Chapter 10 Constructions (Class 10 - Maths NCERT Exemplar Solutions)

Given:

To Find:

The minimum number of points to be marked on ray $AX$.

Solution:

To divide a line segment $AB$ internally in the ratio $m : n$, we follow the standard geometric construction steps. In this case, the ratio is $p : q$.

1. Draw a ray $AX$ making an acute angle with $AB$.

2. Along $AX$, mark points $A_1, A_2, A_3, \dots, A_n$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots$ and so on.

3. The total number of points required on the ray $AX$ to divide the segment in ratio $p : q$ is the sum of the parts of the ratio, which is $p + q$.

For example, to divide a line in the ratio $3 : 2$, we mark $3 + 2 = 5$ points on the ray.

Thus, the minimum number of points is $p + q$.

Hence, the correct option is (B).

Given:

To Find:

The angle between the two radii.

Solution:

Let $O$ be the centre of the circle. Let $PA$ and $PB$ be two tangents from an external point $P$ to the circle at points $A$ and $B$ respectively.

We know that the radius is perpendicular to the tangent at the point of contact.

In the quadrilateral $OAPB$, the sum of all interior angles is $360^\circ$:

$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ$

$\angle AOB + 90^\circ + 35^\circ + 90^\circ = 360^\circ$

$\angle AOB + 215^\circ = 360^\circ$

$\angle AOB = 360^\circ - 215^\circ$

$\angle AOB = 145^\circ$

The angle between the radii must be $145^\circ$.

Hence, the correct option is (D).

Alternate Solution:

The angle between two tangents and the angle between the radii at the points of contact are supplementary.

$\text{Angle between radii} + \text{Angle between tangents} = 180^\circ$

$\text{Angle between radii} + 35^\circ = 180^\circ$

$\text{Angle between radii} = 180^\circ - 35^\circ = 145^\circ$

Given:

To Find:

The minimum number of points to be marked on ray $AX$.

Solution:

To divide a line segment $AB$ in the ratio $m : n$, the minimum number of points to be marked on the ray $AX$ is given by the sum of the terms of the ratio, i.e., $m + n$.

In this case, $m = 5$ and $n = 7$.

Minimum number of points = $m + n$

Minimum number of points = $5 + 7$

Minimum number of points = $12$

Therefore, the minimum number of points to be marked on ray $AX$ is 12.

Hence, the correct option is (D).

Given:

To Find:

The point to which $B$ is joined.

Solution:

In the construction of dividing a line segment $AB$ in the ratio $m : n$, we mark $m + n$ points on the ray $AX$.

Here, the total number of points is:

$4 + 7 = 11$

In the final step of this construction method, the last marked point on the ray (which is $A_{m+n}$) is joined to the endpoint $B$ of the line segment.

In this case, the last point is $A_{11}$.

Thus, the point $B$ is joined to $A_{11}$.

Hence, the correct option is (B).

Given:

To Find:

The points on rays $AX$ and $BY$ that need to be joined.

Solution:

This construction uses the Alternate Method (Parallel Ray Method) to divide a line segment.

1. Draw ray $AX$ at an acute angle to $AB$.

2. Draw ray $BY$ parallel to $AX$ (by constructing $\angle ABY = \angle BAX$ on the opposite side of $AB$).

3. Mark $m$ points on ray $AX$ and $n$ points on ray $BY$ at equal distances.

4. To divide $AB$ in the ratio $m : n$, join the point $A_m$ to the point $B_n$.

Since the ratio is $5 : 6$, we have:

Therefore, we must join $A_5$ and $B_6$.

Hence, the correct option is (A).

Given:

To Find:

The point on ray $BX$ that should be joined to vertex $C$.

Solution:

In the construction of a similar triangle with a scale factor of $\frac{m}{n}$:

1. We locate $\max(m, n)$ points on the ray $BX$. Here, $\max(3, 7) = 7$ points ($B_1$ to $B_7$) are located.

2. The denominator ($n$) of the scale factor tells us which point on the ray should be joined to the endpoint of the base of the original triangle.

3. Since the scale factor is $\frac{3}{7}$, the denominator is 7.

Therefore, the next step is to join the point $B_7$ to $C$. This forms the base for the larger triangle, and the similar triangle will be constructed inside it since the scale factor is less than 1.

Hence, the correct option is (C).

Given:

To Find:

Minimum number of points to be located on ray $BX$.

Solution:

To construct a triangle similar to a given triangle with a scale factor of $\frac{m}{n}$, the number of points to be marked on the ray is equal to the greater of the two numbers $m$ and $n$.

In the given scale factor $\frac{8}{5}$:

Numerator ($m$) = 8

Denominator ($n$) = 5

Comparing the two values, $8 > 5$.

Thus, the minimum number of points to be located on ray $BX$ is 8.

Hence, the correct option is (B).

Given:

To Find:

Angle between the two radii.

Solution:

Let $PA$ and $PB$ be the two tangents to a circle with centre $O$ from an external point $P$. The points of contact are $A$ and $B$.

In quadrilateral $OAPB$, the sum of angles is $360^\circ$:

$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ$

$\angle AOB + 90^\circ + 60^\circ + 90^\circ = 360^\circ$

$\angle AOB + 240^\circ = 360^\circ$

$\angle AOB = 360^\circ - 240^\circ$

$\angle AOB = 120^\circ$

Therefore, the angle between the two radii is $120^\circ$.

Hence, the correct option is (D).

Alternate Solution:

The angle between the radii ($ \theta_r $) and the angle between the tangents ($ \theta_t $) are supplementary:

$\theta_r + \theta_t = 180^\circ$

$\theta_r + 60^\circ = 180^\circ$

$\theta_r = 180^\circ - 60^\circ = 120^\circ$

Statement: False

Given:

Reason/Solution:

To divide a line segment by the standard geometrical construction method, the ratio $m : n$ must be such that both $m$ and $n$ are positive integers.

Let us simplify the given ratio by rationalizing it:

$\text{Ratio} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}$

Multiplying the numerator and denominator by the conjugate $(2 + \sqrt{3})$:

$\frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{(2)^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})}{(2)^2 - (\sqrt{3})^2}$

$= \frac{4 + 3 + 4\sqrt{3}}{4 - 3} = \frac{7 + 4\sqrt{3}}{1}$

Since $7 + 4\sqrt{3}$ is an irrational number, it cannot be expressed as a ratio of two positive integers. In the standard construction, we mark a finite number of points ($m+n$) at equal distances. We cannot mark a non-integer number of points.

Hence, the division of a line segment in this ratio is not possible by the standard construction method taught in the curriculum.

Statement: True

Given:

Reason/Solution:

Let us simplify the given ratio to see if it can be expressed as a ratio of positive integers:

$\text{Ratio} = \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}$

$\text{Ratio} = \sqrt{3} \times \sqrt{3}$

$\text{Ratio} = 3$

This can be written as:

$\text{Ratio} = \frac{3}{1} = 3 : 1$

Since 3 and 1 are positive integers, we can divide the line segment by marking $3 + 1 = 4$ points at equal distances on the ray $AX$.

Therefore, the construction is possible.

Statement: False

Given:

Reason/Solution:

To construct a triangle with a scale factor of $\frac{m}{n}$ where $m > n$ (enlargement):

1. Draw a ray $BX$ and mark $m$ points (the greater of the two). Here, $\max(7, 3) = 7$ points, so $B_1$ to $B_7$ are correctly marked.

2. Join the $n^{th}$ point (the denominator) to the vertex $C$. Since the denominator is 3, joining $B_3$ to $C$ is the correct step.

3. The next step is to draw a line from the $m^{th}$ point (the numerator) parallel to the line $B_3C$.

In the given question, it is stated that $B_6C'$ is drawn parallel to $B_3C$. This is incorrect. To obtain a triangle $\frac{7}{3}$ times the size, we must draw $B_7C'$ parallel to $B_3C$.

Thus, the error lies in joining $B_6$ instead of $B_7$.

Statement: False

Given:

Reason/Solution:

A pair of tangents can be drawn to a circle from an external point only if the distance of the point from the centre of the circle is greater than its radius ($d > r$).

In this case, we compare the distance and the radius:

$3\text{ cm} < 3.5\text{ cm}$

Since the distance from the centre is less than the radius, the point $P$ lies inside the circle.

Geometrically, it is impossible to draw a tangent to a circle from a point located inside it, as any line passing through such a point will be a secant (intersecting the circle at two points) rather than a tangent.

Statement: True

Given:

Reason/Solution:

To construct a pair of tangents inclined at an angle $\theta$, we first need to draw two radii of the circle such that the angle between them is $(180^\circ - \theta)$.

Let the angle between the radii be $\alpha$:

$\alpha = 180^\circ - 170^\circ$

$\alpha = 10^\circ$

Since $10^\circ$ is a valid positive angle ($0^\circ < \alpha < 180^\circ$), we can draw two radii with this central angle and then construct perpendiculars (tangents) at their endpoints. These tangents will eventually meet at an external point, forming an angle of $170^\circ$.

A pair of tangents can be constructed for any angle of inclination $\theta$ as long as $0^\circ < \theta < 180^\circ$.

Given:

1. An equilateral triangle $\triangle ABC$ with side length = $4\text{ cm}$.

2. Scale factor for the similar triangle = $\frac{3}{5}$.

Construction Required:

To construct a triangle $\triangle A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{3}{5}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction:

1. Draw a line segment $BC = 4\text{ cm}$. With $B$ and $C$ as centres and radius $4\text{ cm}$, draw arcs intersecting at $A$. Join $AB$ and $AC$ to get the equilateral $\triangle ABC$.

2. Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.

3. Along $BX$, locate 5 points (the greater of 3 and 5 in the ratio $\frac{3}{5}$) $B_1, B_2, B_3, B_4$ and $B_5$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

4. Join $B_5$ (the denominator) to $C$.

5. Through $B_3$ (the numerator), draw a line parallel to $B_5C$ intersecting $BC$ at $C'$.

6. Through $C'$, draw a line parallel to the line $CA$ intersecting $BA$ at $A'$.

7. $\triangle A'BC'$ is the required triangle.

Solution:

By construction, $A'C' \parallel AC$. Therefore, $\triangle A'BC' \sim \triangle ABC$ by AA similarity criterion (as $\angle B$ is common and $\angle BC'A' = \angle BCA$ being corresponding angles).

In similar triangles, the corresponding angles are equal.

Since all the angles of the new triangle $\triangle A'BC'$ are $60^\circ$, it is also an equilateral triangle.

Also, the sides of the new triangle are:

Yes, the new triangle is also equilateral.

Given:

1. Length of the line segment $AB = 7$ cm.

2. Ratio in which point $P$ divides the segment = $3 : 5$.

To Find:

The location of point $P$ on the line segment $AB$ such that $AP : PB = 3 : 5$.

Construction Required:

1. Draw a line segment $AB = 7$ cm using a ruler.

2. Draw a ray $AX$ making an acute angle ($\angle BAX$) with the line segment $AB$.

3. Locate $3 + 5 = 8$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7$ and $A_8$ on the ray $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_7A_8$.

4. Join the last point $A_8$ to $B$.

5. Through the point $A_3$ (since the first part of the ratio is 3), draw a line parallel to $A_8B$ by constructing an angle equal to $\angle AA_8B$ at $A_3$.

6. Let this parallel line intersect $AB$ at point $P$.

Solution:

By the Basic Proportionality Theorem (Thales Theorem), in $\triangle AA_8B$, since $A_3P \parallel A_8B$:

By construction, $AA_3$ consists of 3 equal units and $A_3A_8$ consists of $8 - 3 = 5$ equal units.

Therefore:

Thus, point $P$ divides the segment $AB$ in the ratio $3 : 5$.

To find the actual lengths of $AP$ and $PB$:

Conclusion: Point $P$ is located at a distance of 2.625 cm from point $A$ on the segment $AB$.

Given:

1. Right-angled $\triangle ABC$ where $BC = 12$ cm, $AB = 5$ cm and $\angle B = 90^\circ$.

2. Scale factor for the similar triangle = $\frac{2}{3}$.

To Construct:

A triangle $A'BC'$ similar to $\triangle ABC$ with sides $\frac{2}{3}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction:

1. Draw a line segment $BC = 12$ cm.

2. At point $B$, construct an angle of $90^\circ$ and cut off a length $BA = 5$ cm. Join $AC$ to obtain the $\triangle ABC$.

3. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.

4. Along $BX$, mark 3 points (the greater of 2 and 3 in the scale factor $\frac{2}{3}$) $B_1, B_2$ and $B_3$ such that $BB_1 = B_1B_2 = B_2B_3$.

5. Join $B_3$ (the denominator) to the point $C$.

6. Through $B_2$ (the numerator), draw a line parallel to $B_3C$ to intersect $BC$ at $C'$.

7. Through $C'$, draw a line parallel to $CA$ to intersect $AB$ at $A'$.

8. $\triangle A'BC'$ is the required similar triangle.

Solution:

In $\triangle ABC$ and $\triangle A'BC'$:

Therefore, $\triangle A'BC' \sim \triangle ABC$ by the AA similarity criterion.

To check if the new triangle is a right triangle, we look at $\angle A'BC'$:

From (i) and (ii), we have:

Since the new triangle $\triangle A'BC'$ has an angle of $90^\circ$ at $B$, it is also a right triangle.

The sides of the new triangle are:

Yes, the new triangle is also a right triangle.

Given:

1. A triangle $\triangle ABC$ with side lengths $BC = 6$ cm, $CA = 5$ cm, and $AB = 4$ cm.

2. Scale factor for the similar triangle = $\frac{5}{3}$.

To Construct:

A triangle $A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{5}{3}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction:

1. Draw a line segment $BC = 6$ cm. Using a compass, draw an arc of radius $4$ cm from $B$ and an arc of radius $5$ cm from $C$. Let the arcs intersect at $A$. Join $AB$ and $AC$ to form $\triangle ABC$.

2. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.

3. Mark 5 points (the greater of 5 and 3) $B_1, B_2, B_3, B_4$ and $B_5$ on ray $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

4. Join $B_3$ (the denominator of the scale factor) to the point $C$.

5. Draw a line through $B_5$ (the numerator of the scale factor) parallel to $B_3C$. To do this, extend $BC$ to $C'$ such that the line through $B_5$ intersects the extended line at $C'$.

6. Draw a line through $C'$ parallel to $AC$. To do this, extend $BA$ to $A'$ such that the line through $C'$ intersects the extended line at $A'$.

7. $\triangle A'BC'$ is the required similar triangle.

Verification/Proof:

By construction, $B_5C' \parallel B_3C$. In $\triangle BB_5C'$, by the Basic Proportionality Theorem:

Similarly, since $A'C' \parallel AC$, we have $\triangle A'BC' \sim \triangle ABC$ by AA Similarity Criterion.

Substituting the value from equation (i):

Thus, all the sides of the new triangle $A'BC'$ are $\frac{5}{3}$ times the corresponding sides of the original triangle $\triangle ABC$.

Given:

1. Radius of the circle ($r$) = $4$ cm.

2. Distance of the external point ($P$) from the centre ($O$) = $6$ cm.

To Construct:

A pair of tangents from point $P$ to the circle with centre $O$.

Steps of Construction:

1. Take a point $O$ as centre and draw a circle of radius $4$ cm using a compass.

2. Mark a point $P$ outside the circle such that the distance $OP = 6$ cm.

3. Join $OP$ and construct its perpendicular bisector. Let $M$ be the midpoint of $OP$.

4. Taking $M$ as centre and $MO$ (or $MP$) as radius, draw another circle (helper circle).

5. Let this second circle intersect the original circle at two points, $Q$ and $R$.

6. Join $PQ$ and $PR$.

7. $PQ$ and $PR$ are the required pair of tangents.

Verification (Proof):

Join $OQ$. We need to prove that $PQ$ is a tangent to the circle.

In the circle with centre $M$, $OP$ is the diameter.

$\angle OQP$ is an angle in a semicircle.

We know that an angle in a semicircle is a right angle ($90^\circ$).

This means that $OQ \perp PQ$.

Since $OQ$ is the radius of the original circle and the line $PQ$ is perpendicular to it at the point of contact $Q$:

Similarly, $PR$ is also a tangent to the circle.

Calculation of Tangent Length:

In right-angled $\triangle OQP$:

Taking $\sqrt{5} \approx 2.236$:

Conclusion: The length of the tangent is approximately 4.47 cm.

Given:

To Find:

Whether $AB'C'D'$ is a rhombus and provide the justification.

Construction Required:

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

2. At $B$, construct $\angle ABC = 60^\circ$ and cut $BC = 4\text{ cm}$.

3. With $A$ and $C$ as centres and radius $4\text{ cm}$, draw arcs intersecting at $D$. Join $AD$ and $CD$ to complete rhombus $ABCD$.

4. Join diagonal $AC$. Since $AB = BC = 4\text{ cm}$ and $\angle B = 60^\circ$, $\Delta ABC$ is an equilateral triangle ($AC = 4\text{ cm}$).

5. Similarly, $\Delta ADC$ is also an equilateral triangle.

6. Construct $\Delta AB'C'$ similar to $\Delta ABC$ with scale factor $\frac{2}{3}$ such that $B'$ lies on $AB$ and $C'$ lies on $AC$.

7. Draw $C'D' \parallel CD$ meeting $AD$ at $D'$.

Solution and Reasoning:

In $\Delta ABC$, since it is equilateral:

Since $\Delta AB'C' \sim \Delta ABC$ with scale factor $\frac{2}{3}$:

Now, consider $\Delta ADC$. We are given $C'D' \parallel CD$. By Basic Proportionality Theorem or Similarity:

$\Delta AC'D' \sim \Delta ACD$

Therefore, the ratios of the sides are equal:

$\frac{AC'}{AC} = \frac{C'D'}{CD} = \frac{AD'}{AD} = \frac{2}{3}$

This gives us the lengths of the remaining segments:

In quadrilateral $AB'C'D'$, we have found:

$AB' = B'C' = C'D' = D'A = \frac{8}{3}\text{ cm}$

Since all four sides are equal and $AB'C'D'$ is a part of the original rhombus (making opposite sides parallel), $AB'C'D'$ is indeed a rhombus.

Conclusion: Yes, $AB'C'D'$ is a rhombus because all its sides are equal to $\frac{8}{3}\text{ cm}$ and opposite sides are parallel.

Given:

To Find:

To locate points $P$ and $Q$ on $AB$ and $AC$ respectively and to measure the length of $PQ$.

Construction Required:

Steps of construction:

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

2. Draw a ray $AZ$ making an angle $\angle BAZ = 60^\circ$ with $AB$.

3. With centre $A$ and radius $7\text{ cm}$, draw an arc cutting the line $AZ$ at $C$.

4. Draw a ray $AX$, making an acute angle $\angle BAX$ below $AB$.

5. Divide $AX$ into four equal parts, namely $AA_1 = A_1A_2 = A_2A_3 = A_3A_4$.

6. Join $A_4B$.

7. Draw $A_3P \parallel A_4B$ meeting $AB$ at $P$.

8. Next, draw a ray $AY$, such that it makes an acute angle $\angle CAY$ on the other side of $AC$.

9. Divide $AY$ into four parts, namely $AB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

10. Join $B_4C$.

11. Draw $B_1Q \parallel B_4C$ meeting $AC$ at $Q$.

12. Join $PQ$ and measure the length using a ruler.

Solution:

By following the steps of construction, we have obtained the points $P$ and $Q$.

The length of $AP$ is calculated as:

$AP = \frac{3}{4} \times 5\text{ cm} = 3.75\text{ cm}$

The length of $AQ$ is calculated as:

$AQ = \frac{1}{4} \times 7\text{ cm} = 1.75\text{ cm}$

Upon measuring the segment $PQ$ with the help of a ruler, we find:

$PQ = 3.25\text{ cm}$

Given:

To Construct:

1. A parallelogram $ABCD$.

2. A triangle $BD'C'$ similar to $\Delta BDC$ with scale factor $\frac{4}{3}$.

3. A line segment $D'A' \parallel DA$ to form $A'BC'D'$.

Construction Required:

Steps of construction:

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

2. At point $B$, draw a ray $BY$ making an angle $\angle CBY = 60^\circ$.

3. From ray $BY$, cut off $AB = 3\text{ cm}$.

4. Since $ABCD$ is a parallelogram, opposite sides are equal and parallel. With $A$ as centre and radius $5\text{ cm}$, draw an arc. With $C$ as centre and radius $3\text{ cm}$, draw another arc to intersect the previous arc at $D$.

5. Join $AD$ and $CD$. $ABCD$ is the required parallelogram. Join diagonal $BD$ to divide it into $\Delta BDC$ and $\Delta ABD$.

6. Below $BC$, draw a ray $BX$ making an acute angle $\angle CBX$.

7. Mark four points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

8. Join $B_3C$. Draw a line through $B_4$ parallel to $B_3C$ to intersect the extended line segment $BC$ at $C'$.

9. From $C'$, draw a line $C'D' \parallel CD$ to intersect the extended diagonal $BD$ at $D'$. $\Delta BD'C'$ is the required similar triangle.

10. From $D'$, draw a line $D'A' \parallel DA$ to intersect the extended side $BA$ at $A'$.

Solution:

We need to determine if $A'BC'D'$ is a parallelogram.

In the original parallelogram $ABCD$:

By construction of similar triangles:

Since $CD \parallel AB$, and $A'$ lies on the extension of $BA$:

From (i) and (ii), we can conclude:

Now, it is given/constructed that:

Since $DA \parallel BC$ (from the original parallelogram) and $C'$ lies on the extension of $BC$:

From (iv) and (v), we can conclude:

In quadrilateral $A'BC'D'$, from equations (iii) and (vi), we see that both pairs of opposite sides are parallel ($C'D' \parallel A'B$ and $D'A' \parallel BC'$).

Yes, $A'BC'D'$ is a parallelogram.

Given:

Construction Required:

1. Draw two concentric circles with centre $O$ and radii $3\text{ cm}$ and $5\text{ cm}$.

2. Mark a point $P$ on the circumference of the outer circle.

3. Join $OP$ and bisect it to find the midpoint $M$.

4. With $M$ as centre and radius $OM = MP$, draw a circle intersecting the inner circle at points $T_1$ and $T_2$.

5. Join $PT_1$ and $PT_2$. These are the required tangents.

Proof and Calculation:

Join $OT_1$. In $\Delta OT_1P$, $\angle OT_1P = 90^\circ$ because the angle in a semi-circle is a right angle.

Therefore, $\Delta OT_1P$ is a right-angled triangle. By Pythagoras Theorem:

$OP^2 = OT_1^2 + PT_1^2$

$25 = 9 + PT_1^2$

$PT_1^2 = 25 - 9$

$PT_1^2 = 16$

$PT_1 = 4\text{ cm}$

Verification: Upon measuring the tangent $PT_1$ with a ruler, the length is found to be $4\text{ cm}$.

Given:

To Find:

The scale factor for construction:

$\text{Scale Factor } (k) = \frac{\text{Side of new triangle}}{\text{Side of old triangle}}$

$k = \frac{PQ}{AB} = \frac{8}{6} = \frac{4}{3}$

Construction Required:

1. Draw the original triangle: Draw $BC = 5\text{ cm}$. With $B$ and $C$ as centres and radius $6\text{ cm}$, draw two arcs intersecting at $A$. Join $AB$ and $AC$.

2. Ray BX: Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.

3. Locate points: Mark 4 points ($B_1, B_2, B_3, B_4$) on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

4. Join Denominator: Join $B_3$ to $C$ (since the scale factor is $4/3$, the denominator is 3).

5. Parallel line for Base: Draw a line through $B_4$ parallel to $B_3C$ to intersect the extended line segment $BC$ at point $R$.

6. Parallel line for Side: Through $R$, draw a line parallel to $CA$ to intersect the extended line segment $BA$ at point $P$.

7. Result: $\Delta PBR$ is the required triangle. In the context of the question, this triangle corresponds to $\Delta PQR$.

Justification:

By construction, $B_4R \parallel B_3C$. In $\Delta BB_4R$:

$\frac{BC}{BR} = \frac{BB_3}{BB_4} = \frac{3}{4}$

Also, since $RP \parallel CA$, $\Delta PBR \sim \Delta ABC$ (by AA similarity).

$\frac{PB}{AB} = \frac{BR}{BC} = \frac{PR}{AC} = \frac{4}{3}$

Since $AB = 6\text{ cm}$, then $PB = \frac{4}{3} \times 6 = 8\text{ cm}$. This satisfies the condition $PQ = 8\text{ cm}$.

Given:

Construction Required:

1. Draw $BC = 6\text{ cm}$. At $B$, construct $\angle ABC = 60^\circ$ and cut $AB = 5\text{ cm}$. Join $AC$.

2. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.

3. Mark 7 points $B_1, B_2, \dots, B_7$ at equal distances on $BX$.

4. Join $B_7$ (the denominator) to $C$.

5. Draw a line from $B_5$ parallel to $B_7C$ to intersect $BC$ at $C'$.

6. Draw a line from $C'$ parallel to $CA$ to intersect $AB$ at $A'$.

7. $\Delta A'BC'$ is the required similar triangle.

Justification:

By construction, $B_5C' \parallel B_7C$. In $\Delta BB_7C$, by Basic Proportionality Theorem:

$\frac{BC'}{BC} = \frac{BB_5}{BB_7} = \frac{5}{7}$

In $\Delta ABC$, since $A'C' \parallel AC$:

$\Delta A'BC' \sim \Delta ABC$ by AA similarity.

Therefore, the ratio of corresponding sides is:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{5}{7}$

Thus, the sides of the new triangle are $\frac{5}{7}$ of the corresponding sides of $\Delta ABC$.

Given:

Construction Required:

1. Draw a circle with centre $O$ and radius $4\text{ cm}$.

2. The angle between the radii at the points of contact is supplementary to the angle between the tangents.

Angle between radii = $180^\circ - 60^\circ = 120^\circ$.

3. Draw any radius $OA$. Construct $\angle AOB = 120^\circ$ such that $B$ is another point on the circle.

4. Construct $90^\circ$ angles at points $A$ and $B$. These perpendicular lines are the tangents.

5. Let the tangents intersect at point $P$. Then $\angle APB = 60^\circ$.

Justification:

In quadrilateral $OAPB$, by the angle sum property:

$\angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ$

$120^\circ + 90^\circ + 90^\circ + \angle APB = 360^\circ$

$300^\circ + \angle APB = 360^\circ$

$\angle APB = 60^\circ$

This confirms the construction is correct.

Measurement and Verification:

To find the distance $OP$ between the centre and the point of intersection:

In right-angled triangle $OAP$, the line $OP$ bisects $\angle APB$. Therefore, $\angle APO = 30^\circ$.

$\sin 30^\circ = \frac{OA}{OP}$

$\frac{1}{2} = \frac{4}{OP}$

$OP = 8\text{ cm}$

Observation: Upon measuring with a ruler, the distance $OP$ is found to be $8\text{ cm}$.

Given:

Construction Required:

1. Draw the original triangle $ABC$ with the given dimensions.

2. Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of $A$.

3. Mark 3 points $B_1, B_2, B_3$ at equal distances on $BX$.

4. Join $B_2$ (denominator) to $C$.

5. Draw $B_3C' \parallel B_2C$, where $C'$ lies on the extended line $BC$.

6. Draw $C'A' \parallel CA$, where $A'$ lies on the extended line $BA$.

7. $\Delta A'BC'$ is the required similar triangle.

Justification:

Since $C'A' \parallel CA$, by AA similarity criterion, $\Delta A'BC' \sim \Delta ABC$.

The ratio of sides is $\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{2} = 1.5$.

The sides of $\Delta A'BC'$ are:

$A'B = 1.5 \times 4 = 6\text{ cm}$

$BC' = 1.5 \times 6 = 9\text{ cm}$

$A'C' = 1.5 \times 9 = 13.5\text{ cm}$

Analysis of Congruency:

For two triangles to be congruent, their scale factor must be 1 ($k=1$).

In this case, $k = 1.5$, so the triangles are not congruent.

Note on sides:

Sides of $\Delta ABC$: 4 cm, 6 cm, 9 cm.

Sides of $\Delta A'BC'$: 6 cm, 9 cm, 13.5 cm.

We can see that two sides of $\Delta ABC$ ($6\text{ cm}$ and $9\text{ cm}$) are equal in length to two sides of $\Delta A'BC'$, and all three angles are equal because they are similar. However, since the third sides ($4\text{ cm}$ vs $13.5\text{ cm}$) are different, the triangles do not satisfy the SSS or SAS criteria for congruency.

Conclusion: No, the triangles are not congruent.