Chapter 10 Practical Geometry (Additional Questions)

The correct option is (A) Ruler and Compass.

Explanation:

To construct a line parallel to a given line through a point not on it using the alternate interior angle property, we need to copy the angle formed by the transversal and the given line at the point outside the line.

The method involves drawing a transversal line intersecting the given line and passing through the external point. Then, using a compass and ruler, we copy the alternate interior angle formed by the transversal and the given line at the external point. Drawing a line through the external point that forms this copied angle with the transversal results in a line parallel to the given line.

Why other options are not suitable for this specific method:

(B) Protractor and Set square: Set squares, often used with a ruler, are primarily for drawing parallel and perpendicular lines using sliding or aligning edges. While a protractor measures angles, the method specified relies on *copying* an angle using arcs, which is typically done with a compass, not measuring with a protractor.

(C) Divider: A divider is used to transfer lengths or measure distances, not angles or for constructing parallel lines based on angle properties.

(D) Only Ruler: A ruler can only draw straight lines. It is impossible to accurately construct a parallel line based on angle properties or copy angles using only a ruler.

The correct option is (B) 3.

Explanation:

To construct a unique triangle, you need to know at least three independent measurements. These measurements can be combinations of sides and angles, but they must be sufficient to fix the shape and size of the triangle uniquely.

The standard criteria for the congruence (and hence constructibility) of triangles are:

• SSS (Side-Side-Side): Knowing the lengths of all three sides.
• SAS (Side-Angle-Side): Knowing the lengths of two sides and the measure of the angle included between them.
• ASA (Angle-Side-Angle): Knowing the measures of two angles and the length of the side included between them.
• AAS (Angle-Angle-Side): Knowing the measures of two angles and the length of a non-included side. (This is equivalent to ASA because if you know two angles, the third angle is fixed since the sum of angles in a triangle is $180^\circ$).
• RHS (Right-angle-Hypotenuse-Side): For a right-angled triangle, knowing the length of the hypotenuse and one side.

In all these cases, the number of independent measurements is three.

Knowing fewer than three measurements (e.g., two sides, two angles, or one side and one angle) is generally not enough to construct a unique triangle. For instance, knowing two sides and a non-included angle (SSA) can sometimes result in two possible triangles (the ambiguous case).

The correct option is (C) ASA.

Explanation:

To construct a unique triangle, we need a set of measurements that fixes both the shape and size of the triangle. The criteria that guarantee the construction of a unique triangle are the triangle congruence criteria:

• SSS (Side-Side-Side)
• SAS (Side-Angle-Side)
• ASA (Angle-Side-Angle)
• AAS (Angle-Angle-Side)
• RHS (Right-angle-Hypotenuse-Side) for right triangles

Let's examine the given options:

(A) SSA (Side-Side-Angle): This criterion is also known as ASS. If you are given two sides and a non-included angle, there can be zero, one, or two distinct triangles that can be formed. This is the ambiguous case and does not guarantee a unique triangle.

(B) AAA (Angle-Angle-Angle): Knowing all three angles determines the shape of the triangle but not its size. Infinitely many triangles can have the same three angles (they would be similar triangles). Therefore, AAA does not construct a unique triangle.

(C) ASA (Angle-Side-Angle): If you are given two angles and the side included between them, this information uniquely determines the triangle. This is a valid congruence criterion, meaning any two triangles with these measurements are congruent, and thus only one unique triangle can be constructed.

(D) ASS (Angle-Side-Side): This is the same as SSA and refers to the ambiguous case. It does not guarantee a unique triangle.

Therefore, among the given options, only ASA can be used to construct a unique triangle.

The correct option is (D) Any one side.

Explanation:

To construct a triangle when the lengths of all three sides (SSS criterion) are given, the standard construction method is as follows:

1. Draw a line segment equal to the length of any one of the given sides.

2. From one endpoint of this line segment, use a compass to draw an arc with a radius equal to the length of a second side.

3. From the other endpoint of the initial line segment, use a compass to draw another arc with a radius equal to the length of the third side.

4. The point where the two arcs intersect is the third vertex of the triangle.

5. Connect the endpoints of the initial line segment to the point of intersection to complete the triangle.

This procedure works correctly regardless of which side is chosen as the first line segment, provided that the sum of the lengths of any two sides is greater than the length of the third side (the triangle inequality theorem). While some may suggest drawing the longest side first for practical reasons (e.g., to keep the construction more compact), it is not a geometric necessity for the construction to be possible or accurate. Therefore, you can start by drawing any one of the three given sides.

The correct option is (B) No.

Explanation:

To determine if a triangle can be constructed with given side lengths, we use the Triangle Inequality Theorem.

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Let the given side lengths be $a = 3\text{ cm}$, $b = 4\text{ cm}$, and $c = 8\text{ cm}$.

We need to check if the following three inequalities hold true:

$a + b > c$

$a + c > b$

$b + c > a$

Let's check the first inequality:

$3\text{ cm} + 4\text{ cm} > 8\text{ cm}$

$7\text{ cm} > 8\text{ cm}$

This statement is False.

Since the sum of the two shorter sides ($3 + 4 = 7$) is not greater than the longest side ($8$), a triangle cannot be formed with these side lengths.

Therefore, a triangle with side lengths 3 cm, 4 cm, and 8 cm cannot be constructed.

The correct option is (A) Draw a side, draw an angle, draw the other side from the angle vertex.

Explanation:

The SAS (Side-Angle-Side) criterion for constructing a triangle requires you to be given the lengths of two sides and the measure of the angle that is included, meaning it is between those two sides.

A standard and common sequence of steps for constructing a triangle using the SAS criterion is as follows:

1. Draw a line segment equal to the length of one of the given sides. Let's call this segment AB.

2. At one endpoint of this segment (say, point A), construct an angle that has the measure of the given included angle. This point A will be the vertex of the included angle.

3. Along the ray of the angle just constructed, starting from the vertex (point A), draw the other side by marking a point C such that the segment AC has the length of the second given side.

4. Finally, join the other endpoint of the first segment (point B) to the endpoint of the second side (point C) to complete the triangle ABC.

This sequence accurately follows the steps described in option (A).

Let's examine the other options:

(B) "Draw an angle, draw the two sides from the vertex of the angle, join the endpoints of the sides." - This also describes a valid method for SAS construction (drawing the angle first, then marking the sides along the rays), but option (A) represents a widely taught approach that starts by establishing one side as a base.

(C) "Draw the included angle, then draw the two sides." - This sequence is incomplete and does not provide enough detail on how the sides relate to the angle.

(D) "Draw one side, draw the other side, then draw the included angle." - This is not a practical or correct sequence for construction using standard tools. You cannot typically 'draw' a specific angle after the two adjacent sides have already been positioned; the angle is determined by their relative position.

Therefore, option (A) provides a clear and correct sequence for constructing a triangle given two sides and the included angle.

The correct option is (B) The included side.

Explanation:

The ASA (Angle-Side-Angle) criterion for constructing a triangle means you are given the measure of two angles and the length of the side that lies between those two angles.

The standard and most efficient way to construct a triangle using the ASA criterion is as follows:

1. Draw a line segment equal to the length of the given included side. Let's call this segment AB.

2. At one endpoint of this segment (say, point A), construct the first given angle.

3. At the other endpoint of the segment (point B), construct the second given angle.

4. The rays extending from A and B for these two angles will intersect at a single point, which is the third vertex of the triangle (let's call it C).

5. The triangle ABC is the unique triangle with the given measurements.

Starting with the included side provides a base of the correct length and position on which to construct the two angles, ensuring they are indeed "included" by that specific side.

Let's consider why other options are not typically the first step:

(A) One of the angles: If you start by drawing an angle, you don't have a specific length established for the included side, making it difficult to proceed to the next steps accurately.

(C) The third angle: While the third angle is determined by the sum of angles in a triangle ($180^\circ$), starting with it doesn't directly use the given included side length to establish the triangle's scale or position.

(D) Any angle: Similar to (A), starting with an angle without the included side length doesn't set up the construction effectively.

Therefore, drawing the included side is the essential first step in constructing a triangle using the ASA criterion.

The correct option is (A) Yes, a unique triangle.

Explanation:

First, let's check if a triangle can be formed with the given angles. The sum of the angles in any triangle must be $180^\circ$.

Sum of angles $= 180^\circ$

Since the sum of the given angles is $180^\circ$, these angles can form a triangle. This eliminates option (C).

Next, we need to determine if the triangle is unique. When only the three angles of a triangle are given (AAA criterion), infinitely many similar triangles can be constructed, differing in size but having the same angles. This would correspond to option (B).

However, the question also provides a side length of 5 cm along with the three angles. When you are given the angles of a triangle (which sum to $180^\circ$) and the length of one side, the triangle is uniquely determined. This falls under the ASA (Angle-Side-Angle) or AAS (Angle-Angle-Side) congruence criteria. If the 5 cm side is included between two of the angles, it's ASA. If the 5 cm side is opposite one of the angles, it's AAS.

Since ASA and AAS are criteria for congruence, they guarantee that any two triangles constructed with these measurements will be congruent, meaning they are identical in shape and size. Therefore, a unique triangle can be constructed.

For example, if the 5 cm side is between the $50^\circ$ and $60^\circ$ angles, we can construct the 5 cm segment, then draw the $50^\circ$ angle at one end and the $60^\circ$ angle at the other end. The intersection of the two rays will form the unique third vertex.

The correct option is (D) Right-angled.

Explanation:

The RHS congruence criterion stands for:

• R - Right Angle ($90^\circ$)
• H - Hypotenuse (the side opposite the right angle)
• S - Side (one of the other two sides)

This criterion is a special case of the SAS or SSA criteria that applies specifically to right-angled triangles. If the hypotenuse and one leg (side) of a right-angled triangle are equal to the hypotenuse and one leg of another right-angled triangle, then the two triangles are congruent.

Therefore, the RHS criterion is inherently used for establishing the congruence (and thus constructibility) of right-angled triangles based on the length of their hypotenuse and one other side.

The correct option is (B) The leg of length 3 cm.

Explanation:

To construct a right-angled triangle using the RHS (Right-angle, Hypotenuse, Side) criterion with a known hypotenuse and one known leg, the most common and straightforward method is:

1. Draw a line segment equal to the length of the given leg (3 cm in this case).

2. At one endpoint of this line segment, construct a perpendicular line or ray (this forms the right angle).

3. From the other endpoint of the initial leg segment, draw an arc with a radius equal to the length of the hypotenuse (5 cm in this case).

4. The point where this arc intersects the perpendicular ray is the third vertex of the triangle.

5. Join the endpoints to complete the triangle.

Starting by drawing the known leg provides a base from which you can accurately construct the right angle at one end and locate the third vertex by swinging an arc from the other end using the hypotenuse length.

The correct option is (C) The sum of the lengths of any two sides must be greater than the third side.

Explanation:

Let's evaluate each statement:

(A) Knowing all three angles is enough to construct a unique triangle.

This statement is False. Knowing all three angles (AAA criterion) determines the shape of a triangle, but not its size. Infinitely many triangles can have the same angles; these are called similar triangles. To construct a unique triangle, you need at least one side length along with the angles (e.g., ASA or AAS).

(B) Knowing two sides and a non-included angle (SSA) always gives a unique triangle.

This statement is False. This is known as the ambiguous case (SSA or ASS). Given two side lengths and an angle that is not between them, it is possible to construct zero, one, or two different triangles, depending on the specific measurements. It does not always result in a unique triangle.

(C) The sum of the lengths of any two sides must be greater than the third side.

This statement is True. This is the fundamental principle of the Triangle Inequality Theorem. For any three line segments to form a triangle, the sum of the lengths of any two sides must be strictly greater than the length of the remaining side. If this condition is not met, the two shorter sides would not be long enough to "reach" each other and form a closed figure with the longest side.

(D) The sum of the angles is not necessarily $180^\circ$.

This statement is False in standard Euclidean geometry, which is typically assumed unless otherwise specified. In Euclidean geometry, the sum of the interior angles of any triangle is always exactly $180^\circ$ ($\pi$ radians).

Therefore, the only true statement among the options is (C).

The correct option is (C) Protractor.

Explanation:

Let's look at the function of each geometric instrument listed:

(A) Ruler: A ruler is a tool used for drawing straight lines and measuring lengths or distances.

(B) Compass: A compass is used for drawing circles or arcs and for transferring measurements (lengths).

(C) Protractor: A protractor is a tool, usually made of transparent plastic or metal, used for measuring angles in degrees or radians, and for drawing angles of a specific size.

(D) Divider: A divider is similar to a compass but has two points instead of a pencil. It is used to transfer distances or compare measurements, but not to measure or draw angles.

Therefore, the instrument specifically designed and used for measuring angles is the Protractor.

The correct option is (C) SSS.

Explanation:

The criteria for constructing a unique triangle are based on combinations of side lengths and angle measures.

• SAS (Side-Angle-Side) requires two side lengths and the measure of the angle included between them.
• ASA (Angle-Side-Angle) requires two angle measures and the length of the side included between them.
• SSS (Side-Side-Side) requires the lengths of all three sides.
• RHS (Right-angle-Hypotenuse-Side) is a special criterion for right-angled triangles, requiring the hypotenuse and one other side.

In the given problem, we are provided with the lengths of all three sides: 5 cm, 6 cm, and 7 cm.

This set of given information directly corresponds to the SSS (Side-Side-Side) criterion.

Provided the triangle inequality theorem holds (which it does here: $5+6>7$, $5+7>6$, $6+7>5$), knowing the lengths of all three sides is sufficient to construct a unique triangle.

The correct option is (C) SAS.

Explanation:

We are given the following information to construct triangle ABC:

• Length of side AB = $6\text{ cm}$
• Measure of angle $\angle\text{A} = 50^\circ$
• Length of side AC = $7\text{ cm}$

Let's analyze the given information in relation to the vertices and sides of the triangle. We have two side lengths, AB and AC, and the angle $\angle\text{A}$ which is the angle formed at the vertex A where the sides AB and AC meet.

Since the given angle ($\angle\text{A}$) is located between the two given sides (AB and AC), this corresponds to the SAS (Side-Angle-Side) criterion for constructing a triangle.

To construct the triangle using the SAS criterion with these measurements, you would:

1. Draw a line segment AB of length $6\text{ cm}$.

2. At point A, construct an angle of $50^\circ$.

3. Along the ray of the $50^\circ$ angle from A, mark a point C such that AC is $7\text{ cm}$ long.

4. Join points B and C to form the triangle ABC.

Let's briefly consider why other options do not apply:

(A) ASA (Angle-Side-Angle): Requires two angles and the side included between them. We are given two sides and one angle.

(B) AAS (Angle-Angle-Side): Requires two angles and a non-included side. We are given two sides and one angle.

(D) SSS (Side-Side-Side): Requires the lengths of all three sides. We are given only two sides.

Therefore, the appropriate criterion for this construction is SAS.

The correct option is (A) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c).

Explanation:

Let's match each construction criterion with the measurements it requires:

(i) SSS (Side-Side-Side): This criterion means you are given the lengths of all three sides of the triangle. This matches with (b) Three sides.

(ii) SAS (Side-Angle-Side): This criterion means you are given the lengths of two sides and the measure of the angle that is included between those two sides. This matches with (d) Two sides and included angle.

(iii) ASA (Angle-Side-Angle): This criterion means you are given the measures of two angles and the length of the side that is included between those two angles. This is a specific instance of (a) Two angles and one side.

(iv) RHS (Right-angle-Hypotenuse-Side): This criterion is specifically for right-angled triangles. It means you are given the presence of a right angle, the length of the hypotenuse, and the length of one other side (a leg). This matches with (c) Right angle, hypotenuse, and one side.

Based on these matches, the correct pairing is:

• (i) - (b)
• (ii) - (d)
• (iii) - (a)
• (iv) - (c)

This corresponds to option (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Explanation:

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): A triangle cannot be constructed with side lengths 2 cm, 3 cm, and 5 cm.

To check if a triangle can be constructed with given side lengths, we apply the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side.

Let the side lengths be $a=2\text{ cm}$, $b=3\text{ cm}$, and $c=5\text{ cm}$. We check the three inequalities:

1. $a + b > c \implies 2 + 3 > 5 \implies 5 > 5$. This is False.

2. $a + c > b \implies 2 + 5 > 3 \implies 7 > 3$. This is True.

3. $b + c > a \implies 3 + 5 > 2 \implies 8 > 2$. This is True.

Since the first condition ($2+3 > 5$) is not met (specifically, $2+3 = 5$), a triangle cannot be formed with these side lengths. Thus, Assertion (A) is True.

Reason (R): The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

This statement is the precise definition of the Triangle Inequality Theorem, which is a fundamental principle in geometry governing the lengths of the sides of any triangle.

Thus, Reason (R) is True.

Now let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that a specific set of side lengths cannot form a triangle. Reason (R) provides the general rule (the Triangle Inequality Theorem) that dictates whether any set of three lengths can form a triangle. The reason why the lengths 2 cm, 3 cm, and 5 cm cannot form a triangle is *because* they violate the condition stated in Reason (R) (specifically, the sum of two sides is not greater than the third side, but equal to it in this case). Therefore, Reason (R) directly explains why Assertion (A) is true.

Conclusion:

• Assertion (A) is True.
• Reason (R) is True.
• Reason (R) is the correct explanation of Assertion (A).

This matches option (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Explanation:

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): You can construct a unique triangle if you are given two angles and any side.

This statement is True. If you are given two angles of a triangle and the length of one side, you can construct a unique triangle. This falls under either the ASA (Angle-Side-Angle) or AAS (Angle-Angle-Side) congruence criteria, both of which guarantee a unique triangle.

For example, if you are given angles $\angle A$ and $\angle B$, and side AC. You can find $\angle C = 180^\circ - (\angle A + \angle B)$. Now you have angles $\angle A$ and $\angle C$ and the included side AC. This is an ASA case.

Reason (R): If two angles are given, the third angle is automatically determined by the angle sum property, reducing it to the ASA case.

This statement is also True. The angle sum property of a triangle states that the sum of the interior angles is $180^\circ$. If you know two angles, say $\angle A$ and $\angle B$, then the third angle, $\angle C$, can be calculated as:

Knowing the third angle means that if you are given any side length in addition to the two initial angles, you can always identify which two angles are at the endpoints of that side. Thus, the problem reduces to knowing two angles and the included side, which is the ASA criterion. The AAS criterion is also equivalent to ASA because if you know two angles and a non-included side, you can find the third angle and then treat it as an ASA case.

Reason (R) provides a correct explanation for why Assertion (A) is true. Knowing two angles implies knowing all three angles. When you combine this with the knowledge of one side length, you effectively have the information required by the ASA (or AAS) criterion, which allows for the construction of a unique triangle.

Therefore, both the Assertion and the Reason are true, and the Reason is the correct explanation for the Assertion.

The correct option is (C) SAS.

Explanation:

In the given problem, the student needs to construct a triangle PQR with the following measurements:

• Side PQ = $5\text{ cm}$
• Side QR = $4\text{ cm}$
• Angle $\angle\text{Q} = 60^\circ$

We are given the lengths of two sides (PQ and QR) and the measure of the angle included between these two sides ($\angle\text{Q}$ is the angle formed at vertex Q by sides PQ and QR).

This set of information corresponds exactly to the SAS (Side-Angle-Side) criterion for constructing a triangle, which requires knowing two side lengths and the measure of the angle included between them.

The construction would involve drawing a segment for one side (say, PQ = 5 cm), constructing the angle at the common vertex ( $\angle\text{Q} = 60^\circ$) at one endpoint (Q), and then marking the length of the other side (QR = 4 cm) along the ray of the angle from Q.

The correct option is (A) Draw line segment PQ = 5 cm.

Explanation:

The problem provides the lengths of two sides (PQ = 5 cm and QR = 4 cm) and the measure of the angle included between them ($\angle\text{Q} = 60^\circ$). This is a SAS (Side-Angle-Side) construction scenario.

A common and systematic approach for constructing a triangle using the SAS criterion is as follows:

1. Draw a line segment equal to the length of one of the given sides. This segment will form one side of the triangle.

2. At the endpoint corresponding to the vertex of the included angle, construct the given angle.

3. Along the ray of the constructed angle, mark the length of the other given side from the vertex.

4. Join the endpoints to complete the triangle.

Based on this sequence, the very first step is to draw one of the line segments representing a side. Options (A) and (B) both represent drawing a side. While drawing either PQ or QR first is a valid initial step, option (A) suggests drawing the side PQ of length 5 cm. This aligns with the standard method of laying down one of the known sides as a base.

Option (C) suggests drawing the angle first. While technically possible by drawing a point and two rays, it is less conventional as the *first* step compared to establishing a side segment first, especially in introductory constructions.

Option (D) is a helpful preliminary step for planning but is not the actual first step of the geometrical construction process itself using tools.

Therefore, drawing one of the given sides, such as line segment PQ = 5 cm, is the appropriate first step for this SAS construction.

The correct option is (C) Consecutive interior angles.

Explanation:

The question specifies using the property that the sum of interior angles on the same side of a transversal is $180^\circ$ to construct a parallel line. This property is the defining characteristic of consecutive interior angles (also known as co-interior angles) formed when a transversal intersects two parallel lines.

To construct a line parallel to a given line 'l' through a point 'P' not on 'l' using this property, you would:

1. Draw a transversal line through P that intersects line 'l'.

2. Identify or measure the interior angle formed on one side of the transversal between the transversal and line 'l'.

3. At point P, on the same side of the transversal as the angle identified in step 2, construct an angle such that its measure, when added to the measure of the angle on line 'l', sums to $180^\circ$. This new angle constructed at P will be the consecutive interior angle to the one on line 'l'.

4. Draw a line through P along the ray that forms this new angle. This line will be parallel to 'l'.

This method directly utilizes the property of consecutive interior angles.

Let's briefly review the other angle types mentioned:

(A) Alternate interior angles: If alternate interior angles are equal, the lines are parallel. Constructing using this involves copying an alternate interior angle at point P.

(B) Corresponding angles: If corresponding angles are equal, the lines are parallel. Constructing using this involves copying a corresponding angle at point P.

(D) Vertically opposite angles: These are formed when two lines intersect and are always equal. They are not used to establish parallelism between two separate lines intersected by a transversal.

Since the question specifically mentions the property of interior angles on the same side of the transversal summing to $180^\circ$, the construction method relies on consecutive interior angles.

The correct option is (D) One side and two angles.

Explanation:

To construct a unique triangle, you need sufficient independent measurements that fix its size and shape. The standard criteria that guarantee the construction of a unique triangle are:

• SSS (Side-Side-Side): You know the lengths of all three sides.
• SAS (Side-Angle-Side): You know the lengths of two sides and the measure of the angle included between them.
• ASA (Angle-Side-Angle): You know the measures of two angles and the length of the side included between them.
• AAS (Angle-Angle-Side): You know the measures of two angles and the length of a non-included side.
• RHS (Right-angle-Hypotenuse-Side): For right triangles, you know the hypotenuse and one side.

The statement provided lists SSS ("three sides") and SAS ("two sides and the included angle"). The blank needs to cover the other general criteria for unique triangle construction involving angles and sides.

Let's evaluate the options:

(A) Two angles and the included side: This describes the ASA criterion, which is a valid criterion for unique construction. This is a correct option, but let's check others.

(B) All three angles: Knowing all three angles (AAA) determines the shape but not the size of the triangle, leading to infinitely many similar triangles. This does not result in a unique triangle.

(C) Two sides and any angle: This describes the SSA (or ASS) case, which is the ambiguous case. It does not always result in a unique triangle; there can be zero, one, or two possible triangles.

(D) One side and two angles: If you know two angles of a triangle, you automatically know the third angle because the sum of angles in a triangle is $180^\circ$. Therefore, knowing two angles and *any* side (whether included between the two known angles or opposite one of them) provides enough information to construct a unique triangle. This general description covers both the ASA and AAS criteria.

Given that the statement lists SSS and SAS, the most comprehensive way to complete the statement by including the remaining angle/side combinations that guarantee uniqueness is the description that covers both ASA and AAS. Option (D) "One side and two angles" does this effectively.

The correct option is (D) Compass.

Explanation:

Let's look at the purpose of each geometric instrument listed:

(A) Ruler: Used for drawing straight lines and measuring linear distances.

(B) Divider: Used for transferring measurements (distances) between points or comparing lengths. It has two points and no pencil lead.

(C) Set square: Used for drawing specific angles (like $90^\circ, 45^\circ, 30^\circ, 60^\circ$) and for drawing parallel or perpendicular lines when used in conjunction with a ruler.

(D) Compass: An instrument with two legs, one with a point and the other usually with a pencil or pen holder. It is specifically used for drawing circles and arcs of varying radii, and also for transferring lengths.

Therefore, the compass is the essential tool for drawing arcs and circles in geometrical constructions.

The correct option is (C) Included between the $40^\circ$ and $80^\circ$ angles.

Explanation:

The ASA construction criterion stands for Angle-Side-Angle.

This criterion states that a unique triangle can be constructed if the measures of two angles and the length of the side included between them are known.

In this question, the two given angles are $40^\circ$ and $80^\circ$. For the ASA criterion, the side whose length is given must be the side that lies between the vertices of these two angles.

Therefore, to construct a unique triangle using the ASA criterion with angles $40^\circ$ and $80^\circ$, you must be given the length of the side that is included between the $40^\circ$ and $80^\circ$ angles.

Let's consider the other options in the context of the ASA criterion:

(A) Opposite to the $40^\circ$ angle: This side is not included between the $40^\circ$ and $80^\circ$ angles.

(B) Opposite to the $80^\circ$ angle: This side is also not included between the $40^\circ$ and $80^\circ$ angles.

(D) Any side: While knowing any side along with two angles is sufficient to construct a unique triangle (this falls under the AAS criterion, which is equivalent to ASA), the question specifically asks for the condition required by the ASA criterion itself. The ASA criterion strictly requires the included side.

Thus, the side must be the one included between the two given angles.

The correct option is (B) Triangle inequality property.

Explanation:

In the SSS (Side-Side-Side) construction of a triangle, you are given the lengths of the three sides. The standard procedure involves:

1. Drawing a line segment equal to the length of one side (let's call this the base).

2. From one endpoint of the base, drawing an arc with a radius equal to the length of the second side.

3. From the other endpoint of the base, drawing an arc with a radius equal to the length of the third side.

4. The point where these two arcs intersect is the location of the third vertex of the triangle.

The success of this construction hinges on whether these two arcs actually intersect. The condition that guarantees the arcs will intersect and form a triangle is the Triangle Inequality Theorem.

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

When you draw arcs with radii equal to the lengths of the two remaining sides from the endpoints of the base, the point of intersection exists only if the sum of the radii of these two arcs is greater than the length of the base. If the sum of the radii is less than or equal to the base length, the arcs will not intersect, and no triangle can be formed.

Therefore, the underlying property that dictates whether the SSS construction using arcs is possible is the Triangle Inequality Property.

The correct option is (B) Infinitely many.

Explanation:

We are given the measures of the three angles of a triangle as $60^\circ, 60^\circ, 60^\circ$.

First, let's check if these angles can form a triangle using the angle sum property:

Sum of angles $= 180^\circ$

Since the sum of the angles is $180^\circ$, a triangle with these angles can be formed.

The criterion given is AAA (Angle-Angle-Angle). Knowing only the three angles of a triangle is sufficient to determine the shape of the triangle (in this case, it is an equilateral triangle, as all angles are equal).

However, the AAA criterion is a condition for similarity, not congruence. This means that any two triangles with angles $60^\circ, 60^\circ, 60^\circ$ will be similar, having the same shape but potentially different sizes.

We can construct an equilateral triangle with side length 1 cm, another with side length 2 cm, another with side length 10 cm, and so on, for any positive side length. All these triangles will have angles $60^\circ, 60^\circ, 60^\circ$.

Therefore, infinitely many different triangles (of different sizes) can be constructed with the angles $60^\circ, 60^\circ, 60^\circ$.

The correct option is (D) RHS.

Explanation:

We are given the following information for triangle ABC:

• $\angle\text{B} = 90^\circ$ (This means it is a right-angled triangle)
• BC = $8\text{ cm}$ (This is a leg or side adjacent to the right angle)
• AC = $10\text{ cm}$ (This is the hypotenuse, as it is opposite the right angle at B)

We have a Right angle, the length of the Hypotenuse, and the length of one Side (leg).

This combination of given information directly corresponds to the RHS (Right-angle, Hypotenuse, Side) criterion for constructing a unique right-angled triangle.

Let's briefly look at why other options are not the primary suitable criteria based on the given information:

(A) SSS (Side-Side-Side): Requires all three side lengths. We only have two side lengths given initially (BC and AC). We could calculate AB using the Pythagorean theorem ($AB = \sqrt{AC^2 - BC^2}$), but the criterion is based on the initially given information.

(B) SAS (Side-Angle-Side): Requires two side lengths and the *included* angle. We have sides BC and AC, and angle $\angle\text{B}$, but $\angle\text{B}$ is included between sides AB and BC. We don't know the length of AB from the start.

(C) ASA (Angle-Side-Angle): Requires two angles and the included side. We only have one angle given initially ($\angle\text{B}$).

Therefore, the most direct and suitable construction criterion based on the provided measurements ($\angle\text{B} = 90^\circ$, BC, and AC) is RHS.

The correct option is (C) Ruler (or Straight edge).

Explanation:

Let's review the primary uses of each geometrical instrument listed:

• (A) Compass: Used to draw circles and arcs, and to transfer lengths.
• (B) Protractor: Used to measure and draw angles.
• (C) Ruler (or Straight edge): A ruler is a tool used for drawing straight lines and for measuring the length of segments based on marked units. A straight edge is a tool used solely for drawing straight lines without measurement marks. The question asks for an instrument used for *both* drawing straight lines and measuring lengths, which is the primary function of a ruler.
• (D) Divider: Used to transfer distances or compare lengths. It does not draw lines or measure lengths directly on a scale.

Based on their functions, the instrument used to draw straight lines and measure lengths is the Ruler.

The correct option is (A) $\angle\text{E}$.

Explanation:

The problem requires constructing triangle DEF using the ASA (Angle-Side-Angle) criterion. The ASA criterion states that a unique triangle can be constructed if you know the measures of two angles and the length of the side included between them.

We are given:

• Side DE = $7\text{ cm}$
• Angle $\angle\text{D} = 70^\circ$
• Angle $\angle\text{F} = 60^\circ$

For side DE to be the included side in the ASA criterion, we need to know the measures of the angles at its endpoints, which are $\angle\text{D}$ and $\angle\text{E}$. We are given $\angle\text{D}$ ($70^\circ$), but we are given $\angle\text{F}$ ($60^\circ$), not $\angle\text{E}$.

However, we can find the measure of the third angle, $\angle\text{E}$, using the angle sum property of a triangle, which states that the sum of the interior angles of a triangle is $180^\circ$.

Substituting the given values:

$130^\circ + \angle\text{E} = 180^\circ$

$\angle\text{E} = 50^\circ$

Once we have found $\angle\text{E} = 50^\circ$, we have the following information:

• $\angle\text{D} = 70^\circ$
• Side DE = $7\text{ cm}$
• $\angle\text{E} = 50^\circ$

Now, we have two angles ($\angle\text{D}$ and $\angle\text{E}$) and the side included between them (DE). This set of information matches the ASA criterion.

Therefore, before using the ASA criterion with the given side DE, you need to find the measure of $\angle\text{E}$.

The correct option is (C) Corresponding angles.

Explanation:

When a transversal intersects two lines, several pairs of angles are formed. Specific relationships between certain pairs of these angles indicate whether the two lines are parallel.

The properties that guarantee the two lines are parallel are:

• If the alternate interior angles are equal.
• If the corresponding angles are equal.
• If the consecutive interior angles (interior angles on the same side of the transversal) are supplementary (their sum is $180^\circ$).

Let's look at the given options:

• (A) Vertically opposite angles: These are always equal when two lines intersect, regardless of whether the lines are parallel or not. So, equality of vertically opposite angles does not guarantee parallelism.
• (B) Adjacent angles on a straight line: These are always supplementary (sum to $180^\circ$). This property applies to angles forming a linear pair on any straight line and does not indicate whether two separate lines are parallel.
• (C) Corresponding angles: Corresponding angles are in the same relative position at each intersection. If the corresponding angles formed by a transversal intersecting two lines are equal, then the two lines are parallel. This is a fundamental theorem used to prove lines are parallel and for constructing parallel lines.
• (D) Complementary angles: Two angles are complementary if their sum is $90^\circ$. This concept is not directly related to the conditions for parallel lines formed by a transversal.

Therefore, among the given options, the equality of corresponding angles is a property that guarantees the lines intersected by the transversal are parallel.

The correct option is (C) SSA.

Explanation:

A unique triangle can be constructed if the given measurements correspond to one of the triangle congruence criteria. These criteria guarantee that only one specific triangle can be formed with those measurements.

Let's review the criteria:

• (A) SSS (Side-Side-Side): If the lengths of three sides are given and they satisfy the triangle inequality theorem, a unique triangle can be constructed.
• (B) ASA (Angle-Side-Angle): If the measures of two angles and the length of the included side are given, a unique triangle can be constructed.
• (C) SSA (Side-Side-Angle): If the lengths of two sides and the measure of a non-included angle are given, this is known as the ambiguous case. This information does not always lead to a unique triangle. Depending on the specific values, there might be zero, one, or two possible triangles that can be constructed. Therefore, SSA is NOT always sufficient to construct a unique triangle.
• (D) SAS (Side-Angle-Side): If the lengths of two sides and the measure of the included angle are given, a unique triangle can be constructed.

The criterion that does not always guarantee the construction of a unique triangle is SSA.

The correct option is (D) All of the above.

Explanation:

The Triangle Inequality Theorem is a fundamental condition that must be satisfied for any three given line segments to form a triangle. It states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.

If the side lengths of a potential triangle are $a, b,$ and $c$, then the Triangle Inequality Theorem requires that all three of the following inequalities must be true simultaneously:

If even one of these inequalities is false (i.e., the sum of two sides is less than or equal to the third side), the three segments cannot form a triangle.

Therefore, for side lengths $a, b, c$ to form a triangle, it must be true that $a+b > c$, $a+c > b$, AND $b+c > a$. This is encompassed by option (D) "All of the above".

The correct option is (B) Ruler and Compass.

Explanation:

The method described involves constructing a line parallel to a given line 'l' through an external point 'P' by creating equal alternate interior angles.

The steps are:

1. Draw a transversal line that passes through point P and intersects line 'l' at a point Q.

2. Identify the alternate interior angle formed by the transversal and line 'l' at point Q.

3. The crucial step is to copy this angle at point P on the opposite side of the transversal such that it becomes the alternate interior angle at P.

To accurately copy an angle in standard geometrical constructions without measuring its degree value, you use a compass and ruler (straight edge). The process involves drawing an arc from the vertex of the original angle, measuring the distance between the points where the arc crosses the angle's rays using the compass, and then replicating this arc and distance from point P to define the ray that forms the copied angle.

Let's consider the other options:

(A) Protractor: A protractor is used to measure angles in degrees and to draw angles of specific degree measures. While you *could* measure the angle at Q with a protractor and then draw the same angle at P, the classical geometric construction method for copying an angle relies on compass and ruler, which is often preferred for accuracy in pure construction exercises and does not require reading degree values.

(C) Set squares: Set squares are useful for drawing specific angles (like $90^\circ, 45^\circ, 30^\circ, 60^\circ$) and for drawing parallel or perpendicular lines using sliding techniques, but they are not designed for copying an arbitrary angle measure.

(D) Divider: A divider is used to transfer or compare linear distances, not angle measures.

Therefore, copying an angle in geometrical construction is typically done using a ruler and compass.

The correct option is (A) Sides: 5 cm, 5 cm, 10 cm.

Explanation:

To determine if a set of measurements can form a triangle, we use either the Triangle Inequality Theorem (for sides) or the Angle Sum Property (for angles).

Let's check each option:

(A) Sides: 5 cm, 5 cm, 10 cm

Apply the Triangle Inequality Theorem: The sum of the lengths of any two sides must be strictly greater than the length of the third side.

$5 + 5 > 10 \implies 10 > 10$ (This is False)

$5 + 10 > 5 \implies 15 > 5$ (True)

$5 + 10 > 5 \implies 15 > 5$ (True)

Since the sum of the two shorter sides ($5+5=10$) is not greater than the longest side ($10$), these lengths cannot form a triangle.

(B) Sides: 6 cm, 8 cm, 10 cm

Apply the Triangle Inequality Theorem:

$6 + 8 > 10 \implies 14 > 10$ (True)

$6 + 10 > 8 \implies 16 > 8$ (True)

$8 + 10 > 6 \implies 18 > 6$ (True)

All conditions are met. These lengths can form a triangle (a right-angled triangle since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$).

(C) Angles: $60^\circ, 60^\circ, 60^\circ$

Apply the Angle Sum Property: The sum of the interior angles must be $180^\circ$.

$60^\circ + 60^\circ + 60^\circ = 180^\circ$ (True)

These angles can form a triangle (an equilateral triangle). Infinitely many triangles can have these angles (similar triangles), but a triangle *can* be formed.

(D) Sides: 3 cm, 4 cm, 5 cm

Apply the Triangle Inequality Theorem:

$3 + 4 > 5 \implies 7 > 5$ (True)

$3 + 5 > 4 \implies 8 > 4$ (True)

$4 + 5 > 3 \implies 9 > 3$ (True)

All conditions are met. These lengths can form a triangle (a right-angled triangle since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$).

Therefore, the set of measurements that cannot form a triangle is (A) 5 cm, 5 cm, 10 cm, because it violates the Triangle Inequality Theorem.

The correct option is (C) ASA.

Explanation:

We are given the following information to construct triangle ABC:

• Angle $\angle\text{B} = 70^\circ$
• Side BC = $5\text{ cm}$
• Angle $\angle\text{C} = 60^\circ$

Let's look at the relationships between the given elements:

We have two angles, $\angle\text{B}$ and $\angle\text{C}$. The side BC is the side that connects vertices B and C. Therefore, the side BC is included between angles $\angle\text{B}$ and $\angle\text{C}$.

This configuration of having two angles and the side included between them matches the definition of the ASA (Angle-Side-Angle) criterion.

The construction using the ASA criterion with this information would typically involve:

1. Drawing the line segment BC of length 5 cm.

2. At point B, constructing an angle of $70^\circ$.

3. At point C, constructing an angle of $60^\circ$.

4. The rays of these two angles will intersect at a point, which is the third vertex A.

Since the given information directly fits the ASA pattern, this is the most direct criterion to use for the construction.

The correct option is (C) Third.

Explanation:

In the SSS (Side-Side-Side) construction of a triangle, you are given the lengths of the three sides. Let these lengths be $s_1, s_2,$ and $s_3$.

The typical construction steps are:

1. Draw a line segment representing the first side (say, length $s_1$). Let the endpoints of this segment be the first two vertices of the triangle (e.g., A and B).

2. From one endpoint (say, A), draw an arc with a radius equal to the length of the second side ($s_2$). This arc represents all possible locations for the third vertex that are a distance of $s_2$ away from A.

3. From the other endpoint (B) of the initial segment, draw an arc with a radius equal to the length of the third side ($s_3$). This arc represents all possible locations for the third vertex that are a distance of $s_3$ away from B.

4. The point (or points) where these two arcs intersect is the location of the third vertex of the triangle (e.g., C). This point C is simultaneously $s_2$ distance from A and $s_3$ distance from B, forming a triangle with sides $s_1, s_2,$ and $s_3$.

Therefore, the intersection of the arcs determines the position of the third vertex of the triangle.

The correct option is (D) AAS.

Explanation:

Let's examine the congruence criteria and how they relate to construction:

• SSS (Side-Side-Side): Given three side lengths, you can directly construct the triangle by drawing one side and then using arcs from the endpoints with the other two side lengths as radii.
• SAS (Side-Angle-Side): Given two side lengths and the included angle, you can directly construct the triangle by drawing one side, constructing the angle at an endpoint, and then marking the second side along the ray of the angle.
• ASA (Angle-Side-Angle): Given two angles and the included side, you can directly construct the triangle by drawing the included side and then constructing the two angles at its endpoints.
• AAS (Angle-Angle-Side): Given two angles and a non-included side. While a direct construction method for AAS exists, the *reason* it is a congruence criterion and allows unique construction is often explained by first using the angle sum property to find the third angle.

If you are given two angles (say $\angle A$ and $\angle B$) and a non-included side (say side AC, which is opposite $\angle B$), you can find the third angle $\angle C = 180^\circ - (\angle A + \angle B)$. Once you know $\angle C$, you have angles $\angle A$ and $\angle C$ and the included side AC. This converts the problem into an ASA case.

Thus, the AAS criterion relies on the angle sum property to find the third angle, effectively reducing it to an ASA case, which is why it is considered less "direct" as a fundamental construction criterion compared to SSS, SAS, and ASA.

The correct option is (D) All of the above.

Explanation:

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.

If the side lengths are denoted as $a, b,$ and $c$, then the following three inequalities must all be true for a triangle to be constructed:

Options (A), (B), and (C) each state one of these necessary inequalities.

For example, if $a, b, c$ are positive values, and we assume without loss of generality that $a \le b \le c$, then the inequalities $a+c > b$ and $b+c > a$ are automatically satisfied because $c$ is the largest or equal largest side. So, $a+c > b$ because $c \ge b$, and $b+c > a$ because $c \ge a$ and $b > 0$. However, the inequality $a+b > c$ is the critical one to check in this case. If the sum of the two shortest sides is not greater than the longest side, a triangle cannot be formed.

Since the Triangle Inequality Theorem requires that the sum of *any* two sides is greater than the third side, all three conditions listed in options (A), (B), and (C) must hold true simultaneously. Therefore, option (D) "All of the above" correctly represents the full requirement of the theorem.

The first step in constructing a line parallel to a given line $l$ through a point P not on $l$ by drawing a transversal line is as follows:

Draw a transversal line passing through point P and intersecting the given line $l$ at some point, say Q.

When constructing a parallel line using the property of corresponding angles, the angles that are made equal are the corresponding angles.

When constructing a parallel line using the property of alternate interior angles, the angles that are made equal are the alternate interior angles.

To construct a unique triangle, at least three minimum measurements are required. The common sets of three measurements that uniquely determine a triangle are:

Three sides (SSS criterion): The lengths of all three sides.

Two sides and the included angle (SAS criterion): The lengths of two sides and the measure of the angle between them.

Two angles and the included side (ASA criterion): The measures of two angles and the length of the side between them.

Two angles and any side (AAS criterion): The measures of two angles and the length of any side. (This is equivalent to ASA since the third angle is determined by the other two).

Right angle, Hypotenuse, and one Side (RHS criterion): For right-angled triangles, the length of the hypotenuse and the length of one other side.

The SSS (Side-Side-Side) criterion for the construction of a triangle states:

If the lengths of the three sides of a triangle are given, then a unique triangle can be constructed using these measurements.

The SAS (Side-Angle-Side) criterion for the construction of a triangle states:

If the lengths of two sides and the measure of the included angle (the angle between the two sides) of a triangle are given, then a unique triangle can be constructed using these measurements.

The ASA (Angle-Side-Angle) criterion for the construction of a triangle states:

If the measures of two angles and the length of the included side (the side between the two angles) of a triangle are given, then a unique triangle can be constructed using these measurements.

The RHS (Right angle-Hypotenuse-Side) criterion for the construction of a right-angled triangle states:

If the length of the hypotenuse and the length of one side of a right-angled triangle are given, then a unique right-angled triangle can be constructed using these measurements.

No, a triangle with side lengths $2$ cm, $3$ cm, and $6$ cm cannot be constructed.

This is because the side lengths must satisfy the triangle inequality theorem. The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Let the side lengths be $a = 2$ cm, $b = 3$ cm, and $c = 6$ cm. We must check if the following inequalities hold:

$a + b > c$

$a + c > b$

$b + c > a$

Checking the first inequality:

$2 + 3 > 6$

$5 > 6$

This inequality is false.

Since the sum of the two shorter sides ($2$ cm and $3$ cm) is not greater than the longest side ($6$ cm), the triangle inequality theorem is not satisfied. Therefore, a triangle with these side lengths cannot be constructed.

To construct a triangle with angles $50^\circ, 60^\circ,$ and $70^\circ$, three pieces of information are given: the measures of the three angles.

No, you cannot construct a unique triangle with only the measures of the three angles.

While the sum of the given angles is $50^\circ + 60^\circ + 70^\circ = 180^\circ$, which is valid for the angles of a triangle, knowing only the angles determines the shape of the triangle but not its size.

Any triangle with angles measuring $50^\circ, 60^\circ,$ and $70^\circ$ will be similar to any other triangle with the same angle measures. Infinitely many such triangles exist, each being a scaled version of the others. To construct a unique triangle, you need at least one side length in addition to the angles (ASA or AAS criterion).

In the construction of a triangle using the SAS (Side-Angle-Side) criterion, the position of the given angle is very important.

The given angle must be the included angle. This means the angle must be located between the two given sides whose lengths are provided.

If the angle is not the included angle (i.e., it is adjacent to only one of the given sides), the SAS criterion is not being used, and a unique triangle may not be constructible (this situation is related to the SSA case, which can be ambiguous).

In the construction of a triangle using the ASA (Angle-Side-Angle) criterion, the position of the given side is crucial.

The given side must be the included side.

This means the side must be situated between the two given angles. It is the side that connects the vertices where the two given angles are located.

Knowing the two angles and the side *between* them uniquely determines the triangle's shape and size, enabling its construction using the ASA criterion.

The tools typically used in classical geometric constructions are:

Ruler (or Straightedge): Used for drawing straight lines or line segments. It is important to note that a ruler in geometric construction is often referred to as a straightedge, meaning it is used only to draw straight lines, not to measure lengths.

Compass: Used for drawing circles or arcs of circles with a given radius and centre, and for transferring lengths.

While other tools like protractors (for measuring angles) and set squares are used in drawing, pure geometric constructions traditionally rely only on the straightedge and compass.

To construct $\triangle PQR$ with the given information PQ = $5$ cm, $\angle P = 45^\circ$, and $\angle Q = 60^\circ$, you will use the ASA (Angle-Side-Angle) criterion.

This is because you are given the measures of two angles ($\angle P$ and $\angle Q$) and the length of the side (PQ) that is included between these two angles.

To construct $\triangle ABC$ with the given information AB = $6$ cm, BC = $8$ cm, and CA = $10$ cm, you will use the SSS (Side-Side-Side) criterion.

This is because you are given the lengths of all three sides of the triangle.

To construct $\triangle XYZ$ with the given information XY = $7$ cm, $\angle X = 50^\circ$, and XZ = $9$ cm, you will use the SAS (Side-Angle-Side) criterion.

This is because you are given the lengths of two sides (XY and XZ) and the measure of the angle ($\angle X$) that is included between these two sides.

To construct a right-angled triangle with hypotenuse $10$ cm and one leg $6$ cm, you will use the RHS (Right angle-Hypotenuse-Side) criterion.

This is because you are given the measure of the right angle (implied by "right-angled triangle"), the length of the hypotenuse, and the length of one leg (side) of the triangle.

Yes, you can generally construct a unique triangle if you are given two angles and one side.

The crucial condition is primarily on the given angles. Their sum must be strictly less than $180^\circ$. If the sum of the two given angles is $180^\circ$ or more, no triangle can be formed.

Assuming the angles satisfy the condition (sum less than $180^\circ$), knowing two angles allows you to determine the third angle since the sum of angles in a triangle is $180^\circ$.

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

Once all three angles are known, the shape of the triangle is determined. Knowing the length of any one side then fixes the scale of the triangle, thereby making it unique.

The given side can be either the included side (between the two given angles, using the ASA criterion) or a non-included side (opposite one of the given angles, using the AAS criterion). Both ASA and AAS criteria guarantee the construction of a unique triangle, provided the angles are valid.

Therefore, the condition on the side is that it must have a positive length, and the sum of the two given angles must be less than $180^\circ$. If these hold, the position of the side does not prevent the construction of a unique triangle.

When constructing a parallel line to a given line $l$ through a point P not on $l$ using the corresponding angles property, you ensure the angles are equal by copying the angle formed by a transversal intersecting the given line $l$ at the point P.

Here's the general method using a compass and straightedge to copy the corresponding angle:

1. Draw a transversal line through point P that intersects the given line $l$ at a point, say Q.

2. Place the point of the compass at Q and draw an arc that intersects the transversal and line $l$. Let the intersection points be A (on the transversal) and B (on line $l$). This arc defines the angle at Q.

3. With the same compass opening used in step 2, place the compass point at P and draw a similar arc that intersects the transversal (on the side opposite to where point A was on the first arc relative to Q). Let this intersection point be C.

4. Open the compass to the distance between points A and B (the points where the first arc intersected the transversal and line $l$).

5. With the compass point at C (the point where the second arc intersected the transversal), draw an arc that intersects the second arc you drew in step 3. Let this new intersection point be D.

6. Draw a straight line passing through points P and D.

The line passing through P and D is parallel to line $l$. The angles formed, for example, $\angle DPC$ and $\angle AQB$ (assuming A, Q, C are collinear on the transversal and B is on $l$ such that $\angle AQB$ is the corresponding angle to $\angle DPC$), are made equal by this construction process. The compass is used to transfer the 'opening' or measure of the angle from point Q to point P.

In geometric constructions, drawing arcs with a compass serves several key purposes:

1. Creating Circles or Parts of Circles: The most fundamental use is to draw a set of points that are equidistant from a central point (the radius of the compass). This forms a circle or an arc (a portion of a circle).

2. Transferring Lengths: A compass allows you to set a specific distance (radius) and then mark off that distance from any point. This is essential for copying segment lengths from one part of the construction to another.

3. Locating Points: By drawing intersecting arcs from two different points, you can locate a third point that is at a specific distance from both original points. This is crucial for constructing triangles (SSS criterion), perpendicular bisectors, angle bisectors, etc.

4. Marking Angles: Arcs are used to 'measure' or replicate angles by capturing the spread between two rays. By drawing an arc across an angle and then transferring the chord length of that arc to another location, you can copy the angle.

Essentially, the compass provides the ability to work with distances and define geometric loci (sets of points satisfying a condition), which are fundamental to carrying out accurate constructions.

Yes, if you are given the perimeter of an equilateral triangle, you can construct a unique triangle.

An equilateral triangle is a triangle with all three sides of equal length.

The perimeter of an equilateral triangle is the sum of the lengths of its three equal sides. If the perimeter is given as $P$ and the length of each side is $s$, then:

$P = s + s + s = 3s$

From this equation, we can find the length of each side:

$s = \frac{P}{3}$

Since you can calculate the exact length of each side using the given perimeter, you now have the lengths of all three sides of the triangle.

With the lengths of all three sides known, you can construct a unique triangle using the SSS (Side-Side-Side) criterion, provided the side lengths satisfy the triangle inequality (which $s+s > s$ always does for $s > 0$).

Therefore, the perimeter provides sufficient information to determine the side length, and knowing the side length allows for the construction of a unique equilateral triangle.

No, you cannot construct a unique triangle if you are given the measures of only its three angles.

Here's the reason why:

While knowing the three angles of a triangle determines its shape (since the sum of the angles in any triangle is always $180^\circ$), it does not determine its size.

For example, many triangles can have angles $50^\circ, 60^\circ,$ and $70^\circ$. A small triangle with these angles and a large triangle with these angles are both valid triangles. They are similar triangles – they have the same angles but different side lengths.

To construct a unique triangle, you need at least one piece of information related to its size, such as the length of one or more sides (as required by the SSS, SAS, ASA, or RHS criteria).

Here are the steps to construct a line parallel to a given line $l$ through a point P not on $l$ using the property of alternate interior angles:

Given: A line $l$ and a point P not on $l$.

To Construct: A line $m$ passing through P such that $m || l$.

Construction Steps:

1. Draw the given line $l$. Mark a point P outside the line $l$.

2. Choose any point, say Q, on the line $l$. Join P to Q. The line segment PQ acts as a transversal intersecting line $l$.

3. Place the compass point at Q and draw an arc that intersects line $l$ at point A and segment PQ at point B. This arc defines an angle, $\angle \text{PQA}$.

4. With the same compass opening, place the compass point at P and draw an arc that intersects the segment PQ at point C (on the opposite side of PQ compared to point B). This arc should be long enough to potentially intersect the new parallel line.

5. Open the compass to the distance between points A and B (the points where the first arc intersected line $l$ and segment PQ). This distance represents the 'width' of the angle at Q.

6. With the compass point at C (the point where the second arc intersected segment PQ), draw an arc that intersects the arc drawn in step 4. Let the intersection point be D.

7. Draw a straight line passing through points P and D. Extend this line on both sides.

Let this new line be $m$. The angle $\angle \text{QPD}$ (or $\angle \text{CPD}$) has been constructed such that its measure is equal to the measure of $\angle \text{PQA}$.

These two angles, $\angle \text{PQA}$ and $\angle \text{QPD}$, are alternate interior angles formed by the transversal PQ intersecting lines $l$ and $m$.

Since the constructed alternate interior angles are equal, the line $m$ is parallel to the line $l$.

(Note: The actual drawing with points and lines cannot be displayed in this text format, but the steps describe the process using standard geometric tools like a ruler and compass.)

Here are the steps to construct a line parallel to a given line $m$ through a point Q not on $m$ using the property of corresponding angles:

Given: A line $m$ and a point Q not on $m$.

To Construct: A line $n$ passing through Q such that $n || m$.

Construction Steps:

1. Draw the given line $m$. Mark a point Q outside the line $m$.

2. Choose any point, say R, on the line $m$. Join Q to R. The line segment QR acts as a transversal intersecting line $m$. Extend the transversal QR beyond R if necessary.

3. Identify one of the corresponding angles formed by the transversal QR and the line $m$. For instance, consider the angle formed below line $m$ and to the right of transversal QR. Let's call a point on $m$ to the right of R as A, so the angle is $\angle \text{QRA}$.

4. Place the compass point at R and draw an arc that intersects line $m$ at point A and the transversal QR (or its extension) at point B. This arc captures the size of $\angle \text{QRA}$.

5. With the same compass opening used in step 4, place the compass point at Q and draw a similar arc that intersects the transversal QR (on the same side of the transversal as the corresponding angle you chose in step 3). Let this intersection point be C.

6. Open the compass to the distance between points A and B (the points where the first arc intersected line $m$ and the transversal QR).

7. With the compass point at C (the point where the second arc intersected the transversal), draw an arc that intersects the arc drawn in step 5. Let this new intersection point be D.

8. Draw a straight line passing through points Q and D. Extend this line on both sides.

Let this new line be $n$. The angle $\angle \text{RQD}$ has been constructed such that its measure is equal to the measure of the corresponding angle $\angle \text{QRA}$ (assuming A is on line $m$ and D is positioned such that $\angle \text{RQD}$ and $\angle \text{QRA}$ are corresponding angles).

Since the constructed corresponding angles ($\angle \text{RQD}$ and $\angle \text{QRA}$) are equal, the line $n$ is parallel to the line $m$.

(Note: The actual drawing with points and lines cannot be displayed in this text format, but the steps describe the process using standard geometric tools like a ruler and compass.)

Given: A triangle $\triangle ABC$ with side lengths AB = $7$ cm, BC = $5$ cm, and CA = $6$ cm.

To Construct: $\triangle ABC$ with the given side lengths.

Construction Steps:

1. Draw a line segment AB of length $7$ cm.

2. With A as the center and a radius equal to $6$ cm (length of CA), draw an arc above the line segment AB.

3. With B as the center and a radius equal to $5$ cm (length of BC), draw another arc above the line segment AB, such that it intersects the previously drawn arc.

4. Mark the point of intersection of the two arcs as C.

5. Join A to C and B to C.

The resulting triangle $\triangle ABC$ is the required triangle with the given side lengths.

The congruence criterion used for this construction is the SSS (Side-Side-Side) criterion, as the lengths of all three sides are given, which uniquely determines the triangle.

Given: A triangle $\triangle PQR$ with side lengths PQ = $6$ cm, PR = $8$ cm, and the included angle $\angle P = 70^\circ$.

To Construct: $\triangle PQR$ with the given measurements.

Construction Steps:

1. Draw a line segment PQ of length $6$ cm.

2. At point P, construct an angle of $70^\circ$ using a protractor or compass and ruler. Draw a ray PX such that $\angle \text{QPX} = 70^\circ$.

3. With P as the center and a radius equal to $8$ cm (length of PR), draw an arc that intersects the ray PX at point R.

4. Join Q to R.

The resulting triangle $\triangle PQR$ is the required triangle with the given measurements.

The congruence criterion used for this construction is the SAS (Side-Angle-Side) criterion, as the lengths of two sides (PQ and PR) and the measure of the included angle ($\angle P$) are given, which uniquely determines the triangle.

Given: A triangle $\triangle XYZ$ with $\angle X = 50^\circ$, side XY = $7$ cm, and $\angle Y = 60^\circ$.

To Construct: $\triangle XYZ$ with the given measurements.

Construction Steps:

1. Draw a line segment XY of length $7$ cm.

2. At point X, construct an angle of $50^\circ$ using a protractor or compass and ruler. Draw a ray XA such that $\angle \text{YXA} = 50^\circ$.

3. At point Y, construct an angle of $60^\circ$ using a protractor or compass and ruler. Draw a ray YB such that $\angle \text{XYB} = 60^\circ$. Note that ray YB should be drawn on the same side of XY as ray XA.

4. The rays XA and YB will intersect at a point. Mark this point as Z.

The resulting triangle $\triangle XYZ$ is the required triangle with the given measurements.

The congruence criterion used for this construction is the ASA (Angle-Side-Angle) criterion, as the measures of two angles ($\angle X$ and $\angle Y$) and the length of the included side (XY) are given, which uniquely determines the triangle.

Given: A right-angled triangle $\triangle ABC$ with $\angle B = 90^\circ$, hypotenuse AC = $9$ cm, and side AB = $5$ cm.

To Construct: $\triangle ABC$ with the given measurements.

Construction Steps:

1. Draw a line segment AB of length $5$ cm.

2. At point B, construct a ray BY perpendicular to AB using a compass and ruler (or a protractor to draw a $90^\circ$ angle).

3. With A as the center and a radius equal to $9$ cm (length of hypotenuse AC), draw an arc that intersects the ray BY.

4. Mark the point of intersection of the arc and the ray BY as C.

5. Join A to C.

The resulting triangle $\triangle ABC$ is the required right-angled triangle with the given measurements.

The congruence criterion used for this construction is the RHS (Right angle-Hypotenuse-Side) criterion, as the right angle, the length of the hypotenuse, and the length of one side (leg) are given, which uniquely determines the right-angled triangle.

Given: An isosceles triangle $\triangle ABC$ with AB = $7$ cm, AC = $7$ cm, and BC = $5$ cm.

To Construct: $\triangle ABC$ and its altitude from A to BC. Measure the length of the altitude.

Construction Steps for $\triangle ABC$:

1. Draw a line segment BC of length $5$ cm.

2. With B as the center and a radius equal to $7$ cm (length of AB), draw an arc above the line segment BC.

3. With C as the center and a radius equal to $7$ cm (length of AC), draw another arc above the line segment BC, such that it intersects the previously drawn arc.

4. Mark the point of intersection of the two arcs as A.

5. Join B to A and C to A.

The resulting triangle $\triangle ABC$ is the required isosceles triangle.

Construction Steps for the Altitude from A to BC:

1. With A as the center and any suitable radius (large enough to intersect BC), draw an arc that intersects the line segment BC at two points, say M and N.

2. With M as the center and a radius greater than half the length of MN, draw an arc below the line segment BC.

3. With N as the center and the same radius used in step 2, draw another arc below the line segment BC, such that it intersects the arc drawn in step 2.

4. Mark the point of intersection of these two arcs as D.

5. Draw a straight line segment joining A to D.

The line segment AD is perpendicular to BC and passes through A. Since $\triangle ABC$ is isosceles with AB = AC, the altitude from A to BC is also the median and angle bisector, and it will indeed meet BC at its midpoint (which the construction of perpendicular bisector would yield if A were the center). In this case, AD is the altitude.

Measurement of the Altitude:

Measure the length of the segment AD using a ruler.

(Note: The length of the altitude can also be calculated using the Pythagorean theorem. In $\triangle ABD$, which is a right-angled triangle at D, AB = $7$ cm and BD = BC/2 = $5/2 = 2.5$ cm.

$AD^2 + BD^2 = AB^2$

$AD^2 + (2.5)^2 = 7^2$

$AD^2 + 6.25 = 49$

$AD^2 = 49 - 6.25 = 42.75$

$AD = \sqrt{42.75} \approx 6.538$ cm.

Your measured value from the construction should be close to this value.)

Given: A triangle $\triangle LMN$ with side LM = $8$ cm, $\angle L = 40^\circ$, and $\angle N = 75^\circ$.

To Construct: $\triangle LMN$ with the given measurements.

Finding $\angle M$:

The sum of the angles in a triangle is $180^\circ$. Therefore, in $\triangle LMN$:

$\angle L + \angle M + \angle N = 180^\circ$

Substitute the given values:

$40^\circ + \angle M + 75^\circ = 180^\circ$

$\angle M + 115^\circ = 180^\circ$

$\angle M = 180^\circ - 115^\circ$

$\angle M = 65^\circ$

Now we have LM = $8$ cm, $\angle L = 40^\circ$, and $\angle M = 65^\circ$. This fits the ASA criterion as we have two angles and the included side.

Construction Steps:

1. Draw a line segment LM of length $8$ cm.

2. At point L, construct an angle of $40^\circ$ using a protractor or compass and ruler. Draw a ray LX such that $\angle \text{MLX} = 40^\circ$.

3. At point M, construct an angle of $65^\circ$ using a protractor or compass and ruler. Draw a ray MY such that $\angle \text{LMY} = 65^\circ$. Note that ray MY should be drawn on the same side of LM as ray LX.

4. The rays LX and MY will intersect at a point. Mark this point as N.

The resulting triangle $\triangle LMN$ is the required triangle with the given measurements.

Given: A triangular park with actual side lengths $10$ m, $12$ m, and $15$ m.

Scale: $1$ cm represents $2$ meters.

Scale Conversion:

To draw the scaled diagram, we need to convert the actual lengths from meters to centimeters using the given scale.

$1 \text{ cm} \iff 2 \text{ m}$

So, $1 \text{ m} \iff \frac{1}{2} \text{ cm} = 0.5 \text{ cm}$

Scaled length of the first side ($10$ m):

$10 \text{ m} \times \frac{1 \text{ cm}}{2 \text{ m}} = \frac{10}{2} \text{ cm} = 5 \text{ cm}$

Scaled length of the second side ($12$ m):

$12 \text{ m} \times \frac{1 \text{ cm}}{2 \text{ m}} = \frac{12}{2} \text{ cm} = 6 \text{ cm}$

Scaled length of the third side ($15$ m):

$15 \text{ m} \times \frac{1 \text{ cm}}{2 \text{ m}} = \frac{15}{2} \text{ cm} = 7.5 \text{ cm}$

Thus, the scaled triangle to be constructed will have side lengths of $5$ cm, $6$ cm, and $7.5$ cm.

Construction Steps for the Scaled Diagram:

We will construct a triangle with side lengths $5$ cm, $6$ cm, and $7.5$ cm using the SSS criterion.

1. Draw a line segment of length $7.5$ cm. Label the endpoints as A and B. This segment represents the scaled $15$ m side.

2. With A as the center and a radius equal to $5$ cm (representing the scaled $10$ m side), draw an arc above the line segment AB.

3. With B as the center and a radius equal to $6$ cm (representing the scaled $12$ m side), draw another arc above the line segment AB, such that it intersects the previously drawn arc.

4. Mark the point of intersection of the two arcs as C.

5. Join A to C and B to C.

The resulting triangle $\triangle ABC$ is the scaled diagram of the triangular park. The side lengths are AB = $7.5$ cm, AC = $5$ cm, and BC = $6$ cm, representing the actual lengths of $15$ m, $10$ m, and $12$ m respectively according to the scale.

(Note: The actual drawing cannot be displayed here, but the steps describe the process to create it.)

Given: A line segment AB = $6$ cm, $\angle A = 60^\circ$, and $\angle B = 45^\circ$.

To Construct: $\triangle ABC$ using the given measurements.

Construction Steps:

1. Draw a line segment AB of length $6$ cm.

2. At point A, construct an angle of $60^\circ$ using a protractor or compass and ruler. Draw a ray AX such that $\angle \text{BAX} = 60^\circ$.

3. At point B, construct an angle of $45^\circ$ using a protractor or compass and ruler. Draw a ray BY such that $\angle \text{ABY} = 45^\circ$. Ensure that ray BY is drawn on the same side of AB as ray AX.

4. The rays AX and BY will intersect at a point. Mark this point as C.

The resulting triangle $\triangle ABC$ is the required triangle with the given measurements.

The criterion used to form this triangle is the ASA (Angle-Side-Angle) criterion, as the measures of two angles ($\angle A$ and $\angle B$) and the length of the included side (AB) are given, which uniquely determines the triangle.

It is not possible to construct a triangle with angles measuring $80^\circ, 70^\circ,$ and $40^\circ$ because the sum of these angles violates the fundamental angle sum property of a triangle.

The angle sum property of a triangle states that the sum of the interior angles of any triangle is always exactly $180^\circ$.

Let's find the sum of the given angles:

Sum of angles $= 80^\circ + 70^\circ + 40^\circ$

Sum of angles $= 150^\circ + 40^\circ$

Sum of angles $= 190^\circ$

Comparing this sum to the angle sum property:

$190^\circ \neq 180^\circ$

Since the sum of the given angles ($190^\circ$) is not equal to $180^\circ$, a closed three-sided figure (a triangle) with these specific angle measures cannot exist. The angles would either not meet to form a closed shape or would overlap.

Given: A line segment PQ of length $7.5$ cm.

To Construct: The perpendicular bisector of PQ. Take a point R on the bisector and verify that RP = RQ.

Construction Steps:

1. Draw a line segment PQ of length $7.5$ cm.

2. With P as the center and a radius greater than half the length of PQ, draw two arcs, one above the line segment PQ and one below it.

3. With Q as the center and the same radius (used in step 2), draw two other arcs, one above PQ and one below PQ, such that these arcs intersect the previously drawn arcs.

4. Mark the points of intersection of the arcs as A (above PQ) and B (below PQ).

5. Draw a straight line passing through points A and B.

The line AB is the required perpendicular bisector of the line segment PQ. It intersects PQ at a point, let's call it M.

Explanation of the Geometric Property (RP = RQ):

Let R be any point on the perpendicular bisector AB. Consider $\triangle \text{RMP}$ and $\triangle \text{RMQ}$.

We know that the line AB is the perpendicular bisector of PQ. This means:

$\angle \text{PMR} = \angle \text{QMR} = 90^\circ$

PM = MQ (since M is the midpoint of PQ)

RM is common to both triangles.

By the SAS (Side-Angle-Side) congruence criterion, $\triangle \text{RMP}$ is congruent to $\triangle \text{RMQ}$ ($\text{PM} = \text{MQ}$, $\angle \text{PMR} = \angle \text{QMR}$, and $\text{RM} = \text{RM}$).

Since the triangles are congruent, their corresponding parts are equal (CPCT - Corresponding Parts of Congruent Triangles).

Therefore, RP = RQ.

This property states that any point on the perpendicular bisector of a line segment is equidistant from the endpoints of the segment.

Verification by Measurement:

After constructing the perpendicular bisector, choose any point R on the line AB (other than M). Use a ruler to measure the length of the segment RP and the length of the segment RQ. You will find that the measured length of RP is equal to the measured length of RQ, confirming the property.