R$. 7. Join OR. Thus, $\angle ROA=\angle ROQ=40^{\circ}+80^{\circ}=120^{\circ}$.

Solution

To construct: a triangle $ABC$ in which $AB=3.6 ~cm, AC=3.0 ~cm$ and $BC=4.8 ~cm$, use the following steps.

1. Draw a line $BC=4.8 ~cm$. 2. From B, point $A$ is at a distance of $3.6 ~cm$. So, having $B$ as centre, draw an arc of radius $3.6 ~cm$. 3. From $C$, point $A$ is at a distance of $3 ~cm$. So, having $C$ as centre, draw an arc of radius $3 ~cm$ which intersect previous arc at $A$. 4. Join $AB$ and $AC$. Therefore, $ABC$ is the required triangle.

Where, angle B is smallest, as AC is the smallest side.

To direct angle B, we use the following steps.

1. With $B$ as centre, draw an arc intersecting $AB$ and $BC$ at $D$ and $E$, respectively. 2. With $D$ and $E$ as centres and draw arcs intersecting at $P$. 3. Joining BP, we obtain angle bisector of $\angle B$. 4. Here, $\angle ABC=39^{\circ}$

Thus, $\angle ABD=\angle DBC=\dfrac{1}{2} \times 139^{\circ}=19.5^{\circ}$

Solution

Given, in $\triangle ABC, BC=5 ~cm, \angle B=60^{\circ}$ and $AC+AB=7.5 ~cm$.

To construct: the triangle $ABC$ use the following steps.

1. Draw the base $BC=5 ~cm$. 2. At the point $B$ make an $\angle XBC=60^{\circ}$. 3. Cut a line segment $BD$ equal to $AB+AC=7.5 ~cm$ from the ray $BX$. 4. Join DC. 5. Make an $\angle DCY=\angle BDC$. 6. Let $CY$ intersect $BX$ at $A$.

Therefore, $ABC$ is the required triangle.

Solution

As we know that, each angle of a square is right angle that is $90^{\circ}$.

To construct: a square of side $3 ~cm$, use the following steps.

1. Take $AB=3 ~cm$. 2. Now, draw an angle of $90^{\circ}$ at points $A$ and $B$ and plot the parallel lines $AX$ and $BY$ at these points. 3. Cut $AD$ and $SC$ of length $3 ~cm$ from $AX$ and $BY$, respectively. 4. Draw $90^{\circ}$ at any one of the point $C$ or $D$ and join both points by $CD$ of length $3 ~cm$. Therefore, $ABCD$ is the required square of side, $3 ~cm$.

Solution

As, we know that, each angle of a rectangle is right angle that is $90^{\circ}$ and its opposite sides are equal and parallel.

To construct: a rectangle whose adjacent sides are of lengths $5 ~cm$ and $3.5 ~cm$, use the 1 following steps

1. Take $BC=5 ~cm$. 2. Draw $90^{\circ}$ at points $B$ and $C$ of the line segment $BC$ and plot the parallel lines $BX$ and $CY$ at these points. 3. Cut $A B$ and $C D$ of length $3.5 ~cm$ from $B X$ and $C Y$, respectively. 4. Draw an angle $90^{\circ}$ at one of the point $A$ or $D$ and join both points by a line segment $AD$ of length $5 ~cm$.

Therefore, $ABCD$ is the required rectangle with adjacent sides of length $5 ~cm$ and 3.5 $~cm$.

As, we know that, in rhombus all sides are equal.

To construct a rhombus whose side is of length $3.4 ~cm$ and one of its angle is $45^{\circ}$.

1. Taking $AB=3.4 ~cm$. 2. AT point $A$ and $B$, construct $\angle B A M=45^{\circ}$ and $\angle T B P=45^{\circ}$, respectively. 3. From $AM$ cut off $AD=3.4 ~cm$ and from $BP$ cut off $BC=3.4 ~cm$. 4. Join $AD, DC$ and $BC . ABCD$ is the required rhombus.

Solution

Let $ABC$ be a triangle.

Given: Perimeter $=10.4 ~cm$ that is, $AB+BC+CA=10.4 ~cm$ and two angles are $45^{\circ}$ and $120^{\circ}$.

So, $\angle B=45^{\circ}$ and $\angle C=120^{\circ}$

Now, to construct the triangle $ABC$ use the following steps.

1. Draw $XY$ and equal to perimeter that is, $AB+BC+CA=10.4 ~cm$.

2. Draw $\angle LXY=\angle B=45^{\circ}$ and $\angle MYX=\angle C=120^{\circ}$.

3. Bisect $\angle LXY$ and $\angle MYX$ and let these bisectors intersect at a point $A$.

4. Draw perpendicular bisectors $PQ$ and $RS$ of $AX$ and $AY$, respectively.

5. Let $PQ$ intersect $XY$ at $B$ and $RS$ intersect $XY$ at $C$. Join $AB$ and $AC$. Therefore, $ABC$ is the required triangle.

Justification:

$B$ lies on the perpendicular bisector $PQ$ of $AX$.

Thus, $AB+BC+CA=XB+BC+CY=XY$

Again, $\angle BAX=\angle AXB[As, AB=XB] \ldots(I)$

Also, $\angle ABC=\angle BAX+\angle AXB \quad[\angle ABC$ is an exterior angle of $\triangle AXB]$

$=\angle AXB+\angle AXB[$ From equation $(I)]$ $=2 \angle AXB=\angle LXY[AX$ is a bisector of $\angle LXB]$

Also, $\angle CAY=\angle AYC[$ As, $AC=CY]$

$\angle ACB=\angle CAY+\angle AYC[\angle ACB$ is an exterior angle of triangle $AYC]$

$=\angle CAY+\angle CAY$

$=2 \angle CAY=\angle MYX[AY$ is a bisector of $\angle MYX]$

Therefore, our construction is justified.

Solution

Given: in triangle $PQR, QR=3 ~cm, \angle PQR=45^{\circ}$ and $QP-PR=2 ~cm$

As, $C$ lies on the perpendicular bisector RS of AY.

To construct: a triangle $PQR$.

Use the following steps of construction.

1. Draw $QR=3 ~cm$.

2. Make an angle $XQR=45^{\circ}$ at point $Q$ of base $QR$.

3. Cut the line segment $QS=QP-PR=2 ~cm$ from the ray $QX$.

4. Join SR and draw the perpendicular bisector of SR say AB.

5. Let bisector $AB$ intersect $QX$ at $P$. Join $PR$ Thus, $\triangle PQR$ is the required triangle.

Justification:

Base $QR$ and $\angle PQR$ are drawn as given.

As, the point $P$ lies on the perpendicular bisector of SR.

$PS=PR$

Now, $QS=PQ-PS$

$ =PQ-PR $

Therefore, our construction is justified.

Solution

In the right triangle $ABC$, given $BC=3.5 ~cm, \angle B=90^{\circ}$ and sum of other side and hypotenuse be, $AB+AC=5.5 ~cm$.

To construct a triangle $ABC$ use the following steps

1. Draw $BC=3.5 ~cm$

2. Make an angle $XBC=90^{\circ}$ at the point $B$ of base $BC$.

3. Cut the line segment $BD$ equal to $AB+AC$ i.e., $5.5 ~cm$ from the ray $XB$.

4. Join $DC$ and make an $\angle DCY$ equal to $\angle BDC$.

5. Let $Y$ intersect $BX$ at $A$.

Therefore, $ABC$ is the required triangle.

Justification:

Base $BC$ and $\angle B$ are drawn as given.

In $\triangle ACD, \angle ACD=\angle ADC$ [by construction]

$AD=AC \ldots$..(I) [sides opposite to equal angles are equal]

Now, $AB=BD-AD=BD-AC[$ From equation $(I)]$

$BD=AB+AC$

Hence, our construction is justified.

Solution

As, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of $60^{\circ}$.

Given: altitude of an equilateral triangle say $ABC$ is $3.2 ~cm$.

To construct the triangle $ABC$ use the following steps.

1. Draw a line PQ.

2. Take a point $D$ on $PQ$ and draw $DE \perp PQ$.

3. Cut $AD$ of length $3.2 ~cm$ from $DE$.

4. Make angles equal to $30^{\circ}$ at $A$ on both sides of $AD$ say $\angle CAD$ and $\angle BAD$, where $B$ and $C$ lie on $PQ$.

5. Cut $DC$ from $PQ$ such that $DC=BD$

Join $AC$

Hence, $ABC$ is the required triangle.

Justification:

Here, $\angle A=\angle BAD+\angle CAD$

$ =30^{\circ}+30^{\circ} $

$ =60^{\circ} \text{. } $

Also, $AD \perp SC$

So, $\angle ADS=90^{\circ}$.

In triangle $ABD, \angle BAD+\angle DBA=180^{\circ}$ [angle sum property]

$30^{\circ}+90^{\circ}+\angle DBA=180^{\circ}[\angle BAD=30^{\circ}.$, by construction $]$

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

Also, $\angle DCA=60^{\circ}$

Therefore, $\angle A=\angle B=\angle C=60^{\circ}$

Hence, $ABC$ is an equilateral triangle.

Solution

As, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are $4 ~cm$ and $6 ~cm$. Use the following steps.

1. Draw $AC=4 ~cm$

2. With $A$ and $C$ as centres and radius more than $\dfrac{1}{2} AC$ draw arcs on both sides of $AC$ to intersect each other.

3. Cut both arcs intersect each other at $P$ and $Q$, then join $P Q$.

4. Let $PQ$ intersect $AC$ at the point $O$. So, $PQ$ is perpendicular bisector of $AC$.

5. Cut off $3 ~cm$ lengths from $OP$ and $OQ$, then we get points $B$ and $D$.

6. Join $AB, BC, CD$, and $DA$.

Hence, $ABCD$ is the required rhombus.

Justification:

$D$ and $B$ lie on perpendicular bisector of $AC$.

$DA=DC$ and $BA=BC \ldots$ (I)

[since, every point on perpendicular bisector of line segment is equidistant from end points of line segment]

Now, $\angle DOC=90^{\circ}$

Also, $OD=OB=3 ~cm$

Thus, $AC$ is perpendicular bisector or $BD$.

$CD=CB \ldots$. (II)

$AB=BC=CD=DA$

Now, from equation (I) and (II):

$ABCD$ is a rhombus.