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

Given:

1. A line segment $AB$.

2. A ratio $p : q$ in which the line segment is to be divided (where $p$ and $q$ are positive integers).

3. A ray $AX$ making an acute angle with $AB$.

To Find:

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

Construction Required:

1. Draw the line segment $AB$.

2. Draw a ray $AX$ making an acute angle $\angle BAX$.

3. Mark points $A_1, A_2, A_3, \dots, A_n$ on $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_{n-1}A_n$.

4. Join $A_n$ to $B$ and draw a line parallel to $A_nB$ to locate the division point.

To divide a line segment $AB$ internally in the ratio $p : q$, we use the following steps according to the Indian NCERT curriculum (Standard Geometric Constructions):

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

2. We need to locate points on $AX$ such that we can use the Basic Proportionality Theorem (Thales Theorem). To divide the segment into $p$ parts and $q$ parts (total $p+q$ parts), we must have $p+q$ equal divisions on the ray $AX$.

3. Let the points be $A_1, A_2, A_3, \dots, A_{p+q}$.

4. Join the last point $A_{p+q}$ to $B$.

5. Through the point $A_p$, draw a line parallel to $A_{p+q}B$ (by making an angle equal to $\angle AA_{p+q}B$) at $A_p$.

6. Suppose this parallel line intersects $AB$ at point $C$. Then, by the property of similar triangles ($\triangle AA_pC \sim \triangle AA_{p+q}B$):

Thus, the minimum number of points required is the sum of the ratio parts, i.e., $p + q$.

Comparing this with the given options:

(A) greater of $p$ and $q$

(B) $p + q$

(C) $p + q - 1$

(D) $pq$

The correct option is (B).

Alternate Solution:

There is another method to divide a line segment where we draw two parallel rays: ray $AX$ from $A$ and ray $BY$ from $B$ (on opposite sides of $AB$). In that specific method, we mark $p$ points on $AX$ and $q$ points on $BY$. However, the question specifically mentions marking points only on "ray AX". Therefore, for the single-ray method, the sum $p + q$ is the only correct requirement.

Given:

1. Angle of inclination between the pair of tangents = $35^\circ$.

2. The tangents are drawn from an external point to a circle.

To Find:

The angle between the two radii at the end points of which the tangents are drawn.

Construction:

Let $O$ be the centre of the circle. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle at points $A$ and $B$ respectively. Join $OA$ and $OB$ to form the radii.

Solution:

Let $PA$ and $PB$ be the two tangents to the circle with centre $O$.

According to the theorem, the radius is perpendicular to the tangent at the point of contact.

The angle between the tangents is given as:

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

Substituting the values from above:

Thus, the angle between the two radii is $145^\circ$.

Comparing this with the given options:

(A) $105^\circ$

(B) $70^\circ$

(C) $140^\circ$

(D) $145^\circ$

The correct option is (D).

Alternate Solution:

We can use the direct property that in a circle, the angle between two tangents drawn from an external point and the angle subtended by the line segments joining the points of contact at the centre are supplementary.

This property provides a faster way to find the required angle.

Given:

1. A line segment $AB$.

2. The ratio in which the segment is to be divided is $5:7$.

3. A ray $AX$ making an acute angle with $AB$.

To Find:

The minimum number of points to be marked on the ray $AX$ at equal distances.

Solution:

To divide a line segment $AB$ internally in a given ratio $m:n$ (where $m$ and $n$ are positive integers), the standard geometric construction method used in the Indian NCERT curriculum involves the following steps:

1. Draw a ray $AX$ making an acute angle $\angle BAX$.

2. Mark $k$ points $A_1, A_2, A_3, \dots, A_k$ on the ray $AX$ such that the distances between the points are equal, i.e., $AA_1 = A_1A_2 = A_2A_3 = \dots = A_{k-1}A_k$.

3. In this method, the total number of points $k$ must be equal to the sum of the parts of the ratio.

Here, the given ratio is $5:7$. Therefore:

The minimum number of points required on ray $AX$ is given by the sum $m + n$:

After marking these 12 points, we join $A_{12}$ to $B$. Then, a line is drawn through $A_5$ parallel to $A_{12}B$, which divides the segment $AB$ in the ratio $5:7$.

Comparing our result with the given options:

(A) 8

(B) 10

(C) 11

(D) 12

The correct option is (D).

Alternate Solution:

Another method to divide a line segment is the "Parallel Ray Method" where two rays $AX$ and $BY$ are drawn parallel to each other. In that case, we mark 5 points on $AX$ and 7 points on $BY$. However, the question specifically asks for points marked on a single ray AX. For the single-ray construction method, the number of equal divisions required is always the sum of the ratio terms, which is $5 + 7 = 12$.

Given:

1. A line segment $AB$.

2. The ratio in which the segment is to be divided is $4 : 7$.

3. A ray $AX$ making an acute angle with $AB$.

To Find:

Which point on the ray $AX$ is joined to the point $B$.

Construction Required:

1. Draw the line segment $AB$.

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

3. Mark $m + n$ points on $AX$ at equal distances.

Solution:

To divide a line segment $AB$ internally in the ratio $m : n$, we follow the standard geometric construction steps:

1. Let the given ratio be $m : n$. Here, $m = 4$ and $n = 7$.

2. The total number of points to be marked on the ray $AX$ at equal distances is the sum of the parts of the ratio.

3. These points are named $A_1, A_2, A_3, \dots, A_{11}$.

4. According to the construction procedure, the last point marked on the ray $AX$, which is $A_{m+n}$, must be joined to the endpoint $B$ of the line segment.

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

Comparing this with the given options:

(A) $A_{12}$

(B) $A_{11}$

(C) $A_{10}$

(D) $A_9$

The correct option is (B).

Alternate Solution:

The construction is based on the Basic Proportionality Theorem. By joining $A_{11}$ to $B$ and then drawing a line through $A_4$ parallel to $A_{11}B$ (which meets $AB$ at $C$), we create two similar triangles $\triangle AA_4C \sim \triangle AA_{11}B$. The ratio of the sides is:

This confirms that $A_{11}$ is the point that must be joined to $B$ to complete the framework for the division.

Given:

1. A line segment $AB$.

2. The ratio in which the line segment is to be divided is $5:6$.

3. Two parallel rays $AX$ and $BY$ are drawn on opposite sides of $AB$.

To Find:

The points on ray $AX$ and ray $BY$ that must be joined to divide the segment.

Construction Required:

1. Draw the line segment $AB$.

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

3. Draw another ray $BY$ parallel to $AX$ by constructing an angle at $B$ equal to $\angle BAX$ (Alternate interior angles).

4. Locate points $A_1, A_2, A_3, A_4, A_5$ on $AX$ and $B_1, B_2, B_3, B_4, B_5, B_6$ on $BY$ such that $AA_1 = A_1A_2 = \dots = BB_1 = B_1B_2 = \dots$

Solution:

There are two primary methods for dividing a line segment in a given ratio $m:n$ as per the Indian NCERT curriculum. The method described in this question is the Parallel Ray Method.

In the parallel ray method:

1. To divide $AB$ in the ratio $m:n$, we mark $m$ points on the first ray ($AX$) and $n$ points on the second parallel ray ($BY$).

2. Here, the ratio is $5:6$. Thus, we have $m = 5$ and $n = 6$.

3. This means we mark 5 points on $AX$ (labeled $A_1, A_2, A_3, A_4, A_5$) and 6 points on $BY$ (labeled $B_1, B_2, B_3, B_4, B_5, B_6$).

4. To find the point of division on $AB$, we must join the $m^{th}$ point of the first ray to the $n^{th}$ point of the second ray.

5. Let the line $A_5B_6$ intersect $AB$ at point $C$. By the Alternate Interior Angles and AA Similarity, $\triangle AA_5C \sim \triangle BB_6C$.

Therefore, the points joined are $A_5$ and $B_6$.

Comparing this with the given options:

(A) $A_5$ and $B_6$

(B) $A_6$ and $B_5$

(C) $A_4$ and $B_5$

(D) $A_5$ and $B_4$

The correct option is (A).

Alternate Solution:

If the question had asked to divide the segment in the ratio $6:5$, then the points joined would have been $A_6$ and $B_5$. However, since the first part of the ratio corresponds to the segment starting from $A$, we always take the $m$ value on the ray starting from $A$. Hence, $A_5$ must be paired with $B_6$.

Given:

1. A triangle $\triangle ABC$.

2. Scale factor for the similar triangle = $\frac{3}{7}$.

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

To Find:

The specific point on ray $BX$ that should be joined to point $C$ in the next step of construction.

Construction Required:

1. Draw $\triangle ABC$.

2. Draw ray $BX$ such that $\angle CBX$ is acute.

3. Mark points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.

Solution:

To construct a triangle similar to a given $\triangle ABC$ with a scale factor of $\frac{m}{n}$ (where $m < n$), we follow these standard geometric steps:

1. Identify the greater of the two numbers $m$ and $n$. In the ratio $\frac{3}{7}$, the greater number is $7$.

2. Mark $7$ points ($B_1, B_2, \dots, B_7$) on the ray $BX$ at equal distances.

3. The rule of construction: The point corresponding to the denominator of the scale factor is always joined to the endpoint of the base of the original triangle.

4. Here, the denominator is $7$. Therefore, the point $B_7$ must be joined to $C$.

5. In the subsequent step, a line is drawn through $B_3$ (the numerator) parallel to $B_7C$ to intersect $BC$ at a new point $C'$.

Comparing this with the given options:

(A) $B_{10}$ to $C$

(B) $B_3$ to $C$

(C) $B_7$ to $C$

(D) $B_4$ to $C$

The correct option is (C).

Alternate Solution:

If the scale factor were an improper fraction (e.g., $\frac{7}{3}$), we would still mark $7$ points (the maximum of the two), but we would join the point $B_3$ (the denominator) to $C$ and then extend $BC$ to meet a parallel line drawn from $B_7$. However, for a scale factor of $\frac{3}{7}$, the original triangle is the "7-part" reference, so we join $B_7$ to $C$.

Given:

1. A triangle $\triangle ABC$.

2. The scale factor (ratio of sides) for the similar triangle is $\frac{8}{5}$.

3. A ray $BX$ is drawn making an acute angle with $BC$.

To Find:

The minimum number of points to be located at equal distances on ray $BX$.

Construction Required:

1. Draw the original triangle $\triangle ABC$.

2. Draw a ray $BX$ such that $\angle CBX$ is an acute angle and $X$ is on the opposite side of $A$ with respect to $BC$.

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

Solution:

To construct a triangle similar to a given triangle with a scale factor of $\frac{m}{n}$, the standard geometric procedure followed is as follows:

1. Identify the scale factor, which is $\frac{8}{5}$ in this case.

2. The rule for marking points on the ray $BX$ is that the number of points must be equal to the greater of the numerator ($m$) and the denominator ($n$).

3. Here, $m = 8$ and $n = 5$.

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

5. Specifically, these points are $B_1, B_2, B_3, B_4, B_5, B_6, B_7, B_8$.

In the construction process, we join $B_5$ (the denominator) to $C$ and then draw a line through $B_8$ (the numerator) parallel to $B_5C$ to intersect the extended line $BC$.

Comparing this with the given options:

(A) 5

(B) 8

(C) 13

(D) 3

The correct option is (B).

Alternate Solution:

One might confuse this with the construction of dividing a line segment in a ratio, where the sum of the ratio ($8+5=13$) is required. However, for similar triangles, we only need to mark points up to the largest value present in the fraction to ensure we can create the necessary parallel intercepts. Since we need to reach the 8th part to enlarge the triangle, the minimum points required must be 8.

Given:

1. Angle of inclination between the pair of tangents = $60^\circ$.

2. The tangents are drawn to a circle from an external point.

To Find:

The angle between the two radii at whose end points the tangents are drawn.

Construction Required:

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

Solution:

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

The angle between the tangents is given as:

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

Substituting the known values:

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

Comparing this with the given options:

(A) $135^\circ$

(B) $90^\circ$

(C) $60^\circ$

(D) $120^\circ$

The correct option is (D).

Alternate Solution:

According to a standard theorem in geometry (often cited in 10th-grade Indian mathematics), the angle between two tangents drawn from an external point and the angle subtended by the radii at the centre are supplementary.

This property allows us to find the angle between the radii directly without calculating the full quadrilateral sum.

Given:

The ratio in which a line segment is to be divided is $(2 + \sqrt{3}) : (2 - \sqrt{3})$.

To Verify:

Whether it is possible to divide a line segment in the given ratio by geometrical construction.

Solution:

The statement is False.

According to the standard geometric construction procedure for dividing a line segment $AB$ in a ratio $m : n$, the values of $m$ and $n$ must be positive integers. This is because the construction process involves the following steps:

1. Drawing a ray $AX$ at an acute angle to $AB$.

2. Marking $m + n$ points at equal distances on the ray $AX$.

3. Since we are marking a physical count of points ($A_1, A_2, A_3, \dots$), the number of points must be a whole number, which requires $m$ and $n$ to be integers (or at least have a ratio that simplifies to a ratio of integers).

Let us simplify the given ratio $(2 + \sqrt{3}) : (2 - \sqrt{3})$ by rationalizing it:

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

Since $\sqrt{3}$ is an irrational number, the value $7 + 4\sqrt{3}$ is also irrational. An irrational ratio cannot be expressed as $m : n$ where $m$ and $n$ are integers. Therefore, we cannot determine an exact number of equal segments to mark on the ray $AX$ to divide the line segment precisely in this ratio using the standard ruler and compass method taught in the Indian NCERT curriculum.

Conclusion: It is not possible to divide a line segment in the ratio $(2 + \sqrt{3}) : (2 - \sqrt{3})$ by the taught geometrical construction methods.

Alternate Solution:

If a ratio can be simplified into rational numbers, it is constructible. For example, consider the ratio $\sqrt{3} : \frac{1}{\sqrt{3}}$.

$\frac{\sqrt{3}}{1/\sqrt{3}} = \sqrt{3} \times \sqrt{3} = 3 = \frac{3}{1}$.

In the case of $3 : 1$, $m=3$ and $n=1$ are integers, so the construction is possible by marking $3+1=4$ points. However, as shown above, $(2 + \sqrt{3}) : (2 - \sqrt{3})$ does not simplify to a rational ratio, making it non-constructible by standard methods.

Given:

The ratio in which the line segment is to be divided is $\sqrt{3} : \frac{1}{\sqrt{3}}$.

To Verify:

Whether it is possible to divide a line segment in the given ratio by geometrical construction.

Solution:

The statement is True.

According to the geometric construction principles for dividing a line segment, the ratio $m : n$ must be expressible such that $m$ and $n$ are positive integers. This is necessary because the construction involves marking a discrete number of points at equal distances on a ray.

Let us simplify the given ratio $\sqrt{3} : \frac{1}{\sqrt{3}}$:

To simplify the complex fraction, we multiply the numerator by the reciprocal of the denominator:

This can be written as a ratio of two positive integers:

Since the ratio simplifies to $3 : 1$, and both $3$ and $1$ are positive integers, the line segment can be divided using the standard ruler and compass method.

Construction Procedure (Brief):

1. Draw the line segment $AB$.

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

3. Mark $3 + 1 = 4$ points ($A_1, A_2, A_3, A_4$) on $AX$ at equal distances.

4. Join $A_4$ to $B$.

5. Draw a line through $A_3$ parallel to $A_4B$ to intersect $AB$ at point $C$.

Point $C$ will divide $AB$ in the ratio $3 : 1$, which is equivalent to the given ratio $\sqrt{3} : \frac{1}{\sqrt{3}}$.

Conclusion: Therefore, it is possible to divide the line segment in the given ratio.

Given:

1. Scale factor for the similar triangle = $\frac{7}{3}$.

2. Construction steps: Mark points $B_1$ to $B_7$, join $B_3$ to $C$, draw $B_6C' \parallel B_3C$, and $A'C' \parallel AC$.

To Verify:

Whether the described construction steps are correct to obtain a triangle with sides $\frac{7}{3}$ of the corresponding sides of $\triangle ABC$.

Solution:

The statement is False.

To construct a triangle similar to a given $\triangle ABC$ with a scale factor $\frac{m}{n} = \frac{7}{3}$, we follow these standard geometric construction rules:

1. Locate $k$ points on ray $BX$, where $k$ is the greater of the numerator ($m$) and the denominator ($n$).

Thus, locating points $B_1, B_2, \dots, B_7$ is correct.

2. Join the point corresponding to the denominator ($n$) to the vertex $C$. Since the denominator is 3, joining $B_3$ to $C$ is correct.

3. Draw a line through the point corresponding to the numerator ($m$) parallel to the line $B_nC$. In this case, the numerator is 7.

4. However, the description states that a line segment $B_6C'$ is drawn parallel to $B_3C$. Using point $B_6$ would result in a triangle whose sides are in the ratio $\frac{6}{3} = 2$ times the original sides, which does not satisfy the required scale factor of $\frac{7}{3}$.

Conclusion: The construction is incorrect because the parallel line must be drawn from $B_7$ to achieve the $\frac{7}{3}$ ratio.

Alternate Solution:

By the Basic Proportionality Theorem, in $\triangle BB_7C'$ where $B_3C \parallel B_7C'$:

This confirms that the 7th point ($B_7$) is essential to get the ratio $\frac{7}{3}$. The mention of $B_6$ in the input description makes the statement mathematically incorrect.

The statement is False.

To construct a pair of tangents from a point P to a circle, the point P must be located outside the circle.

A point P is outside the circle if its distance from the center of the circle is greater than the radius of the circle.

In this question, the radius of the circle is given as 3.5 cm.

The distance of the point P from the centre is given as 3 cm.

Comparing the distance of the point P from the centre with the radius:

Since $3 \text{ cm} < 3.5 \text{ cm}$, the distance of point P from the centre is less than the radius.

This means that the point P is located inside the circle.

It is not possible to draw any tangent to a circle from a point lying inside the circle.

Therefore, a pair of tangents cannot be constructed from a point P situated at a distance of 3 cm from the centre to a circle of radius 3.5 cm.

Given:

The angle of inclination between the pair of tangents = $170^\circ$.

To Verify:

Whether it is possible to construct a pair of tangents to a circle inclined at an angle of $170^\circ$.

Construction Required:

To construct tangents inclined at an angle $\theta$, we first draw two radii of the circle such that the angle between them is $(180^\circ - \theta)$. Then, we draw perpendiculars at the endpoints of these radii.

Solution:

The statement is True.

A pair of tangents can be constructed to a circle from an external point as long as the angle between them, say $\theta$, satisfies the condition $0^\circ < \theta < 180^\circ$.

Let $PA$ and $PB$ be two tangents to a circle with centre $O$, where $A$ and $B$ are points of contact. We know that the angle between the tangents and the angle subtended by the radii at the centre are supplementary.

Given the angle of inclination $\angle APB = 170^\circ$:

Since $10^\circ$ is a valid positive angle, we can draw two radii $OA$ and $OB$ inclined at $10^\circ$ to each other. By constructing perpendiculars (tangents) at $A$ and $B$, they will eventually intersect at an external point $P$ making an angle of $170^\circ$.

As the angle between the tangents increases towards $180^\circ$, the external point $P$ moves closer to the circle. Since $170^\circ < 180^\circ$, the point $P$ still lies outside the circle, making the construction possible.

Conclusion: It is mathematically and geometrically possible to construct such a pair of tangents.

Alternate Solution:

From an Indian perspective (following the NCERT curriculum), the only condition for the non-existence of a pair of tangents from a point $P$ is if the point $P$ lies inside the circle. If the angle between the tangents is $170^\circ$, the distance of point $P$ from the centre $O$ is given by $OP = r \cdot \text{cosec}(170^\circ/2) = r \cdot \text{cosec}(85^\circ)$. Since $\sin(85^\circ) < 1$, its reciprocal $\text{cosec}(85^\circ) > 1$.

Thus, $OP > r$, which confirms point $P$ is outside the circle, and tangents can be drawn.

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.

Alternate Solution:

We can verify the equilateral nature using the scale factor. The sides of $\triangle A'BC'$ are $a' = \frac{3}{5}a$. Since $a = 4\text{ cm}$ for all three sides of $\triangle ABC$, then $a' = 2.4\text{ cm}$ for all three sides of $\triangle A'BC'$. Since all three sides of the new triangle are equal, it must be equilateral by definition.

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 $AB$.

Alternate Solution:

We can also use the Parallel Ray Method:

1. Draw $AB = 7$ cm.

2. Draw ray $AX$ at an acute angle and ray $BY$ parallel to $AX$ (on the opposite side of $AB$).

3. Mark 3 equal divisions on $AX$ ($A_1, A_2, A_3$) and 5 equal divisions on $BY$ ($B_1, B_2, B_3, B_4, B_5$).

4. Join $A_3$ to $B_5$. The intersection of $A_3B_5$ and $AB$ is the required point $P$.

This method utilizes the similarity of $\triangle AA_3P$ and $\triangle BB_5P$ to achieve the same ratio $3 : 5$.

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.

Alternate Solution:

We can verify the similarity by calculating the hypotenuse of the original triangle first. In $\triangle ABC$:

$AC^2 = AB^2 + BC^2 = 5^2 + 12^2 = 25 + 144 = 169$

$AC = 13$ cm

The sides of the new triangle $\triangle A'BC'$ are $a' = \frac{2}{3}a$:

$A'B = 3.33$ cm, $BC' = 8$ cm, and $A'C' = \frac{2}{3} \times 13 = 8.67$ cm.

Using Pythagoras Theorem: $(3.33)^2 + (8)^2 \approx 11.11 + 64 = 75.11$.

Also, $(8.67)^2 \approx 75.16$.

Since $(A'B)^2 + (BC')^2 = (A'C')^2$, the new triangle must be right-angled by the Converse of Pythagoras Theorem.

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 $\triangle ABC$.

Alternate Solution:

The scale factor $\frac{5}{3}$ is an improper fraction (greater than 1), which indicates that the resulting triangle will be an enlarged version of the original triangle. We can verify the side lengths of the new triangle as follows:

By constructing a triangle with these specific dimensions, we would achieve the same result as the geometric parallel-line method.

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:

1. A rhombus $ABCD$ with side length $AB = 4$ cm.

2. $\angle ABC = 60^\circ$.

3. Rhombus is divided into two triangles $\triangle ABC$ and $\triangle ADC$.

4. $\triangle AB'C' \sim \triangle ABC$ with a scale factor of $\frac{2}{3}$.

5. $C'D' \parallel CD$ where $D'$ lies on $AD$.

To Find:

Is $AB'C'D'$ a rhombus? Provide reasons for the answer.

Construction:

1. Draw the rhombus $ABCD$ with $AB = 4$ cm and $\angle B = 60^\circ$.

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

3. Through $C'$, draw a line parallel to $CD$ to intersect $AD$ at $D'$.

Solution:

In the original rhombus $ABCD$:

In $\triangle ABC$, we have $AB = BC = 4$ cm and $\angle ABC = 60^\circ$. Since two sides are equal, the base angles are also equal.

Therefore, $\triangle ABC$ is an equilateral triangle. Thus, $AC = 4$ cm.

Similarly, since $AD = CD = 4$ cm and $\angle ADC = \angle ABC = 60^\circ$, $\triangle ADC$ is also an equilateral triangle.

Now, $\triangle AB'C' \sim \triangle ABC$ with a scale factor of $\frac{2}{3}$:

From this ratio, we calculate the lengths of the sides of $\triangle AB'C'$:

Since $C'D' \parallel CD$ and $D'$ lies on $AD$, in $\triangle ADC$, by the Basic Proportionality Theorem or similarity ($\triangle AC'D' \sim \triangle ACD$):

Using the values of $CD = 4$ cm and $AD = 4$ cm:

From equations (ii) and (iii), we observe the sides of the quadrilateral $AB'C'D'$:

Since all four sides of the quadrilateral $AB'C'D'$ are equal, it satisfies the definition of a rhombus.

Furthermore, because $B'C' \parallel BC$ and $BC \parallel AD$, it follows that $B'C' \parallel AD'$. Similarly, $AB' \parallel C'D'$ because $AB \parallel CD$.

Yes, $AB'C'D'$ is a rhombus.

Reasons:

1. Equality of Sides: By the property of similar triangles and the given scale factor, all four sides ($AB', B'C', C'D',$ and $D'A$) are equal to $\frac{8}{3}$ cm.

2. Parallelism: By construction and similarity, $B'C' \parallel BC$. In a rhombus, $BC \parallel AD$. Thus, $B'C' \parallel AD'$. Similarly, $C'D' \parallel CD$ and $AB \parallel CD$, implying $AB' \parallel C'D'$. A quadrilateral with all sides equal and opposite sides parallel is a rhombus.

Alternate Solution:

We can consider the homothety (scaling) with centre $A$. Since a rhombus is defined by its side length and one angle, scaling a rhombus by a factor $k$ around one of its vertices results in another rhombus. Here, the entire figure $ABCD$ is scaled by $k = \frac{2}{3}$ with respect to vertex $A$. Since the original figure $ABCD$ is a rhombus, the scaled figure $AB'C'D'$ must also be a rhombus with sides $k \times AB$.

Given:

1. Line segment $AB = 5$ cm.

2. Line segment $AC = 7$ cm.

3. Included angle $\angle BAC = 60^\circ$.

4. Point $P$ lies on $AB$ such that $AP = \frac{3}{4} AB$.

5. Point $Q$ lies on $AC$ such that $AQ = \frac{1}{4} AC$.

To Find:

Locate points $P$ and $Q$, join them, and find the length of $PQ$.

Construction Required:

1. Draw a line segment $AB = 5$ cm.

2. At point $A$, construct an angle $\angle BAC = 60^\circ$ using a protractor or compass and draw $AC = 7$ cm.

3. To locate P: Draw a ray $AX$ making an acute angle with $AB$. Mark 4 equal points $A_1, A_2, A_3, A_4$. Join $A_4B$ and draw a line through $A_3$ parallel to $A_4B$ to intersect $AB$ at $P$.

4. To locate Q: Draw a ray $AY$ making an acute angle with $AC$. Mark 4 equal points $A'_1, A'_2, A'_3, A'_4$. Join $A'_4C$ and draw a line through $A'_1$ parallel to $A'_4C$ to intersect $AC$ at $Q$.

5. Join $PQ$ and measure its length.

Solution:

First, we calculate the lengths of $AP$ and $AQ$ based on the given ratios:

In $\triangle APQ$, we have $AP = 3.75$ cm, $AQ = 1.75$ cm and $\angle PAQ = 60^\circ$. Using the Law of Cosines to find $PQ$:

Substituting the values:

Hence, the measured length of $PQ$ is 3.25 cm.

Alternate Solution:

The position of $P$ and $Q$ can also be found by direct measurement using a ruler. Since $AP = 3.75$ cm and $AQ = 1.75$ cm, one can mark these points directly from $A$ on the respective lines. After joining $P$ and $Q$, a ruler can be used to verify that the distance is indeed $3.25$ cm. The geometrical construction method using rays and parallel lines is used to ensure precision without relying on the markings of a standard ruler.

Given:

1. Parallelogram $ABCD$ where $BC = 5$ cm, $AB = 3$ cm and $\angle ABC = 60^\circ$.

2. Diagonal $BD$ divides it into $\triangle BCD$ and $\triangle ABD$.

3. Scale factor for similar triangle $\triangle BD'C'$ = $\frac{4}{3}$.

4. $D'A' \parallel DA$ where $A'$ lies on extended side $BA$.

To Find:

Locate the points, construct the figure, and determine if $A'BC'D'$ is a parallelogram.

Construction Required:

1. Draw a line segment $BC = 5$ cm.

2. At $B$, draw an angle $\angle XBC = 60^\circ$ and cut off $BA = 3$ cm. Complete the parallelogram $ABCD$ by drawing $AD \parallel BC$ and $CD \parallel AB$.

3. Join diagonal $BD$.

4. Draw a ray $BY$ making an acute angle with $BC$ on the opposite side of $D$. Mark 4 equal points $B_1, B_2, B_3, B_4$ on $BY$.

5. Join $B_3$ to $C$. Draw a line through $B_4$ parallel to $B_3C$ to intersect the extended line $BC$ at $C'$.

6. Through $C'$, draw a line parallel to $CD$ to intersect the extended diagonal $BD$ at $D'$. This gives $\triangle BD'C' \sim \triangle BDC$.

7. Through $D'$, draw a line parallel to $DA$ to intersect the extended line $BA$ at $A'$.

Solution:

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

Reasons:

In a quadrilateral, if both pairs of opposite sides are parallel, it is a parallelogram.

1. First pair of opposite sides ($A'D'$ and $BC'$):

From (i) and (ii), since $BC'$ is an extension of $BC$:

2. Second pair of opposite sides ($D'C'$ and $A'B$):

By construction, $\triangle BD'C' \sim \triangle BDC$ with scale factor $\frac{4}{3}$. In similar triangles, corresponding angles are equal:

Since these are alternate interior angles for lines $D'C'$ and $DC$ with transversal $BD$:

From (iii) and (iv), since $A'B$ is an extension of $AB$:

Since both pairs of opposite sides ($A'D' \parallel BC'$ and $D'C' \parallel A'B$) are parallel, quadrilateral $A'BC'D'$ is a parallelogram.

The sides of the new parallelogram are:

Alternate Solution:

A transformation that scales a figure from a point (homothety) preserves the shape and parallelism of the lines. Here, parallelogram $ABCD$ is scaled from point $B$ by a factor of $k = \frac{4}{3}$. Any scaling of a parallelogram results in another parallelogram. Since $A', C', D'$ are the scaled images of $A, C, D$ respectively from vertex $B$, the resulting figure $A'BC'D'$ must be a parallelogram by the property of geometric similarity.

Given:

1. Two concentric circles with common centre $O$.

2. Radius of the inner circle ($r$) = $3$ cm.

3. Radius of the outer circle ($R$) = $5$ cm.

To Construct:

A pair of tangents to the inner circle from a point $P$ lying on the outer circle.

Steps of Construction:

1. Draw two concentric circles with centre $O$ and radii $3$ cm and $5$ cm.

2. Take any point $P$ on the circumference of the outer circle.

3. Join $OP$ and construct its perpendicular bisector. Let it intersect $OP$ at point $M$.

4. Taking $M$ as the centre and $OM$ (or $MP$) as the radius, draw a helper circle.

5. Let this helper circle intersect the inner circle at two points, $Q$ and $R$.

6. Join $PQ$ and $PR$.

7. $PQ$ and $PR$ are the required pair of tangents from the outer circle to the inner circle.

Measurement:

On measuring the length of the tangents $PQ$ and $PR$ with a ruler, we find:

Verification by Actual Calculation:

Join $OQ$. In the inner circle, $OQ$ is the radius and $PQ$ is the tangent at point $Q$.

According to the theorem, the radius is perpendicular to the tangent at the point of contact.

Now, in right-angled $\triangle OQP$:

By Pythagoras Theorem:

Here, $OP$ is the radius of the outer circle and $OQ$ is the radius of the inner circle.

Substituting values from (i) and (ii) into the Pythagoras equation:

Since the measured length ($4$ cm) matches the calculated length ($4$ cm), the construction is verified.

Alternate Solution:

We can use the property of a chord of a larger circle touching the smaller circle. The tangent $PQ$ is actually part of a chord $QR'$ of the larger circle that is tangent to the smaller circle. In such a configuration, the length of the tangent is given by the formula:

$L = \sqrt{R^2 - r^2}$

Substituting $R = 5$ and $r = 3$:

$L = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \text{ cm}$.

This confirms that the geometry of the concentric circles naturally dictates a tangent length of exactly 4 cm.

Given:

1. In $\triangle ABC$, $AB = AC = 6 \text{ cm}$ and $BC = 5 \text{ cm}$.

2. $\triangle PQR \sim \triangle ABC$ such that $PQ = 8 \text{ cm}$.

To Construct:

A triangle $PQR$ similar to $\triangle ABC$ with side $PQ = 8 \text{ cm}$.

Construction Required:

First, we determine the scale factor. Since $PQ$ corresponds to $AB$:

Thus, we need to construct a triangle whose sides are $\frac{4}{3}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction:

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

2. With $B$ and $C$ as centres and radius $6 \text{ cm}$, draw two arcs intersecting at point $A$. Join $AB$ and $AC$ to form $\triangle ABC$.

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

4. Mark 4 equal points (the greater of 4 and 3) $B_1, B_2, B_3$ and $B_4$ on ray $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

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

6. Through $B_4$ (the numerator), draw a line parallel to $B_3C$ to intersect the extended line $BC$ at point $R$.

7. Through $R$, draw a line parallel to $CA$ to intersect the extended line $BA$ at point $P$.

8. Let point $B$ be renamed as $Q$. Then, $\triangle PQR$ is the required triangle.

Justification:

By construction, $B_4R \parallel B_3C$. In $\triangle BB_4R$, by the Basic Proportionality Theorem:

Also, by construction, $PR \parallel AC$. Therefore, $\triangle PBR \sim \triangle ABC$ by AA similarity criterion (as $\angle B$ is common and $\angle BRP = \angle BCA$ being corresponding angles).

Since $\triangle PBR \sim \triangle ABC$ and $B$ is same as $Q$, we have $\triangle PQR \sim \triangle ABC$. The ratio of their corresponding sides is:

Now, we verify the length of $PQ$:

Since $PQ = 8 \text{ cm}$ and $\triangle PQR \sim \triangle ABC$, the construction is justified.

Alternate Solution:

We can calculate the required lengths of the new triangle directly:

1. $PQ = 8 \text{ cm}$ (Given)

2. $QR = \frac{4}{3} \times BC = \frac{4}{3} \times 5 = \frac{20}{3} \approx 6.67 \text{ cm}$

3. $PR = \frac{4}{3} \times AC = \frac{4}{3} \times 6 = 8 \text{ cm}$

Since $PQ = PR = 8 \text{ cm}$, the resulting triangle $PQR$ is also an isosceles triangle. We can construct this triangle directly using these calculated side lengths by drawing a base $QR = 6.67 \text{ cm}$ and arcs of $8 \text{ cm}$ from $Q$ and $R$. This would yield the same result as the geometric scaling method.

Given:

1. In $\triangle ABC$, $AB = 5$ cm, $BC = 6$ cm and $\angle ABC = 60^\circ$.

2. Scale factor for the similar triangle = $\frac{5}{7}$.

To Construct:

A triangle $A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{5}{7}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction:

1. Draw a line segment $BC = 6$ cm.

2. At point $B$, draw a ray $BY$ making an angle of $60^\circ$ with $BC$.

3. From ray $BY$, cut off a line segment $BA = 5$ cm. Join $AC$ to form $\triangle ABC$.

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

5. Locate 7 points (since 7 is the greater of 5 and 7) $B_1, B_2, B_3, B_4, B_5, B_6$ and $B_7$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.

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

7. Through $B_5$ (the numerator), draw a line parallel to $B_7C$ to intersect $BC$ at point $C'$.

8. Through $C'$, draw a line parallel to $AC$ to intersect $AB$ at point $A'$.

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

Justification:

By construction, $B_5C' \parallel B_7C$. Therefore, in $\triangle BB_7C$, by the Basic Proportionality Theorem:

Now, in $\triangle ABC$ and $\triangle A'BC'$:

Since the triangles are similar, their corresponding sides are proportional:

Substituting the value from equation (i):

This shows that the sides of the newly constructed triangle are $\frac{5}{7}$ times the corresponding sides of the original triangle.

Alternate Solution:

The scale factor $\frac{5}{7}$ is a proper fraction (less than 1), which indicates that the resulting triangle $A'BC'$ will be smaller than the original triangle and will lie within it. One can also justify the construction by measuring the side lengths. If $BC = 6$ cm, then $BC'$ should measure:

Similarly, $A'B$ should measure:

Geometric construction via parallel lines as per the NCERT method ensures these ratios are maintained accurately without needing to perform manual division during construction.

Given:

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

2. Angle of inclination between the pair of tangents = $60^\circ$.

To Construct:

1. A pair of tangents to the circle such that the angle between them is $60^\circ$.

2. Measurement of the distance between the centre ($O$) and the point of intersection of tangents ($P$).

Construction Required:

To construct tangents inclined at $60^\circ$, the angle between the two radii at the points of contact must be the supplement of $60^\circ$.

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

Steps of Construction:

1. Draw a circle with centre $O$ and radius $4$ cm.

2. Draw any radius $OQ$.

3. At centre $O$, construct an angle $\angle QOR = 120^\circ$ such that $OR$ is another radius of the circle.

4. At point $Q$, construct a ray perpendicular to $OQ$ (using a compass to draw $90^\circ$).

5. At point $R$, construct a ray perpendicular to $OR$ (using a compass to draw $90^\circ$).

6. Let these two perpendicular rays intersect at point $P$.

7. $PQ$ and $PR$ are the required pair of tangents inclined at $60^\circ$.

Justification:

In quadrilateral $OQPR$:

Using the angle sum property of a quadrilateral:

This justifies that the tangents are inclined at $60^\circ$.

Measurement and Calculation:

On measuring the distance $OP$ with a ruler, we find:

Verification by Calculation:

Join $OP$. We know that the line joining the external point to the centre bisects the angle between the tangents.

In right-angled $\triangle OQP$:

The calculated value matches the measured value.

Alternate Solution:

Alternatively, we could first calculate the distance $OP$ using trigonometry ($OP = r \cdot \text{cosec } 30^\circ = 8 \text{ cm}$). Then, draw $OP = 8$ cm and construct the helper circle with $OP$ as the diameter. The points where the helper circle intersects the original circle will be the points of contact $Q$ and $R$. This method is useful if the angle between the radii is difficult to construct with a compass.

Given:

1. In $\triangle ABC$, $AB = 4$ cm, $BC = 6$ cm and $AC = 9$ cm.

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

To Construct:

A triangle $A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{3}{2}$ of the corresponding sides of $\triangle ABC$.

Steps of Construction:

1. Draw a line segment $BC = 6$ cm.

2. With $B$ as centre and radius $4$ cm, draw an arc. With $C$ as centre and radius $9$ cm, draw another arc intersecting the first arc at $A$. Join $AB$ and $AC$ to get $\triangle ABC$.

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

4. Along $BX$, mark 3 equal points (the greater of 3 and 2) $B_1, B_2$ and $B_3$ such that $BB_1 = B_1B_2 = B_2B_3$.

5. Join $B_2$ (the denominator) to $C$.

6. Through $B_3$ (the numerator), draw a line parallel to $B_2C$ to intersect the extended line $BC$ at point $C'$.

7. Through $C'$, draw a line parallel to $AC$ to intersect the extended line $BA$ at point $A'$.

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

Justification:

By construction, $B_3C' \parallel B_2C$. In $\triangle BB_3C'$, by the Basic Proportionality Theorem:

Since $A'C' \parallel AC$, $\triangle A'BC' \sim \triangle ABC$ by AA similarity criterion. Therefore, the corresponding sides are proportional:

Substituting the value from equation (i):

This shows that the sides of the new triangle are $1.5$ times the corresponding sides of $\triangle ABC$.

Analysis of Congruence:

To determine if the triangles are congruent, we compare their side lengths.

Sides of original $\triangle ABC$: 4 cm, 6 cm, and 9 cm.

Sides of the new $\triangle A'BC'$:

Comparing the two triangles:

1. All three angles are equal because the triangles are similar.

2. Two sides of the two triangles are equal: $BC'$ and $A'B$ of the new triangle are equal to $AC$ and $BC$ of the old triangle ($9$ cm and $6$ cm respectively).

3. However, the third sides are different ($4$ cm and $13.5$ cm).

For two triangles to be congruent, all corresponding sides must be equal. Since the corresponding side ratio is $1.5 \neq 1$, the triangles are not congruent.

Conclusion: No, the two triangles are not congruent.

Alternate Solution:

The scale factor $k$ determines congruence. If $k=1$, the triangles are congruent. If $k \neq 1$, the triangles are of different sizes. In this case, $k = 1.5$, which means the new triangle is a magnified version of the original. Even though the sets of side lengths share two common values $\{6, 9\}$, they do not describe identical triangles.