Chapter 11 Constructions (Class 9 - Maths NCERT Exemplar Solutions)

Given: Options for angles: $35^\circ$, $40^\circ$, $37.5^\circ$, and $47.5^\circ$.

To Find: Which angle is possible to construct using a ruler and a compass.

Solution:

In geometry, using a ruler and a compass, we can construct angles that are multiples of $15^\circ$ (such as $15^\circ, 30^\circ, 45^\circ, 60^\circ, 75^\circ, 90^\circ,$ etc.). Furthermore, any angle that is obtained by the repeated bisection of these angles can also be constructed.

Let us analyze the given options:

Option (A): $35^\circ$ is not a multiple of $15^\circ$ and cannot be reached by bisecting $15^\circ, 30^\circ,$ etc.

Option (B): $40^\circ$ is not a multiple of $15^\circ$.

Option (C): $37.5^\circ$

Since $75^\circ$ can be constructed (by bisecting the angle between $60^\circ$ and $90^\circ$), its half, $37.5^\circ$, can also be constructed.

Option (D): $47.5^\circ$ cannot be obtained through these methods.

Thus, the correct option is (C).

Given: Side $AB = 4 \textsf{ cm}$ and $\angle A = 60^\circ$.

To Find: The condition under which the construction of $\triangle ABC$ is not possible regarding the difference $|BC - AC|$.

Solution:

According to the triangle inequality theorem, for a triangle to be possible, the difference between the lengths of any two sides must always be strictly less than the length of the third side.

In $\triangle ABC$, the third side is given as $AB = 4 \textsf{ cm}$. Therefore, for the triangle to exist, the following condition must be satisfied:

Substituting the value of $AB$:

Now let us check the given options:

(A) $3.5 \textsf{ cm} < 4 \textsf{ cm}$ (Possible)

(B) $4.5 \textsf{ cm} > 4 \textsf{ cm}$ (Not Possible)

(C) $3 \textsf{ cm} < 4 \textsf{ cm}$ (Possible)

(D) $2.5 \textsf{ cm} < 4 \textsf{ cm}$ (Possible)

Since $4.5 \textsf{ cm}$ is greater than $4 \textsf{ cm}$, the construction is not possible for option (B).

The correct option is (B).

To Find: The angle that cannot be constructed using a ruler and a compass.

Solution:

Angles that can be constructed with a ruler and compass are those that are multiples of $15^\circ$ or their bisectors. Let's evaluate the options:

(A) $37.5^\circ$: As shown previously, $37.5^\circ = \frac{75^\circ}{2}$. Since $75^\circ$ is constructible, $37.5^\circ$ is possible.

(B) $40^\circ$: $40^\circ$ is not a multiple of $15^\circ$, nor can it be obtained by bisecting $60^\circ, 90^\circ, 120^\circ,$ etc. Therefore, it is not possible.

(C) $22.5^\circ$: This is obtained by bisecting $45^\circ$.

Since $45^\circ$ is constructible, $22.5^\circ$ is possible.

(D) $67.5^\circ$: This is the bisector of $135^\circ$.

Since $135^\circ$ (between $120^\circ$ and $150^\circ$) is constructible, $67.5^\circ$ is possible.

The only angle that cannot be constructed is $40^\circ$.

The correct option is (B).

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

To Find: The value of $|AB - AC|$ for which construction is not possible.

Solution:

A triangle $ABC$ can be constructed only if the difference between the lengths of any two sides is strictly less than the length of the third side.

Here, the third side is $BC = 6 \textsf{ cm}$. Thus, the condition for existence is:

Substituting $BC = 6 \textsf{ cm}$:

Let's check the options:

(A) $6.9 \textsf{ cm}$: Here $6.9 > 6$. This violates the triangle inequality.

(B) $5.2 \textsf{ cm}$: Here $5.2 < 6$. (Possible)

(C) $5.0 \textsf{ cm}$: Here $5.0 < 6$. (Possible)

(D) $4.0 \textsf{ cm}$: Here $4.0 < 6$. (Possible)

Therefore, construction is not possible when the difference is $6.9 \textsf{ cm}$.

The correct option is (A).

Given: Side $BC = 3 \textsf{ cm}$ and $\angle C = 60^\circ$.

To Find: The value of $|AB - AC|$ for which construction is possible.

Solution:

For $\triangle ABC$ to be constructible, the difference of the two sides $AB$ and $AC$ must be less than the third side $BC$.

Substituting the given value of $BC$:

Let's evaluate the options against the condition $|AB - AC| < 3$:

(A) $3.2 \textsf{ cm}$: $3.2 \not< 3$ (Not possible)

(B) $3.1 \textsf{ cm}$: $3.1 \not< 3$ (Not possible)

(C) $3 \textsf{ cm}$: $3$ is not strictly less than $3$ (Not possible)

(D) $2.8 \textsf{ cm}$: $2.8 < 3$ (Possible)

Thus, the construction is possible only when the difference is $2.8 \textsf{ cm}$.

The correct option is (D).

Statement: An angle of $67.5^\circ$ can be constructed.

Result: True

Reason:

An angle can be constructed using a ruler and compass if it is a multiple of $15^\circ$ or can be obtained by repeated bisection of such angles.

We know that $135^\circ$ can be constructed as it is the bisector of the angle between $90^\circ$ and $180^\circ$ (or $120^\circ$ and $150^\circ$).

Since $135^\circ$ is constructible, its half, $67.5^\circ$, can also be constructed by bisecting the angle of $135^\circ$.

Statement: An angle of $52.5^\circ$ can be constructed.

Result: True

Reason:

An angle is constructible if it can be expressed in the form of $\frac{15^\circ \times n}{2^k}$.

Consider the angle $105^\circ$. It can be constructed by bisecting the angle between $90^\circ$ and $120^\circ$.

Since $105^\circ$ is constructible, its bisector will give $52.5^\circ$. Therefore, it is possible to construct this angle.

Statement: An angle of $42.5^\circ$ can be constructed.

Result: False

Reason:

For an angle to be constructible using a ruler and compass, it must be a multiple of $7.5^\circ$ (which is $15^\circ / 2$) or further bisections like $3.75^\circ$.

Let us check if $42.5$ is a multiple of $7.5$:

$\frac{42.5}{7.5} = \frac{425}{75} = \frac{17}{3} = 5.66...$

As the result is not a whole number or a power-of-two division of constructible angles, $42.5^\circ$ cannot be constructed.

Since $85^\circ$ is not a multiple of $15^\circ$, $42.5^\circ$ is not constructible.

Given: In $\triangle ABC$, side $AB = 5 \textsf{ cm}$, $\angle A = 45^\circ$, and sum of other two sides $BC + AC = 5 \textsf{ cm}$.

Result: False

Reason:

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.

Substituting the given values:

Since the sum of sides $BC$ and $AC$ is equal to the third side $AB$ and not greater than it, the triangle cannot be formed. The points $A, C,$ and $B$ would be collinear.

Given: In $\triangle ABC$, $BC = 6 \textsf{ cm}$, $\angle C = 30^\circ$, and the difference of sides $AC - AB = 4 \textsf{ cm}$.

Result: True

Reason:

For a triangle to be constructible when the difference of two sides is given, the difference must be strictly less than the third side.

Given $AC - AB = 4 \textsf{ cm}$ and $BC = 6 \textsf{ cm}$.

Since the condition for the triangle inequality is satisfied, the construction of $\triangle ABC$ is possible.

Given: $\angle B = 105^\circ$ and $\angle C = 90^\circ$.

Result: False

Reason:

According to the Angle Sum Property of a triangle, the sum of all internal angles of a triangle must be exactly $180^\circ$.

Summing the given angles:

Since the sum of just two angles ($195^\circ$) is already greater than $180^\circ$, it is impossible for such a triangle to exist.

Given: $\angle B = 60^\circ$, $\angle C = 45^\circ$, and perimeter $AB + BC + AC = 12 \textsf{ cm}$.

Result: True

Reason:

A triangle can be constructed if the sum of the given angles is less than $180^\circ$.

Since $105^\circ < 180^\circ$, the third angle $\angle A$ can exist:

Also, the perimeter is given as a positive value ($12 \textsf{ cm}$). Under these conditions, a unique triangle can be constructed using the standard construction method for a triangle given its perimeter and base angles.

Given: In $\triangle ABC$, base $BC = 7.5 \textsf{ cm}$, $\angle B = 45^\circ$ and the difference of sides $AB - AC = 4 \textsf{ cm}$.

Construction Required: To construct $\triangle ABC$ with the given parameters.

Steps of Construction:

1. Draw a line segment $BC = 7.5 \textsf{ cm}$.

2. At point $B$, construct an angle $\angle CBX = 45^\circ$ using a ruler and compass.

3. From the ray $BX$, cut off a line segment $BD = 4 \textsf{ cm}$ (since $AB > AC$).

4. Join the points $C$ and $D$.

5. Draw the perpendicular bisector of the line segment $CD$.

6. Let this perpendicular bisector intersect the ray $BX$ at point $A$.

7. Join $AC$.

8. $\triangle ABC$ is the required triangle.

Proof:

Point $A$ lies on the perpendicular bisector of $CD$.

Now, from the figure:

Given $BD = 4 \textsf{ cm}$, therefore $AB - AC = 4 \textsf{ cm}$.

Given: An angle of $110^\circ$.

Construction Required: To draw and bisect an angle of $110^\circ$.

Steps of Construction:

1. Draw a ray $OA$.

2. Place the protractor on $OA$ and mark a point $B$ such that $\angle AOB = 110^\circ$. Join $OB$.

3. With $O$ as center and any convenient radius, draw an arc intersecting $OA$ at $P$ and $OB$ at $Q$.

4. With $P$ as center and radius more than half of $PQ$, draw an arc.

5. With $Q$ as center and the same radius, draw another arc intersecting the previous arc at point $R$.

6. Join $OR$ and extend it to form a ray. $OR$ is the angle bisector of $\angle AOB$.

Measurement:

By measuring with a protractor, we find:

Since $\frac{110^\circ}{2} = 55^\circ$, the bisection is accurate.

Given: Line segment $AB = 4 \textsf{ cm}$.

Construction Required: To draw perpendiculars at $A$ and $B$ and check if they are parallel.

Steps of Construction:

1. Draw a line segment $AB = 4 \textsf{ cm}$.

2. At point $A$, use a compass to construct a $90^\circ$ angle. Let the perpendicular line be $L_1$.

3. At point $B$, use a compass to construct a $90^\circ$ angle. Let the perpendicular line be $L_2$.

Reasoning:

Let the angles formed at $A$ and $B$ with segment $AB$ be $\angle 1$ and $\angle 2$.

Sum of interior angles on the same side of the transversal $AB$:

Since the sum of consecutive interior angles is $180^\circ$, the lines $L_1$ and $L_2$ must be parallel.

Conclusion: Yes, the lines are parallel.

Given: An initial angle $\angle AOB = 80^\circ$ drawn using a protractor.

(i) Construction of $40^\circ$ angle

To Construct: An angle of $40^\circ$ by bisecting the given $80^\circ$ angle.

Steps of Construction:

1. Draw $\angle AOB = 80^\circ$ using a protractor where $OA$ is the base ray.

2. With $O$ as center and any convenient radius, draw an arc intersecting $OA$ at $P$ and $OB$ at $Q$.

3. With $P$ as center and radius more than half of $PQ$, draw an arc in the interior of $\angle AOB$.

4. With $Q$ as center and the same radius, draw another arc intersecting the previous arc at point $C$.

5. Join $OC$. Ray $OC$ is the angle bisector.

(ii) Construction of $160^\circ$ angle

To Construct: An angle of $160^\circ$ using the given $80^\circ$ angle.

Steps of Construction:

1. Draw the base ray $OA$ and the given angle $\angle AOB = 80^\circ$.

2. With $O$ as center and radius $OP$ (where $P$ is on $OA$), draw a large circular arc $PQ$ such that $Q$ is on $OB$.

3. Measure the distance of arc segment $PQ$ using a compass.

4. With $Q$ as center and the same radius $PQ$, draw an arc intersecting the extension of the previous large arc at point $D$.

5. Join $OD$. The angle $\angle AOD$ is the required angle.

(iii) Construction of $120^\circ$ angle

To Construct: An angle of $120^\circ$.

Steps of Construction:

1. Use the construction from part (ii) where we have $\angle AOB = 80^\circ$ and $\angle AOD = 160^\circ$.

2. The angle between ray $OB$ and ray $OD$ is $80^\circ$. To get $120^\circ$, we need to add $40^\circ$ to the initial $80^\circ$.

3. Bisect the angle $\angle BOD$ (which is $80^\circ$). Let the bisector be ray $OE$.

4. Now, consider the angle $\angle AOE$:

5. Thus, $\angle AOE$ is the required $120^\circ$ angle.

Given: Sides of $\triangle ABC$ are $a = 3.6 \textsf{ cm}$, $b = 3.0 \textsf{ cm}$, and $c = 4.8 \textsf{ cm}$.

To Find: The smallest angle and its bisected parts.

Solution:

In any triangle, the smallest angle is always opposite to the shortest side. Here, the shortest side is $3.0 \textsf{ cm}$. Let this side be $AC$. Thus, the smallest angle is $\angle B$.

Steps of Construction:

1. Draw the longest side $BC = 4.8 \textsf{ cm}$.

2. With $B$ as center and radius $3.6 \textsf{ cm}$, draw an arc.

3. With $C$ as center and radius $3.0 \textsf{ cm}$, draw another arc intersecting the previous arc at $A$.

4. Join $AB$ and $AC$.

5. To bisect the smallest angle ($\angle C$ in this orientation where $AB$ is the side opposite to it), draw arcs from $C$ intersecting $AC$ and $BC$.

6. Bisect the angle using the standard bisection method.

Measurement:

Upon measuring the smallest angle with a protractor, let it be $\theta$. After bisection, each part will measure $\frac{\theta}{2}$.

Given: In $\triangle ABC$, base $BC = 5 \textsf{ cm}$, $\angle B = 60^\circ$ and the sum of the other two sides $AB + AC = 7.5 \textsf{ cm}$.

To Construct: Triangle $ABC$.

Steps of Construction:

1. Draw a line segment $BC = 5 \textsf{ cm}$.

2. At point $B$, construct an angle $\angle CBX = 60^\circ$.

3. From the ray $BX$, cut off a line segment $BD = 7.5 \textsf{ cm}$ (which is equal to $AB + AC$).

4. Join $CD$.

5. Draw the perpendicular bisector of $CD$. Let it intersect $BD$ at a point $A$.

6. Join $AC$.

7. $\triangle ABC$ is the required triangle.

Proof:

Since point $A$ lies on the perpendicular bisector of $CD$, the distance of $A$ from $C$ and $D$ must be equal.

From the construction, we have:

Substituting the value from equation (i) into the above:

Hence, $\triangle ABC$ is the required triangle.

Given: Side of the square $= 3 \textsf{ cm}$.

To Construct: A square $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 3 \textsf{ cm}$.

2. At point $A$, construct an angle of $90^\circ$ using a compass and ruler. Extend this as ray $AX$.

3. At point $B$, construct an angle of $90^\circ$. Extend this as ray $BY$.

4. From ray $AX$, cut an arc of $3 \textsf{ cm}$ and mark it as point $D$.

5. From ray $BY$, cut an arc of $3 \textsf{ cm}$ and mark it as point $C$.

6. Join $CD$.

7. $ABCD$ is the required square of side $3 \textsf{ cm}$.

Proof:

In the quadrilateral $ABCD$:

Since all sides are equal and angles are $90^\circ$, $ABCD$ is a square.

Given: Adjacent sides of a rectangle are $5 \textsf{ cm}$ and $3.5 \textsf{ cm}$.

To Construct: A rectangle $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 5 \textsf{ cm}$.

2. At point $A$, construct an angle of $90^\circ$ and cut an arc of $AD = 3.5 \textsf{ cm}$ on the perpendicular line.

3. At point $B$, construct an angle of $90^\circ$ and cut an arc of $BC = 3.5 \textsf{ cm}$ on the perpendicular line.

4. Join $CD$.

5. $ABCD$ is the required rectangle.

Alternate Solution (Using Arcs):

1. Draw $AB = 5 \textsf{ cm}$ and construct a perpendicular at $B$.

2. Cut $BC = 3.5 \textsf{ cm}$ on the perpendicular.

3. With $C$ as center and radius $5 \textsf{ cm}$, draw an arc.

4. With $A$ as center and radius $3.5 \textsf{ cm}$, draw another arc intersecting the previous arc at $D$.

5. Join $AD$ and $CD$.

Given: Side of rhombus $= 3.4 \textsf{ cm}$ and one angle $= 45^\circ$.

To Construct: A rhombus $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 3.4 \textsf{ cm}$.

2. At point $A$, construct an angle $\angle BAX = 45^\circ$ using a ruler and compass (bisect $90^\circ$).

3. From ray $AX$, cut a segment $AD = 3.4 \textsf{ cm}$.

4. With $D$ as center and radius $3.4 \textsf{ cm}$, draw an arc.

5. With $B$ as center and radius $3.4 \textsf{ cm}$, draw another arc intersecting the previous arc at point $C$.

6. Join $BC$ and $CD$.

7. $ABCD$ is the required rhombus.

Proof:

In quadrilateral $ABCD$:

Since all sides are equal, $ABCD$ is a rhombus.

Given: Altitude of an equilateral triangle is $6 \textsf{ cm}$.

Construction Required: To construct an equilateral triangle $ABC$ with altitude $AD = 6 \textsf{ cm}$.

Steps of Construction:

1. Draw a line $PQ$ and take a point $D$ on it.

2. At point $D$, construct a ray $DX$ perpendicular to $PQ$.

3. From $DX$, cut off a line segment $DA = 6 \textsf{ cm}$.

4. At point $A$, construct $\angle DAC = 30^\circ$ and $\angle DAB = 30^\circ$ on either side of $AD$, where $B$ and $C$ lie on $PQ$.

5. $\triangle ABC$ is the required equilateral triangle.

Justification:

In $\triangle ABD$ and $\triangle ACD$:

Thus, $\triangle ABD \cong \triangle ACD$ by ASA congruence criterion.

Now, in $\triangle ABD$:

Similarly, $\angle C = 60^\circ$.

And, $\angle BAC = \angle DAB + \angle DAC = 30^\circ + 30^\circ = 60^\circ$.

Since all angles are $60^\circ$, $\triangle ABC$ is an equilateral triangle with altitude $6 \textsf{ cm}$.

Given: Perimeter $(AB+BC+CA) = 10.4 \textsf{ cm}$, $\angle B = 45^\circ$, and $\angle C = 120^\circ$.

Construction Required: To construct $\triangle ABC$.

Steps of Construction:

1. Draw a line segment $XY = 10.4 \textsf{ cm}$.

2. At $X$, construct $\angle YXL = \frac{45^\circ}{2} = 22.5^\circ$.

3. At $Y$, construct $\angle XYM = \frac{120^\circ}{2} = 60^\circ$.

4. Let the rays $XL$ and $YM$ intersect at point $A$.

5. Draw perpendicular bisectors of $AX$ and $AY$.

6. Let the bisector of $AX$ intersect $XY$ at $B$ and the bisector of $AY$ intersect $XY$ at $C$.

7. Join $AB$ and $AC$. $\triangle ABC$ is the required triangle.

Justification:

Since $B$ lies on the perpendicular bisector of $AX$:

Similarly, since $C$ lies on the perpendicular bisector of $AY$:

Perimeter $= AB + BC + AC = XB + BC + CY = XY = 10.4 \textsf{ cm}$.

In $\triangle ABX$, $XB = AB$, so $\angle BAX = \angle BXA = 22.5^\circ$.

Exterior $\angle ABC = \angle BAX + \angle BXA = 22.5^\circ + 22.5^\circ = 45^\circ$.

Similarly, Exterior $\angle ACB = \angle CAY + \angle CYA = 60^\circ + 60^\circ = 120^\circ$.

Given: Base $QR = 3 \textsf{ cm}$, $\angle Q = 45^\circ$, and $PQ - PR = 2 \textsf{ cm}$.

Construction Required: To construct $\triangle PQR$.

Steps of Construction:

1. Draw $QR = 3 \textsf{ cm}$.

2. At $Q$, construct $\angle RQX = 45^\circ$.

3. From ray $QX$, cut off $QD = 2 \textsf{ cm}$.

4. Join $RD$.

5. Draw perpendicular bisector of $RD$ to intersect ray $QX$ at $P$.

6. Join $PR$. $\triangle PQR$ is the required triangle.

Justification:

Point $P$ lies on the perpendicular bisector of $DR$.

Now, from the construction:

Since $QD = 2 \textsf{ cm}$, then $PQ - PR = 2 \textsf{ cm}$.

Given: Base $BC = 3.5 \textsf{ cm}$, $\angle B = 90^\circ$, and sum of other two sides $AB + AC = 5.5 \textsf{ cm}$.

Construction Required: To construct right-angled $\triangle ABC$.

Steps of Construction:

1. Draw $BC = 3.5 \textsf{ cm}$.

2. At $B$, construct $\angle CBX = 90^\circ$.

3. Cut off $BD = 5.5 \textsf{ cm}$ from ray $BX$.

4. Join $CD$.

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

6. Join $AC$. $\triangle ABC$ is the required triangle.

Justification:

Since $A$ lies on the perpendicular bisector of $CD$:

From construction:

Given: Altitude $h = 3.2 \textsf{ cm}$.

Steps of Construction:

1. Draw a line $L$ and take a point $D$ on it.

2. Construct a perpendicular $DX$ to line $L$ at $D$.

3. Cut $DA = 3.2 \textsf{ cm}$ from $DX$.

4. Construct $\angle DAB = 30^\circ$ and $\angle DAC = 30^\circ$ such that $B$ and $C$ lie on line $L$.

5. $\triangle ABC$ is the required triangle.

Justification:

In $\triangle ABD$, $\angle D = 90^\circ$ and $\angle DAB = 30^\circ$, so $\angle B = 60^\circ$.

In $\triangle ACD$, $\angle D = 90^\circ$ and $\angle DAC = 30^\circ$, so $\angle C = 60^\circ$.

$\angle BAC = 30^\circ + 30^\circ = 60^\circ$.

Since all angles are $60^\circ$, the triangle is equilateral.

Given: Diagonals $AC = 6 \textsf{ cm}$ and $BD = 4 \textsf{ cm}$.

Construction Required: To construct rhombus $ABCD$.

Steps of Construction:

1. Draw $AC = 6 \textsf{ cm}$.

2. Draw the perpendicular bisector of $AC$. Let it intersect $AC$ at point $O$.

3. From $O$, cut off $OB = \frac{4}{2} = 2 \textsf{ cm}$ and $OD = 2 \textsf{ cm}$ on the perpendicular bisector.

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

5. $ABCD$ is the required rhombus.

Justification:

In $ABCD$, diagonals $AC$ and $BD$ bisect each other at $90^\circ$ by construction.

A quadrilateral whose diagonals bisect each other at right angles is a rhombus.