Chapter 11 Constructions

To determine which angle can be constructed with the help of a ruler and a compass, we need to understand the principles of geometric construction. An angle is constructible if its measure can be obtained from basic constructible angles ($60^\circ$ and $90^\circ$) through a finite sequence of additions, subtractions, and bisections.

Basic constructible angles include $60^\circ$ (derived from an equilateral triangle) and $90^\circ$ (derived from a perpendicular bisector). By repeatedly bisecting these angles, we can obtain angles like $30^\circ$, $15^\circ$, $7.5^\circ$, $45^\circ$, $22.5^\circ$, etc.

Also, if angles $\alpha$ and $\beta$ are constructible, then the angle $\alpha + \beta$ and $|\alpha - \beta|$ are also constructible.

Let's examine the given options:

(A) $35^\circ$: $35^\circ = 5^\circ \times 7$. An angle of $5^\circ$ is not constructible using simple combinations of $60^\circ/2^n$ or $90^\circ/2^n$.

(B) $40^\circ$: $40^\circ = 120^\circ / 3$. The trisection of a general angle (like $120^\circ$) is not possible with a ruler and compass.

(C) $37.5^\circ$: We can try to express this angle as a combination of known constructible angles: We know $60^\circ$ is constructible. Bisecting $60^\circ$ gives $30^\circ$, which is constructible. Bisecting $30^\circ$ gives $15^\circ$, which is constructible. Bisecting $15^\circ$ gives $7.5^\circ$, which is constructible. Now, notice that $37.5^\circ = 30^\circ + 7.5^\circ$. Since both $30^\circ$ and $7.5^\circ$ are constructible, their sum $37.5^\circ$ is also constructible. Alternatively, $90^\circ$ is constructible. Bisecting $90^\circ$ gives $45^\circ$, which is constructible. $37.5^\circ = 45^\circ - 7.5^\circ$. Since both $45^\circ$ and $7.5^\circ$ are constructible, their difference is also constructible.

More formally, an angle $\theta$ is constructible if and only if the degree measure $\theta$ is a rational multiple of $360^\circ$ whose denominator, when the fraction $\theta/360$ is reduced to lowest terms, is of the form $2^k \cdot F_{i_1} \cdots F_{i_m}$, where $k \ge 0$ and $F_{i_j}$ are distinct Fermat primes (primes of the form $2^{2^n} + 1$, such as 3, 5, 17, 257, 65537). For $37.5^\circ$: $\frac{37.5^\circ}{360^\circ} = \frac{37.5}{360} = \frac{375}{3600} = \frac{5 \times 75}{5 \times 720} = \frac{75}{720} = \frac{5 \times 15}{5 \times 144} = \frac{15}{144} = \frac{3 \times 5}{3 \times 48} = \frac{5}{48}$. The denominator in the reduced fraction $\frac{5}{48}$ is 48. We prime factorize 48: $48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$. The denominator is of the form $2^k \cdot p_1$ where $k=4$ and $p_1=3$. Since 3 is a Fermat prime ($F_0 = 2^{2^0} + 1 = 2^1 + 1 = 3$), the angle $37.5^\circ$ is constructible.

(D) $47.5^\circ$: $\frac{47.5^\circ}{360^\circ} = \frac{47.5}{360} = \frac{475}{3600} = \frac{95}{720} = \frac{19}{144}$. The denominator in the reduced fraction $\frac{19}{144}$ is 144. $144 = 2^4 \times 3^2$. The denominator contains the prime factor 3 raised to the power of 2, which is not allowed for constructibility (the powers of Fermat primes must be 0 or 1). Thus, $47.5^\circ$ is not constructible.

Therefore, among the given options, only $37.5^\circ$ is constructible using a ruler and compass.

The correct option is (C) 37.5°.

For the construction of a triangle ABC with a given side AB and angle ∠A, and the difference between the other two sides (BC - AC or AC - BC), there is a condition that must be satisfied.

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

In triangle ABC, the relevant inequality for the difference of sides BC and AC with respect to the side AB is:

We are given that $\text{AB = 4 cm}$. So the condition for the construction of triangle ABC to be possible is:

The construction of the triangle is not possible if this condition is violated, i.e., if:

Let's check the given options for the difference of BC and AC:

(A) $\text{|BC - AC|} = 3.5 \text{ cm}$. Here, $3.5 < 4$. The condition is satisfied, so the construction is possible.

(B) $\text{|BC - AC|} = 4.5 \text{ cm}$. Here, $4.5 \ge 4$. The condition is not satisfied, so the construction is not possible.

(C) $\text{|BC - AC|} = 3 \text{ cm}$. Here, $3 < 4$. The condition is satisfied, so the construction is possible.

(D) $\text{|BC - AC|} = 2.5 \text{ cm}$. Here, $2.5 < 4$. The condition is satisfied, so the construction is possible.

Therefore, the construction of triangle ABC is not possible when the difference of BC and AC is equal to 4.5 cm.

The correct option is (B) 4.5 cm.

An angle $\theta$ is constructible with a ruler and compass if and only if the degree measure $\theta$ is a rational multiple of $360^\circ$ whose denominator, when the fraction $\frac{\theta}{360^\circ}$ is reduced to its lowest terms, is of the form $2^k \cdot F_{i_1} \cdots F_{i_m}$, where $k \ge 0$ is an integer and $F_{i_j}$ are distinct Fermat primes. Fermat primes are prime numbers of the form $F_n = 2^{2^n} + 1$. The first few Fermat primes are $F_0 = 2^{2^0} + 1 = 3$, $F_1 = 2^{2^1} + 1 = 5$, $F_2 = 2^{2^2} + 1 = 17$, $F_3 = 2^{2^3} + 1 = 257$, and $F_4 = 2^{2^4} + 1 = 65537$.

Let's examine each option by calculating the fraction $\frac{\theta}{360^\circ}$ and reducing it:

(A) $37.5^\circ$: $\frac{37.5}{360} = \frac{375}{3600} = \frac{5}{48}$ The denominator is $48$. The prime factorization of $48$ is $2^4 \times 3$. Since $3$ is a Fermat prime ($F_0$), the denominator is of the form $2^k \cdot F_0^1$. Thus, $37.5^\circ$ is constructible.

(B) $40^\circ$: $\frac{40}{360} = \frac{4}{36} = \frac{1}{9}$ The denominator is $9$. The prime factorization of $9$ is $3^2$. The denominator must be of the form $2^k$ times a product of *distinct* Fermat primes, each raised to the power of 1. Since the denominator has $3^2$, it is not of the required form. Thus, $40^\circ$ is not constructible.

(C) $22.5^\circ$: $\frac{22.5}{360} = \frac{225}{3600} = \frac{1}{16}$ The denominator is $16$. The prime factorization of $16$ is $2^4$. The denominator is of the form $2^k$. Thus, $22.5^\circ$ is constructible.

(D) $67.5^\circ$: $\frac{67.5}{360} = \frac{675}{3600} = \frac{3}{16}$ The denominator is $16$. The prime factorization of $16$ is $2^4$. The denominator is of the form $2^k$. Thus, $67.5^\circ$ is constructible.

Based on the constructibility criterion, the only angle among the options that is not constructible with a ruler and compass is $40^\circ$.

The correct option is (B) 40°.

To determine when the construction of a triangle ABC, with a given side BC, a given angle ∠B, and the difference between the other two sides (AB and AC), is not possible, we use the fundamental property of triangles known as 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. A consequence of this theorem is that the difference between the lengths of any two sides of a triangle must be less than the length of the third side.

For a triangle ABC to be constructible, the difference between the lengths of sides AB and AC must be less than the length of side BC. Mathematically, this is expressed as:

We are given that $\text{BC = 6 cm}$. Therefore, the condition for the construction of triangle ABC to be possible is:

The construction of the triangle is not possible if this condition is violated. This happens when the difference between AB and AC is greater than or equal to BC.

Now, let's check each given option for the value of the difference of AB and AC against the condition for non-constructibility (inequation (i)):

(A) 6.9 cm: If $\text{|AB - AC|} = 6.9 \text{ cm}$, then $6.9 \ge 6$ is true. This satisfies the condition for non-constructibility.

(B) 5.2 cm: If $\text{|AB - AC|} = 5.2 \text{ cm}$, then $5.2 \ge 6$ is false. This satisfies the condition for constructibility ($5.2 < 6$).

(C) 5.0 cm: If $\text{|AB - AC|} = 5.0 \text{ cm}$, then $5.0 \ge 6$ is false. This satisfies the condition for constructibility ($5.0 < 6$).

(D) 4.0 cm: If $\text{|AB - AC|} = 4.0 \text{ cm}$, then $4.0 \ge 6$ is false. This satisfies the condition for constructibility ($4.0 < 6$).

The construction of triangle ABC is not possible when the difference of AB and AC is equal to a value that is greater than or equal to 6 cm. From the given options, only 6.9 cm satisfies this condition.

The correct option is (A) 6.9 cm.

For the construction of a triangle ABC, given the length of one side (BC), the measure of an adjacent angle (∠C), and the difference between the lengths of the other two sides (AB and AC), a specific condition based on the triangle inequality theorem must be met.

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. An important consequence of this theorem is that the absolute difference between the lengths of any two sides of a triangle must be strictly less than the length of the third side.

In the context of this problem, for the triangle ABC to be constructible, the absolute difference between the lengths of sides AB and AC must be less than the length of side BC. This can be written as:

We are given that $\text{BC = 3 cm}$. Therefore, the condition for the construction of triangle ABC to be possible is:

The construction is possible if and only if this condition (inequation (i)) is satisfied. We need to examine the given options for the difference of AB and AC and see which one satisfies this condition.

Let's check each option:

(A) 3.2 cm: If $\text{|AB - AC|} = 3.2 \text{ cm}$, is $3.2 < 3$? No, $3.2$ is not less than $3$. The condition is not satisfied.

(B) 3.1 cm: If $\text{|AB - AC|} = 3.1 \text{ cm}$, is $3.1 < 3$? No, $3.1$ is not less than $3$. The condition is not satisfied.

(C) 3 cm: If $\text{|AB - AC|} = 3 \text{ cm}$, is $3 < 3$? No, $3$ is equal to $3$, not strictly less than $3$. The condition is not satisfied.

(D) 2.8 cm: If $\text{|AB - AC|} = 2.8 \text{ cm}$, is $2.8 < 3$? Yes, $2.8$ is less than $3$. The condition is satisfied.

The construction of triangle ABC is possible only when the difference of AB and AC is strictly less than the length of BC (3 cm). Among the given options, only 2.8 cm satisfies this condition.

The correct option is (D) 2.8 cm.

The statement is True.

An angle $\theta$ is constructible with a ruler and compass if and only if $\frac{\theta}{360^\circ}$ is a rational number whose denominator, when the fraction is reduced to lowest terms, is a product of a power of 2 and distinct Fermat primes.

Let's check the fraction for $67.5^\circ$:

$\frac{67.5^\circ}{360^\circ} = \frac{67.5}{360}$

Multiply the numerator and denominator by 10 to remove the decimal:

$\frac{675}{3600}$

Now, reduce the fraction to its lowest terms. Both are divisible by 25:

$\frac{675 \div 25}{3600 \div 25} = \frac{27}{144}$

Both are divisible by 9:

$\frac{27 \div 9}{144 \div 9} = \frac{3}{16}$

The reduced fraction is $\frac{3}{16}$. The denominator is 16.

The prime factorization of the denominator 16 is $2^4$.

The denominator is of the form $2^k$ where $k=4$. This satisfies the condition for constructibility (it is a power of 2 multiplied by an empty product of distinct Fermat primes).

Alternatively, we can express $67.5^\circ$ as a combination of constructible angles. $90^\circ$ is constructible. Bisecting it gives $45^\circ$. $45^\circ$ is constructible. Bisecting it gives $22.5^\circ$. $67.5^\circ = 45^\circ + 22.5^\circ$. Since both $45^\circ$ and $22.5^\circ$ are constructible, their sum $67.5^\circ$ is also constructible.

Therefore, an angle of $67.5^\circ$ can be constructed using a ruler and compass.

The statement is True.

An angle $\theta$ is constructible using a ruler and compass if and only if the degree measure $\theta$ is a rational multiple of $360^\circ$ such that when the fraction $\frac{\theta}{360^\circ}$ is reduced to its lowest terms, its denominator is of the form $2^k \cdot F_{i_1} \cdots F_{i_m}$, where $k \ge 0$ is an integer and $F_{i_j}$ are distinct Fermat primes. Fermat primes are primes of the form $2^{2^n} + 1$ (e.g., 3, 5, 17, 257, 65537).

Let's evaluate the fraction $\frac{52.5^\circ}{360^\circ}$:

$\frac{52.5}{360} = \frac{52.5 \times 10}{360 \times 10} = \frac{525}{3600}$

Now, we reduce this fraction to its lowest terms:

$\frac{\cancel{525}^{105}}{\cancel{3600}_{720}} = \frac{\cancel{105}^{21}}{\cancel{720}_{144}} = \frac{\cancel{21}^{7}}{\cancel{144}_{48}} = \frac{7}{48}$

The reduced fraction is $\frac{7}{48}$.

The denominator of the reduced fraction is 48. Let's find the prime factorization of 48:

$48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$

The denominator is $48 = 2^4 \times 3$.

The denominator is in the form $2^k \cdot F_{i_1} \cdots F_{i_m}$, where $k=4$ and $F_{i_1} = 3$. Since 3 is the Fermat prime $F_0 = 2^{2^0} + 1 = 3$, and it appears with power 1, the condition for constructibility is met.

Thus, an angle of $52.5^\circ$ is constructible using a ruler and compass.

The statement is False.

An angle $\theta$ is constructible using a ruler and compass if and only if the degree measure $\theta$ is a rational multiple of $360^\circ$ such that when the fraction $\frac{\theta}{360^\circ}$ is reduced to its lowest terms, its denominator is of the form $2^k \cdot F_{i_1} \cdots F_{i_m}$, where $k \ge 0$ is an integer and $F_{i_j}$ are distinct Fermat primes. Fermat primes are primes of the form $2^{2^n} + 1$ (e.g., $F_0=3$, $F_1=5$, $F_2=17$, $F_3=257$, $F_4=65537$).

Let's evaluate the fraction $\frac{42.5^\circ}{360^\circ}$:

$\frac{42.5}{360} = \frac{42.5 \times 10}{360 \times 10} = \frac{425}{3600}$

Now, we reduce this fraction to its lowest terms. Both numerator and denominator are divisible by 25:

$\frac{425 \div 25}{3600 \div 25} = \frac{17}{144}$

The reduced fraction is $\frac{17}{144}$.

The denominator of the reduced fraction is 144. Let's find the prime factorization of 144:

$\begin{array}{c|cc} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 144 is $2^4 \times 3^2$.

For the angle to be constructible, the denominator in lowest terms must be of the form $2^k$ multiplied by distinct Fermat primes raised to the power of 1.

The denominator $144 = 2^4 \times 3^2$. It includes the prime factor 3 (which is a Fermat prime $F_0$), but it is raised to the power of 2 ($3^2$), not 1. This violates the condition for constructibility.

Therefore, an angle of $42.5^\circ$ cannot be constructed using a ruler and compass.

The statement is False.

For any triangle to be constructed, the lengths of its sides must satisfy the triangle inequality theorem. This 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.

In triangle ABC, the triangle inequality theorem requires the following conditions to hold:

1. $\text{AB + BC} > \text{AC}$

2. $\text{AB + AC} > \text{BC}$

3. $\text{BC + AC} > \text{AB}$

We are given that $\text{AB = 5 cm}$ and $\text{BC + AC = 5 cm}$.

Let's check if these given values satisfy the third condition of the triangle inequality:

Substitute the given values:

This inequality is false, as 5 cm is not strictly greater than 5 cm; it is equal to 5 cm.

When the sum of two sides of a potential triangle is equal to the third side ($\text{BC + AC} = \text{AB}$), the three points A, B, and C are collinear, and C lies on the line segment AB. This forms a degenerate triangle, which is essentially a straight line segment. A non-degenerate triangle (a true triangle with area) cannot be formed under these conditions.

Since the triangle inequality $\text{BC + AC} > \text{AB}$ is not satisfied, a triangle ABC with the given side lengths cannot be constructed. The value of ∠A (45°) is irrelevant if the side length conditions are not met.

The statement is True.

For a triangle to be constructible given one side (BC), an adjacent angle (∠C), and the difference of the other two sides (AC and AB), there is a condition based on the triangle inequality theorem.

The triangle inequality theorem implies that the absolute difference between the lengths of any two sides of a triangle must be strictly less than the length of the third side.

In this case, we are given the difference between sides AC and AB, and the length of side BC. The condition for constructibility is:

We are given that $\text{BC = 6 cm}$ and $\text{AC - AB = 4 cm}$.

Substitute these values into the inequality:

This simplifies to:

The inequality $4 < 6$ is true. The condition for the constructibility of the triangle based on the difference of two sides and the third side is satisfied.

Additionally, the given angle ∠C = 30° is constructible using a ruler and compass (since $30^\circ = 60^\circ / 2$, and $60^\circ$ is constructible, and bisection is a standard construction).

Since both the necessary condition from the triangle inequality is met and the angle is constructible, the triangle ABC can be constructed with the given measurements.

The statement is False.

One of the fundamental properties of a Euclidean triangle is that the sum of the measures of its interior angles is always equal to $180^\circ$.

In a triangle ABC, the sum of the angles is:

We are given the measures of two angles: ∠B = 105° and ∠C = 90°.

Let's find the sum of these two angles:

According to the property of the sum of angles in a triangle (equation (i)), the sum of all three angles must be $180^\circ$.

In this case, the sum of just two angles (∠B and ∠C) is $195^\circ$, which is already greater than $180^\circ$.

If we were to find the third angle ∠A, it would be:

A triangle cannot have a negative angle measure.

Since the sum of the two given angles is greater than $180^\circ$, it is impossible to form a valid triangle with these angle measures. The information about the perimeter (AB + BC + AC = 10 cm) is irrelevant because the geometric condition regarding the sum of angles is not satisfied.

The statement is True.

First, let's check if the given angles are compatible with the properties of a triangle. The sum of the interior angles in any Euclidean triangle must be $180^\circ$.

We are given ∠B = 60° and ∠C = 45°. Let's find the measure of the third angle, ∠A:

Substitute the given values:

Since all three angles ($75^\circ$, $60^\circ$, $45^\circ$) are positive and sum to $180^\circ$, a triangle with these angle measures can exist.

Next, let's consider the constructibility with a ruler and compass.

- An angle of $60^\circ$ is constructible (by constructing an equilateral triangle).

- An angle of $45^\circ$ is constructible ($90^\circ$ can be constructed by drawing a perpendicular, and $45^\circ$ is the bisection of $90^\circ$).

- An angle of $75^\circ$ is constructible because $75^\circ = 45^\circ + 30^\circ$. Both $45^\circ$ and $30^\circ$ ($60^\circ$ bisected) are constructible, and the sum/difference of constructible angles is constructible.

A triangle can be constructed when the perimeter and two base angles are given. The standard method involves drawing a line segment equal to the perimeter, constructing the two given angles at the ends of this segment, and then finding the vertices of the triangle using perpendicular bisectors of the sides formed by connecting the third vertex (where the angle rays meet) to the endpoints of the perimeter segment.

Given that the angles are valid for a triangle, the angles are constructible, and the perimeter is a positive length, a triangle with these specifications can be constructed.

Given: A triangle ABC where:

• BC = 7.5 cm
• ∠B = 45°
• AB – AC = 4 cm

To Construct: Triangle ABC.

Steps of Construction:

1. Draw a line segment BC of length 7.5 cm.

2. At point B, construct an angle $\angle\text{XBC} = 45^\circ$ using a ruler and compass. Draw the ray BX.

3. Along the ray BX, from point B, cut off a line segment BD equal to the difference AB - AC, which is 4 cm.

4. Join point D to point C to form the line segment DC.

5. Draw the perpendicular bisector of the line segment DC. To do this, with D as the center and a radius more than half the length of DC, draw arcs on both sides of DC. With C as the center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of the arcs. This line is the perpendicular bisector of DC.

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

7. Join A to C.

Triangle ABC is the required triangle.

Justification:

The point A lies on the perpendicular bisector of the line segment DC. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.

From the construction, point D was marked on the ray BX such that BD = 4 cm. Since A lies on the ray BX and D is between B and A (because AB > AC implies AB > 4 and BD = 4), we have:

Substitute $\text{AD} = \text{AC}$:

Rearrange the equation:

By construction, $\text{DB} = 4 \text{ cm}$.

Also, by construction, BC = 7.5 cm and $\angle\text{B} = 45^\circ$.

Thus, the triangle ABC constructed satisfies all the given conditions.

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

To Construct:

• Draw an angle of $110^\circ$ using a protractor.
• Bisect the constructed angle using a ruler and compass.
• Measure the angles obtained after bisection.

Steps of Construction:

1. Draw a ray BC with initial point B. This will be one arm of the angle.

2. Place the protractor at point B such that its base line coincides with the ray BC and the center mark is exactly on B.

3. Start reading the inner scale of the protractor from $0^\circ$ along BC. Mark a point A on the paper against the $110^\circ$ mark on the protractor's scale.

4. Remove the protractor and draw a ray BA joining B to A.

The angle $\angle\text{ABC}$ is the required angle of $110^\circ$.

Now, let's bisect the angle $\angle\text{ABC}$.

5. With B as the center and a convenient radius, draw an arc that intersects ray BA at point P and ray BC at point Q.

6. With P as the center and a radius greater than half the length of PQ, draw an arc inside the angle.

7. With Q as the center and the same radius as in step 6, draw another arc that intersects the previous arc at point R.

8. Draw the ray BR joining point B to point R.

The ray BR is the angle bisector of $\angle\text{ABC}$. It divides the angle $\angle\text{ABC}$ into two equal angles, $\angle\text{ABR}$ and $\angle\text{RBC}$.

Measurement of each angle:

Using a protractor, measure the angles $\angle\text{ABR}$ and $\angle\text{RBC}$.

The measure of each bisected angle should be half of the original angle.

Similarly,

Upon measurement, we find that $\angle\text{ABR} = 55^\circ$ and $\angle\text{RBC} = 55^\circ$.

Given: A line segment AB of length 4 cm.

To Construct:

• A line perpendicular to AB passing through A.
• A line perpendicular to AB passing through B.

To Determine: Whether the two constructed lines are parallel.

Steps of Construction:

1. Draw a line segment AB of length 4 cm.

2. Construction of a perpendicular at A:

a. With A as the center and any convenient radius, draw an arc that intersects the line AB at some point, say P. Extend the line segment AB beyond A temporarily if needed to draw the arc.

b. With P as the center and the same radius, draw an arc intersecting the first arc at a point Q.

c. With Q as the center and the same radius, draw an arc intersecting the first arc at a point R.

d. With Q and R as centers and a radius greater than half the distance between Q and R, draw two arcs intersecting each other at a point S.

e. Draw a ray AT passing through A and S. The line AT is perpendicular to AB at A, i.e., $\angle\text{TAB} = 90^\circ$. Let's call this line $l_1$.

3. Construction of a perpendicular at B:

a. With B as the center and any convenient radius, draw an arc that intersects the line AB at some point, say U. Extend the line segment AB beyond B temporarily if needed to draw the arc.

b. With U as the center and the same radius, draw an arc intersecting the first arc at a point V.

c. With V as the center and the same radius, draw an arc intersecting the first arc at a point W.

d. With V and W as centers and a radius greater than half the distance between V and W, draw two arcs intersecting each other at a point Z.

e. Draw a ray BX passing through B and Z. The line BX is perpendicular to AB at B, i.e., $\angle\text{XBA} = 90^\circ$. Let's call this line $l_2$.

Conclusion:

We have constructed line $l_1$ such that $l_1 \perp \text{AB}$ at A, and line $l_2$ such that $l_2 \perp \text{AB}$ at B.

Both lines $l_1$ and $l_2$ are perpendicular to the same line segment AB.

In Euclidean geometry, if two lines are perpendicular to the same line, then they are parallel to each other.

Therefore, the line drawn perpendicular to AB through A and the line drawn perpendicular to AB through B are parallel to each other.

Given: An angle of $80^\circ$.

To Construct: Angles of $40^\circ$, $160^\circ$, and $120^\circ$.

Steps of Construction:

Draw $80^\circ$:

1. Draw a ray OB. Use a protractor to draw ray OA such that $\angle\text{AOB} = 80^\circ$.

(i) Construct $40^\circ$:

$40^\circ = \frac{1}{2} \times 80^\circ$. Bisect the angle $\angle\text{AOB}$.

1. With O as center and a suitable radius, draw an arc intersecting OA at P and OB at Q.

2. With P and Q as centers and a radius greater than $\frac{1}{2}\text{PQ}$, draw arcs intersecting at R.

3. Draw ray OR. $\angle\text{ROB} = 40^\circ$.

(ii) Construct $160^\circ$:

$160^\circ = 2 \times 80^\circ$. Add an $80^\circ$ angle adjacent to $\angle\text{AOB}$.

1. Extend the arc from the initial $80^\circ$ construction.

2. Measure the chord length PQ (from the $80^\circ$ angle $\angle\text{AOB}$).

3. With P (on OA) as center and radius equal to PQ, draw an arc intersecting the extended arc at S.

4. Draw ray OS. $\angle\text{SOB} = \angle\text{SOA} + \angle\text{AOB} = 80^\circ + 80^\circ = 160^\circ$.

(iii) Construct $120^\circ$:

$120^\circ = 80^\circ + 40^\circ$. Add the $40^\circ$ angle (obtained by bisection) adjacent to the $80^\circ$ angle.

1. We have ray OB and $\angle\text{AOB} = 80^\circ$. We also have ray OR bisecting $\angle\text{AOB}$, such that $\angle\text{AOR} = 40^\circ$.

2. Measure the chord length PR (from the bisection construction, representing $40^\circ$).

3. With P (on OA) as center and radius equal to PR, draw an arc.

4. With O as center and the initial radius (from step 5 of 40° construction), draw an arc intersecting the arc from step 3 at T.

5. Draw ray OT. $\angle\text{BOT} = \angle\text{BOA} + \angle\text{AOT} = 80^\circ + 40^\circ = 120^\circ$.

Given: A triangle with side lengths 3.6 cm, 3.0 cm, and 4.8 cm.

To Construct:

• The triangle with the given side lengths.
• The bisector of the smallest angle.
• Measure each part of the bisected angle.

Finding the smallest angle: In any triangle, the angle opposite the smallest side is the smallest angle. The given side lengths are 3.6 cm, 3.0 cm, and 4.8 cm. The smallest side is 3.0 cm. Let's construct the triangle PQR such that PR = 3.0 cm, QR = 3.6 cm, and PQ = 4.8 cm. The smallest angle is the angle opposite the side PR, which is $\angle$Q.

Steps of Construction:

1. Draw a line segment PQ of length 4.8 cm.

2. With P as the center and a radius of 3.0 cm (the length of PR), draw an arc.

3. With Q as the center and a radius of 3.6 cm (the length of QR), draw another arc.

4. Let the two arcs intersect at point R.

5. Join PR and QR. Triangle PQR is the required triangle.

6. Now, we need to bisect the smallest angle, which is $\angle$Q. With Q as the center and a convenient radius, draw an arc intersecting ray QP at point M and ray QR at point N.

7. With M as the center and a radius greater than half the length of MN, draw an arc inside $\angle$PQR.

8. With N as the center and the same radius as in step 7, draw another arc intersecting the previous arc at point S.

9. Draw the ray QS. The ray QS is the angle bisector of $\angle$PQR.

Measurement:

Using a protractor, measure the angle $\angle$PQR (or $\angle$Q). Let's say the measure is approximately $41^\circ$ (using the Law of Cosines, $\cos Q = \frac{3.6^2 + 4.8^2 - 3.0^2}{2 \times 3.6 \times 4.8} = \frac{12.96 + 23.04 - 9}{34.56} = \frac{27}{34.56} \approx 0.78125$, which gives $Q \approx 38.68^\circ$). The actual measured value might vary slightly due to drawing inaccuracies.

Now, measure the angles $\angle$PQS and $\angle$RQS.

Each of these angles should be half the measure of $\angle$PQR.

If $\angle\text{PQR}$ is measured as approximately $41^\circ$, then $\angle\text{PQS}$ and $\angle\text{RQS}$ should each be approximately $20.5^\circ$. If $\angle\text{PQR}$ is measured as approximately $38.7^\circ$, then $\angle\text{PQS}$ and $\angle\text{RQS}$ should each be approximately $19.35^\circ$.

Given: A triangle ABC where:

• BC = 5 cm
• ∠B = 60°
• AB + AC = 7.5 cm

To Construct: Triangle ABC.

Steps of Construction:

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

2. At point B, construct an angle $\angle\text{XBC} = 60^\circ$ using a ruler and compass. Draw the ray BX.

3. Along the ray BX, from point B, cut off a line segment BD equal to the sum of AB and AC, which is 7.5 cm.

4. Join point D to point C to form the line segment DC.

5. Draw the perpendicular bisector of the line segment DC. To do this, with D as the center and a radius more than half the length of DC, draw arcs on both sides of DC. With C as the center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of the arcs. This line is the perpendicular bisector of DC.

6. Let the perpendicular bisector of DC intersect the ray BX at point A.

7. Join A to C.

Triangle ABC is the required triangle.

Justification:

The point A lies on the perpendicular bisector of the line segment DC. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.

From the construction, point D was marked on the ray BX such that BD = 7.5 cm. Since A lies on the ray BX, we have:

Substitute $\text{AD} = \text{AC}$ into this equation:

By construction, $\text{BD} = 7.5 \text{ cm}$.

Also, by construction, BC = 5 cm and $\angle\text{B} = 60^\circ$.

Thus, the triangle ABC constructed satisfies all the given conditions.

Given: A square of side length 3 cm.

To Construct: A square with side length 3 cm.

Properties of a Square: A square is a quadrilateral with four equal sides and four right angles ($90^\circ$).

Steps of Construction:

1. Draw a line segment AB of length 3 cm.

2. At point A, construct a right angle ($\angle\text{PAB} = 90^\circ$). This can be done by:

a. With A as the center and a convenient radius, draw an arc intersecting AB at some point, say X.

b. With X as the center and the same radius, draw an arc intersecting the first arc at Y ($60^\circ$).

c. With Y as the center and the same radius, draw an arc intersecting the first arc at Z ($120^\circ$).

d. With Y and Z as centers and a radius greater than half the distance YZ, draw arcs intersecting each other at a point P.

e. Draw the ray AP. The angle $\angle\text{PAB}$ is $90^\circ$.

3. Along the ray AP, from point A, cut off a line segment AD of length 3 cm.

4. With D as the center and a radius of 3 cm, draw an arc.

5. With B as the center and a radius of 3 cm, draw another arc intersecting the arc from step 4 at point C.

6. Join DC and BC.

ABCD is the required square with side length 3 cm.

Verification:

By construction, AB = AD = 3 cm and $\angle\text{DAB} = 90^\circ$. Also, BC = 3 cm and DC = 3 cm by the radii used in steps 4 and 5. Thus, all four sides are equal to 3 cm. Consider quadrilateral ABCD. We have AB = BC = CD = DA = 3 cm. We know $\angle\text{DAB} = 90^\circ$. In a quadrilateral with all sides equal, if one angle is $90^\circ$, it is a square. Alternatively, we can show that BC is parallel to AD and DC is parallel to AB, making it a parallelogram. Since adjacent sides are equal and one angle is $90^\circ$, it is a square.

Given: A rectangle with adjacent sides of lengths 5 cm and 3.5 cm.

To Construct: A rectangle with adjacent side lengths 5 cm and 3.5 cm.

Properties of a Rectangle: A rectangle is a quadrilateral with opposite sides equal and parallel, and all four angles equal to $90^\circ$. The adjacent sides are perpendicular to each other.

Steps of Construction:

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

2. At point A, construct a right angle ($\angle\text{PAB} = 90^\circ$). This can be done using the standard construction for a $90^\circ$ angle:

a. With A as the center and a convenient radius, draw an arc intersecting AB at X.

b. With X as the center and the same radius, draw an arc intersecting the first arc at Y ($60^\circ$).

c. With Y as the center and the same radius, draw an arc intersecting the first arc at Z ($120^\circ$).

d. With Y and Z as centers and a radius greater than half YZ, draw arcs intersecting at P.

e. Draw the ray AP. The angle $\angle\text{PAB}$ is $90^\circ$.

3. Along the ray AP, from point A, cut off a line segment AD of length 3.5 cm (the length of the adjacent side).

4. With D as the center and a radius of 5 cm (the length of AB), draw an arc.

5. With B as the center and a radius of 3.5 cm (the length of AD), draw another arc intersecting the arc from step 4 at point C.

6. Join DC and BC.

ABCD is the required rectangle.

Verification:

By construction, AB = 5 cm, AD = 3.5 cm, and $\angle\text{DAB} = 90^\circ$. By the radii used in steps 4 and 5, DC = 5 cm and BC = 3.5 cm. Thus, opposite sides are equal (AB=DC, AD=BC). In quadrilateral ABCD, we have adjacent sides perpendicular at A, and opposite sides equal. This ensures that all angles are $90^\circ$ and opposite sides are parallel, satisfying the properties of a rectangle.

Given: A rhombus with side length 3.4 cm and one angle equal to $45^\circ$.

To Construct: A rhombus with side 3.4 cm and one angle $45^\circ$.

Properties of a Rhombus: A rhombus is a quadrilateral with all four sides of equal length. Opposite angles are equal, and adjacent angles are supplementary (sum to $180^\circ$).

If one angle is $45^\circ$, the adjacent angles are $180^\circ - 45^\circ = 135^\circ$. The angles of the rhombus will be $45^\circ, 135^\circ, 45^\circ, 135^\circ$.

Steps of Construction:

1. Draw a line segment AB of length 3.4 cm.

2. At point A, construct an angle of $45^\circ$. This can be done by constructing a $90^\circ$ angle and then bisecting it:

a. Construct a $90^\circ$ angle at A (e.g., ray AP such that $\angle\text{PAB} = 90^\circ$).

b. Bisect the angle $\angle\text{PAB}$ to get a $45^\circ$ angle. Let the angle bisector be ray AQ such that $\angle\text{QAB} = 45^\circ$. (Refer to steps for bisecting an angle if needed).

3. Along the ray AQ, from point A, cut off a line segment AD of length 3.4 cm (since all sides of a rhombus are equal).

4. With D as the center and a radius of 3.4 cm, draw an arc.

5. With B as the center and a radius of 3.4 cm, draw another arc intersecting the arc from step 4 at point C.

6. Join DC and BC.

ABCD is the required rhombus.

Verification:

By construction, AB = AD = BC = DC = 3.4 cm. Thus, all four sides are equal. Also, $\angle\text{DAB} = 45^\circ$ by construction. Since all sides are equal, ABCD is a rhombus. The opposite angle $\angle\text{BCD}$ will also be $45^\circ$. The adjacent angles $\angle\text{ABC}$ and $\angle\text{ADC}$ will be $180^\circ - 45^\circ = 135^\circ$.

Given: An equilateral triangle whose altitude is 6 cm.

To Construct: An equilateral triangle ABC with altitude 6 cm.

Construction Required: We need to construct an equilateral triangle given its altitude. In an equilateral triangle, the altitude from a vertex is also the median and the angle bisector of the angle at that vertex. The angle at each vertex of an equilateral triangle is $60^\circ$. Thus, the altitude bisects the vertex angle into two angles of $30^\circ$. The altitude is also perpendicular to the opposite side.

Steps of Construction:

1. Draw a line $l$.

2. Mark a point D on the line $l$.

3. Construct a perpendicular to the line $l$ at point D. Let the ray perpendicular to $l$ at D be DX. (This represents the line containing the altitude).

4. Along the ray DX, from point D, cut off a line segment DA equal to the given altitude, i.e., AD = 6 cm. Point A will be one vertex of the equilateral triangle.

5. At point A, construct an angle of $30^\circ$ on both sides of the altitude line segment AD. This can be done by:

a. Constructing a $60^\circ$ angle (standard construction) with vertex A and one arm along AD. Let the other arm be a ray, say AE. $\angle\text{DAE} = 60^\circ$.

b. Bisecting the $60^\circ$ angle $\angle\text{DAE}$ to obtain a $30^\circ$ angle. Let the bisector ray be AF. $\angle\text{DAF} = 30^\circ$.

c. Repeating steps 5a and 5b on the other side of AD to construct a ray AG such that $\angle\text{DAG} = 30^\circ$.

6. Extend the rays AF and AG to intersect the line $l$ at points B and C respectively.

7. Join AB and AC.

Triangle ABC is the required equilateral triangle.

Justification:

By construction, AD is perpendicular to BC (line $l$), so $\angle\text{ADB} = \angle\text{ADC} = 90^\circ$.

Also by construction, $\angle\text{BAD} = 30^\circ$ and $\angle\text{CAD} = 30^\circ$.

Consider $\triangle\text{ABD}$ and $\triangle\text{ACD}$.

Therefore, by the ASA congruence criterion, $\triangle\text{ABD} \cong \triangle\text{ACD}$.

From the congruence, we have:

Since BD = CD, D is the midpoint of BC, and AD is the median to BC. In $\triangle\text{ABC}$, the altitude AD is also the median to BC. This property is true for isosceles triangles.

Now, let's find the angles $\angle\text{ABD}$ and $\angle\text{ACD}$.

In $\triangle\text{ABD}$, the sum of angles is $180^\circ$.

Similarly, in $\triangle\text{ACD}$, $\angle\text{ACD} = 60^\circ$.

So, $\angle\text{ABC} = 60^\circ$ and $\angle\text{ACB} = 60^\circ$.

The angle $\angle\text{BAC} = \angle\text{BAD} + \angle\text{CAD} = 30^\circ + 30^\circ = 60^\circ$.

Thus, $\angle\text{BAC} = \angle\text{ABC} = \angle\text{ACB} = 60^\circ$.

Since all three angles of triangle ABC are equal to $60^\circ$, the triangle is equiangular. An equiangular triangle is also equilateral, meaning all its sides are equal: AB = BC = AC.

Finally, the altitude AD is 6 cm by construction, which matches the given condition.

Therefore, the constructed triangle ABC is an equilateral triangle with an altitude of 6 cm.

Given: Triangle ABC with Perimeter = 10.4 cm, ∠B = 45°, ∠C = 120°.

To Construct: Triangle ABC.

Steps of Construction:

1. Draw a line segment PQ = 10.4 cm.

2. At P, construct $\angle\text{XPQ} = \frac{1}{2} \times 45^\circ = 22.5^\circ$. Draw ray PX.

3. At Q, construct $\angle\text{YQP} = \frac{1}{2} \times 120^\circ = 60^\circ$. Draw ray QY.

4. Let rays PX and QY intersect at A.

5. Draw the perpendicular bisector of AP, intersecting PQ at B.

6. Draw the perpendicular bisector of AQ, intersecting PQ at C.

7. Join AB and AC. Triangle ABC is the required triangle.

Justification:

B is on the perpendicular bisector of AP, so $\text{AB} = \text{BP}$.

C is on the perpendicular bisector of AQ, so $\text{AC} = \text{CQ}$.

Perimeter of $\triangle\text{ABC} = \text{AB} + \text{BC} + \text{AC} = \text{BP} + \text{BC} + \text{CQ} = \text{PQ} = 10.4 \text{ cm}$ (since B and C are on PQ).

In $\triangle\text{ABP}$, AB = BP, so $\angle\text{BAP} = \angle\text{BPA} = 22.5^\circ$. By exterior angle theorem, $\angle\text{ABC} = \angle\text{BAP} + \angle\text{BPA} = 22.5^\circ + 22.5^\circ = 45^\circ$.

In $\triangle\text{ACQ}$, AC = CQ, so $\angle\text{CAQ} = \angle\text{CQA} = 60^\circ$. By exterior angle theorem, $\angle\text{ACB} = \angle\text{CAQ} + \angle\text{CQA} = 60^\circ + 60^\circ = 120^\circ$.

The triangle ABC has the given perimeter and two of its angles are $45^\circ$ and $120^\circ$.

Given: A triangle PQR where:

• QR = 3 cm
• ∠PQR = 45°
• QP – PR = 2 cm

To Construct: Triangle PQR.

Steps of Construction:

1. Draw a line segment QR of length 3 cm.

2. At point Q, construct an angle $\angle\text{XQR} = 45^\circ$ using a ruler and compass. Draw the ray QX.

3. Along the ray QX, from point Q, cut off a line segment QD equal to the difference QP – PR, which is 2 cm.

4. Join point D to point R to form the line segment DR.

5. Draw the perpendicular bisector of the line segment DR. To do this, with D as the center and a radius more than half the length of DR, draw arcs on both sides of DR. With R as the center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of the arcs. This line is the perpendicular bisector of DR.

6. Let the perpendicular bisector of DR intersect the ray QX at point P.

7. Join P to R.

Triangle PQR is the required triangle.

Justification:

The point P lies on the perpendicular bisector of the line segment DR. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.

From the construction, point D was marked on the ray QX such that QD = 2 cm. Since P lies on the ray QX and D is between Q and P (because QP > PR implies QP > 2 and QD = 2), we have:

Substitute $\text{DP} = \text{PR}$:

Rearrange the equation:

By construction, $\text{QD} = 2 \text{ cm}$.

Also, by construction, QR = 3 cm and $\angle\text{PQR} = 45^\circ$.

Thus, the triangle PQR constructed satisfies all the given conditions.

Given: A right triangle ABC, with the right angle at B ($\angle\text{B} = 90^\circ$). One side (let's assume it is one of the legs, BC) = 3.5 cm. The sum of the other leg (AB) and the hypotenuse (AC) is 5.5 cm, i.e., AB + AC = 5.5 cm.

To Construct: The right triangle ABC.

Construction Required: We need to construct a right triangle given one leg and the sum of the other leg and the hypotenuse. This is a variation of constructing a triangle given one side, an adjacent angle, and the sum of the other two sides.

Steps of Construction:

1. Draw a line segment BC of length 3.5 cm.

2. At point B, construct a ray BX perpendicular to BC, forming a right angle ($\angle\text{CBX} = 90^\circ$). This ray will contain the side AB.

3. Along the ray BX, from point B, cut off a line segment BD equal to the sum of the other side and the hypotenuse (AB + AC), which is 5.5 cm.

4. Join point D to point C to form the line segment DC.

5. Draw the perpendicular bisector of the line segment DC. To do this, with D as the center and a radius more than half the length of DC, draw arcs on both sides of DC. With C as the center and the same radius, draw arcs intersecting the previous arcs. Join the points of intersection of the arcs. This line is the perpendicular bisector of DC.

6. Let the perpendicular bisector of DC intersect the ray BX at point A.

7. Join A to C.

Triangle ABC is the required right triangle.

Justification:

By construction, BC = 3.5 cm and $\angle\text{ABC} = 90^\circ$ (since A lies on ray BX which is perpendicular to BC).

The point A lies on the perpendicular bisector of the line segment DC. By the property of a perpendicular bisector, any point on it is equidistant from the endpoints of the segment.

From the construction, point D was marked on the ray BX such that BD = 5.5 cm. Since A lies on the ray BX, and given the angle $\angle\text{CBX}=90^\circ$ and segment BC, the perpendicular bisector of DC will intersect ray BX at a point A such that A is between B and D. Therefore,

Substitute $\text{AD} = \text{AC}$ into this equation:

By construction, $\text{BD} = 5.5 \text{ cm}$.

Thus, the constructed triangle ABC is a right triangle with BC = 3.5 cm, $\angle\text{B} = 90^\circ$, and AB + AC = 5.5 cm, satisfying all the given conditions.

Given: An equilateral triangle whose altitude is 3.2 cm.

To Construct: An equilateral triangle ABC with altitude 3.2 cm.

Construction Required: In an equilateral triangle, the altitude from a vertex is also the median and the angle bisector of the vertex angle. The vertex angle is $60^\circ$, so the altitude bisects it into two $30^\circ$ angles. The altitude is also perpendicular to the opposite side.

Steps of Construction:

1. Draw a line $l$.

2. Mark a point D on the line $l$.

3. Construct a perpendicular to the line $l$ at point D. Let the ray perpendicular to $l$ at D be DX.

4. Along the ray DX, from point D, cut off a line segment DA equal to the given altitude, i.e., AD = 3.2 cm. Point A is one vertex of the equilateral triangle.

5. At point A, construct an angle of $30^\circ$ on both sides of the altitude line AD.

a. Construct a $60^\circ$ angle with vertex A and one arm along AD. Bisect this $60^\circ$ angle. Let the bisector ray on one side be AF. $\angle\text{DAF} = 30^\circ$.

b. Repeat the bisection on the other side of AD to construct a ray AG such that $\angle\text{DAG} = 30^\circ$.

6. Extend the rays AF and AG to intersect the line $l$ at points B and C respectively.

7. Join AB and AC.

Triangle ABC is the required equilateral triangle.

Justification:

By construction, AD is perpendicular to BC ($\angle\text{ADB} = \angle\text{ADC} = 90^\circ$), and $\angle\text{BAD} = \angle\text{CAD} = 30^\circ$.

In $\triangle\text{ABD}$ and $\triangle\text{ACD}$:

By ASA congruence, $\triangle\text{ABD} \cong \triangle\text{ACD}$.

Therefore, AB = AC and BD = CD.

In $\triangle\text{ABD}$, $\angle\text{ABD} = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.

Similarly, in $\triangle\text{ACD}$, $\angle\text{ACD} = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.

The angles of $\triangle\text{ABC}$ are $\angle\text{ABC} = 60^\circ$, $\angle\text{ACB} = 60^\circ$, and $\angle\text{BAC} = \angle\text{BAD} + \angle\text{CAD} = 30^\circ + 30^\circ = 60^\circ$.

Since all angles are $60^\circ$, the triangle is equilateral, and thus all sides are equal (AB = BC = AC). AD is the altitude of 3.2 cm by construction.

Given: A rhombus whose diagonals are 4 cm and 6 cm.

To Construct: The rhombus with the given diagonal lengths.

Properties of a Rhombus:

• The diagonals of a rhombus bisect each other at right angles ($90^\circ$).
• The diagonals divide the rhombus into four congruent right-angled triangles.

Let the diagonals be $d_1 = 4$ cm and $d_2 = 6$ cm. When they bisect each other, the half-lengths of the diagonals are $\frac{d_1}{2} = \frac{4}{2} = 2$ cm and $\frac{d_2}{2} = \frac{6}{2} = 3$ cm.

Steps of Construction:

1. Draw a line segment representing one of the diagonals. Let's draw the diagonal AC of length 6 cm.

2. Draw the perpendicular bisector of the line segment AC. This bisector will pass through the midpoint of AC and be perpendicular to AC. Let the midpoint of AC be O.

a. With A as the center and a radius more than half of AC, draw arcs above and below AC.

b. With C as the center and the same radius, draw arcs intersecting the previous arcs.

c. Join the points of intersection. This line is the perpendicular bisector and passes through O, the midpoint of AC.

3. The other diagonal BD lies on this perpendicular bisector and is bisected at O. The length of the other diagonal is 4 cm, so half its length is 2 cm.

4. Along the perpendicular bisector, from point O, cut off line segments OB and OD of length 2 cm each on opposite sides of AC.

5. Join AB, BC, CD, and DA.

ABCD is the required rhombus.

Justification:

By construction, AC is a diagonal of length 6 cm. The perpendicular bisector of AC passes through O, the midpoint of AC. So, AO = OC = $\frac{1}{2} \times 6 = 3$ cm. The diagonal BD lies on the perpendicular bisector and is bisected at O. So, BO = OD = 2 cm. Thus, BD is a diagonal of length $2 + 2 = 4$ cm. The diagonals AC and BD bisect each other at right angles at O.

Consider $\triangle\text{AOB}$. It is a right-angled triangle at O.

Similarly, in $\triangle\text{BOC}$, $\text{BC}^2 = \text{BO}^2 + \text{OC}^2 = 2^2 + 3^2 = 4 + 9 = 13$, so $\text{BC} = \sqrt{13}$ cm. In $\triangle\text{COD}$, $\text{CD}^2 = \text{CO}^2 + \text{OD}^2 = 3^2 + 2^2 = 9 + 4 = 13$, so $\text{CD} = \sqrt{13}$ cm. In $\triangle\text{DOA}$, $\text{DA}^2 = \text{DO}^2 + \text{OA}^2 = 2^2 + 3^2 = 4 + 9 = 13$, so $\text{DA} = \sqrt{13}$ cm.

Since all four sides are equal ($\text{AB} = \text{BC} = \text{CD} = \text{DA} = \sqrt{13}$ cm), and the diagonals bisect each other at right angles, the quadrilateral ABCD is a rhombus with diagonals of length 6 cm and 4 cm.