CBSE Class 9 Constructions Assignment Set 02

11. Constructions

Question 1. Draw an angle of \(135^\circ\) using ruler and compasses only.
Answer: 1. Draw a ray \(OA\).
2. Taking \(O\) as center and any radius, draw an arc intersecting \(OA\) at \(P\).
3. With \(P\) as center and same radius, cut the arc at \(Q\), and then with \(Q\) as center cut the arc at \(R\).
4. With \(Q\) and \(R\) as centers, draw two arcs intersecting each other at \(S\). Join \(OS\) to form a \(90^\circ\) angle with \(OA\). Let this ray be \(OB\).
5. The \(90^\circ\) angle is now divided into two parts at point \(T\) where ray \(OB\) intersects the major arc. With the points corresponding to \(90^\circ\) and \(180^\circ\) on the main arc as centers, draw intersecting arcs to bisect the obtuse angle between \(90^\circ\) and \(180^\circ\).
6. The resulting angle is \(90^\circ + 45^\circ = 135^\circ\).
In simple words: An angle of \(135^\circ\) is constructed by adding \(45^\circ\) to a right angle (\(90^\circ\)).

Exam Tip: Always keep your compass width fixed while marking standard degree arcs (\(60^\circ\), \(120^\circ\)).

Question 2. Draw a line segment of length 8 cm. Bisect it and measure the length of each part.
Answer: 1. Draw a line segment \(AB = 8\) cm using a ruler.
2. With \(A\) as center and radius more than half of \(AB\) (\(> 4\) cm), draw arcs above and below the line segment \(AB\).
3. With \(B\) as center and the same radius, draw arcs above and below the line segment intersecting the previous arcs at \(C\) and \(D\).
4. Join \(CD\) intersecting \(AB\) at point \(M\).
5. \(CD\) is the perpendicular bisector of \(AB\). Measuring each part gives \(AM = MB = 4\) cm.
In simple words: A perpendicular bisector divides a line segment into two equal halves of 4 cm each.

Exam Tip: Ensure the compass radius is strictly greater than half the length of the line segment to ensure the arcs intersect properly.

Question 3. Construct an equilateral triangle whose altitude is 4 cm.
Answer: 1. Draw a line \(XY\) of arbitrary length.
2. Take any point \(D\) on \(XY\) and draw a perpendicular line segment \(AD = 4\) cm at \(D\).
3. Since the altitude of an equilateral triangle also bisects the base at \(30^\circ\) to the sides, construct an angle of \(30^\circ\) at \(D\) on both sides of \(AD\), or construct angles of \(60^\circ\) at the base such that the altitude meets the sides at \(90^\circ\).
4. Alternatively, draw a line segment \(PQ\) and construct angles of \(60^\circ\) at \(P\) and \(Q\), then drop a perpendicular of length 4 cm.
5. Complete triangle \(ABC\) with height 4 cm and all angles equal to \(60^\circ\).
In simple words: Draw a reference base, erect a 4 cm altitude, and construct \(60^\circ\) base angles to form the equilateral triangle.

Exam Tip: Remember that in an equilateral triangle, the altitude is also the median and angle bisector.

Question 4. Construct a triangle ABC in which \(AB = 5.8\) cm, \(BC + CA = 8.4\) cm and \(\angle B = 60^\circ\).
Answer: 1. Draw base segment \(AB = 5.8\) cm.
2. At point \(B\), construct an angle \(\angle ABX = 60^\circ\).
3. From ray \(BX\), cut off a line segment \(BD = 8.4\) cm (\(BC + CA\)).
4. Join \(AD\).
5. Draw the perpendicular bisector of \(AD\) which intersects \(BD\) at point \(C\).
6. Join \(AC\). Triangle \(ABC\) is the required triangle.
In simple words: Use the sum of the two sides to make a large auxiliary triangle, then use perpendicular bisector to find the exact vertex \(C\).

Exam Tip: The cut-off length on the ray must equal the exact sum given (\(8.4\) cm).

Question 5. Construct a triangle ABC in which \(BC = 3.4\) cm, \(AB - AC = 1.5\) cm and \(\angle B = 45^\circ\).
Answer: 1. Draw base segment \(BC = 3.4\) cm.
2. At point \(B\), construct an angle \(\angle XBC = 45^\circ\).
3. Cut off line segment \(BD = 1.5\) cm (\(AB - AC\)) from ray \(BX\).
4. Join \(CD\).
5. Draw the perpendicular bisector of \(CD\) intersecting ray \(BX\) at point \(A\).
6. Join \(AC\). Triangle \(ABC\) is the required triangle.
In simple words: When the difference of two sides is given, mark the difference along the base angle ray and use perpendicular bisector to locate the vertex.

Exam Tip: Since \(AB > AC\), the difference is marked directly on the ray containing the angle.

Question 6. Construct an equilateral triangle whose altitude is 5 cm.
Answer: 1. Draw a baseline \(XY\).
2. Choose a point \(D\) on \(XY\) and draw a perpendicular line segment \(AD = 5\) cm.
3. At point \(D\), construct angles of \(30^\circ\) on both the left and right sides, making \(30^\circ\) with \(AD\), which creates \(60^\circ\) angles with the base line \(XY\).
4. The rays from \(A\) intersect \(XY\) at \(B\) and \(C\).
5. Triangle \(ABC\) is the required equilateral triangle with altitude 5 cm.
In simple words: Erect a 5 cm perpendicular and construct \(30^\circ\) lines from the top to meet the base, forming a \(60^\circ\) equilateral triangle.

Question 7. Construct a triangle ABC in which \(AB = 5.8\) cm, \(BC + CA = 8.4\) cm and \(\angle B = 45^\circ\).
Answer: 1. Draw line segment \(AB = 5.8\) cm.
2. At \(B\), construct an angle \(\angle ABX = 45^\circ\).
3. Cut off \(BD = 8.4\) cm from ray \(BX\).
4. Join \(AD\).
5. Construct the perpendicular bisector of \(AD\) to intersect \(BD\) at \(C\).
6. Join \(AC\). Triangle \(ABC\) is formed.
In simple words: Follow standard sum-of-sides construction steps using a \(45^\circ\) base angle.

Exam Tip: Ensure accuracy when bisecting segment \(AD\) to locate \(C\).

Question 8. Construct a right-angled triangle whose base is 5 cm and sum of its hypotenuse and other side is 8 cm. Marks (4)
Answer: 1. Draw base \(BC = 5\) cm.
2. At point \(B\), construct a right angle \(\angle XBC = 90^\circ\).
3. From ray \(BX\), cut off segment \(BD = 8\) cm (sum of hypotenuse and other side).
4. Join \(CD\).
5. Draw the perpendicular bisector of \(CD\) intersecting \(BD\) at point \(A\).
6. Join \(AC\). Triangle \(ABC\) is the required right-angled triangle.
In simple words: Construct a \(90^\circ\) base angle, mark the sum of sides, and use perpendicular bisector to find the third vertex.

Exam Tip: The base angle must be exactly \(90^\circ\) for right-angled triangle constructions.

Question 9. Construct a triangle ABC in which \(BC = 3.4\) cm, \(AB - AC = 1.5\) cm and \(\angle B = 30^\circ\). Marks (4)
Answer: 1. Draw base line segment \(BC = 3.4\) cm.
2. At point \(B\), construct an angle \(\angle XBC = 30^\circ\).
3. Cut off \(BD = 1.5\) cm from ray \(BX\).
4. Join \(CD\).
5. Draw the perpendicular bisector of \(CD\) intersecting ray \(BX\) at point \(A\).
6. Join \(AC\). Triangle \(ABC\) is completed.
In simple words: Use the difference method with a \(30^\circ\) base angle.

Exam Tip: Construct \(30^\circ\) by bisecting a \(60^\circ\) angle at vertex \(B\).

Question 10. Write the steps of constructions for a triangle ABC whose perimeter and two base angles \(\angle B\) and \(\angle C\) are given. Marks (4)
Answer: 1. Draw a line segment \(XY\) equal to the given perimeter (\(AB + BC + CA\)).
2. Construct an angle equal to \(\angle B\) at \(X\) (let us call it \(\angle LXY\)) and an angle equal to \(\angle C\) at \(Y\) (let us call it \(\angle MYX\)).
3. Bisect both angles \(\angle LXY\) and \(\angle MYX\). Let the bisectors intersect at point \(A\).
4. Draw the perpendicular bisectors of \(AX\) and \(AY\), intersecting \(XY\) at \(B\) and \(C\) respectively.
5. Join \(AB\) and \(AC\). Triangle \(ABC\) is the required triangle.
In simple words: Start with the total perimeter line, make half the base angles at the ends, find vertex \(A\), and fold back to find base points \(B\) and \(C\).

Exam Tip: Remember to bisect the base angles before constructing perpendicular bisectors.

Question 11. Using ruler and compasses only, construct a triangle ABC from the following data \(AB + BC + CA = 12\) cm, \(\angle B = 45^\circ\) and \(\angle C = 60^\circ\). Marks (4)
Answer: 1. Draw a line segment \(XY = 12\) cm representing the total perimeter.
2. At point \(X\), construct an angle of \(45^\circ\). Bisect this angle.
3. At point \(Y\), construct an angle of \(60^\circ\). Bisect this angle.
4. Let the two angle bisectors intersect at point \(A\).
5. Draw the perpendicular bisector of \(AX\) to intersect \(XY\) at \(B\), and the perpendicular bisector of \(AY\) to intersect \(XY\) at \(C\).
6. Join \(AB\) and \(AC\). Triangle \(ABC\) is constructed.
In simple words: Use the perimeter line method with half of \(45^\circ\) and half of \(60^\circ\) angles at the ends.

Exam Tip: Bisecting \(45^\circ\) gives \(22.5^\circ\) and bisecting \(60^\circ\) gives \(30^\circ\).

Most Important Questions

Question 1. Construct an angle \(45^\circ\) at the initial point of a line segment PQ of length 6 cm.
Answer: 1. Draw line segment \(PQ = 6\) cm.
2. At point \(P\), construct a \(90^\circ\) angle using compasses.
3. Bisect the \(90^\circ\) angle to obtain an angle of \(45^\circ\).
In simple words: Construct a right angle at \(P\) and bisect it to get \(45^\circ\).

Exam Tip: Always make the \(90^\circ\) construction lines dotted or thin so the final \(45^\circ\) ray stands out.

Question 2. Construct an angle \(30^\circ\) at the initial point of a line segment PQ of length 4 cm.
Answer: 1. Draw line segment \(PQ = 4\) cm.
2. At point \(P\), construct a \(60^\circ\) angle using compasses.
3. Bisect the \(60^\circ\) angle to get \(30^\circ\).
In simple words: Make a \(60^\circ\) angle and cut it in half to get \(30^\circ\).

Exam Tip: Bisecting the angle between the baseline and the \(60^\circ\) arc gives the required \(30^\circ\) ray.

Question 3. Construct an angle \(15^\circ\) at the initial point of a line segment PQ of length 6 cm.
Answer: 1. Draw line segment \(PQ = 6\) cm.
2. Construct a \(60^\circ\) angle at \(P\).
3. Bisect \(60^\circ\) to get \(30^\circ\).
4. Bisect the \(30^\circ\) angle to get \(15^\circ\).
In simple words: Successively bisect \(60^\circ\) down to \(30^\circ\) and finally to \(15^\circ\).

Exam Tip: Keep all intermediate arc markings visible for full credit.

Question 4. Construct an angle \(105^\circ\) at the initial point of a line segment PQ of length 4 cm.
Answer: 1. Draw line segment \(PQ = 4\) cm.
2. At point \(P\), construct a \(90^\circ\) angle and a \(120^\circ\) angle.
3. Bisect the region between \(90^\circ\) and \(120^\circ\) to get \(105^\circ\) (\(90^\circ + 15^\circ\)).
In simple words: Construct \(105^\circ\) halfway between \(90^\circ\) and \(120^\circ\).

Question 5. Construct an angle \(135^\circ\) at the initial point of a line segment PQ of length 5 cm.
Answer: 1. Draw line segment \(PQ = 5\) cm.
2. Construct a \(90^\circ\) angle and extend the line backwards or construct a \(180^\circ\) straight angle.
3. Bisect the angle between \(90^\circ\) and \(180^\circ\) to get \(135^\circ\).
In simple words: \(135^\circ\) is the bisector between \(90^\circ\) and \(180^\circ\).

Exam Tip: Use the standard \(90^\circ\) and \(180^\circ\) reference points to construct the bisector.

Question 6. Construct an angle \(22\frac{1}{2}^\circ\) at the initial point of a line segment PQ of length 7 cm.
Answer: 1. Draw line segment \(PQ = 7\) cm.
2. Construct a \(90^\circ\) angle at \(P\).
3. Bisect \(90^\circ\) to get \(45^\circ\).
4. Bisect \(45^\circ\) to get \(22.5^\circ\) (\(22\frac{1}{2}^\circ\)).
In simple words: Halve a right angle twice to get \(22.5^\circ\).

Exam Tip: Step 1: \(90^\circ \rightarrow 45^\circ \rightarrow 22.5^\circ\).

Question 7. Construct an angle \(75^\circ\) at the initial point of a line segment PQ of length 5 cm.
Answer: 1. Draw line segment \(PQ = 5\) cm.
2. Construct a \(60^\circ\) angle and a \(90^\circ\) angle at \(P\).
3. Bisect the angle between \(60^\circ\) and \(90^\circ\) to get \(75^\circ\).
In simple words: \(75^\circ\) is exactly halfway between \(60^\circ\) and \(90^\circ\).

Exam Tip: Use the \(60^\circ\) and \(90^\circ\) arc intersections as centers for the final bisecting arcs.

Question 8. Construct a triangle PQR, in which \(PQ = 7\) cm, \(\angle P = 60^\circ\) and \(PR + RQ = 13\) cm.
Answer: 1. Draw base \(PQ = 7\) cm.
2. At point \(P\), construct an angle \(\angle QPX = 60^\circ\).
3. From ray \(PX\), cut off \(PD = 13\) cm.
4. Join \(DQ\).
5. Draw the perpendicular bisector of \(DQ\) intersecting \(PD\) at point \(R\).
6. Join \(RQ\). Triangle \(PQR\) is formed.
In simple words: Standard sum-of-sides triangle construction using \(60^\circ\) and a 13 cm cut-off.

Exam Tip: Ensure your ruler scale is aligned correctly when measuring the 13 cm segment.

Question 9. Construct a triangle PQR, in which \(PQ = 6\) cm, \(\angle P = 45^\circ\) and \(PR + RQ = 10\) cm.
Answer: 1. Draw base \(PQ = 6\) cm.
2. At point \(P\), construct a \(45^\circ\) angle \(\angle QPX\).
3. Cut off \(PD = 10\) cm from ray \(PX\).
4. Join \(DQ\).
5. Draw the perpendicular bisector of \(DQ\) intersecting \(PD\) at \(R\).
6. Join \(RQ\). Triangle \(PQR\) is constructed.
In simple words: Construct \(45^\circ\) at \(P\), mark 10 cm, and use perpendicular bisector to find \(R\).

Question 10. Construct a triangle PQR, in which \(PQ = 8\) cm, \(\angle P = 45^\circ\) and \(PR - RQ = 3\) cm.
Answer: 1. Draw base \(PQ = 8\) cm.
2. At point \(P\), construct a \(45^\circ\) angle \(\angle QPX\).
3. Cut off \(PD = 3\) cm from ray \(PX\).
4. Join \(DQ\).
5. Draw the perpendicular bisector of \(DQ\) intersecting ray \(PX\) at point \(R\).
6. Join \(RQ\). Triangle \(PQR\) is formed.
In simple words: Use the difference method (\(PR - RQ = 3\) cm) along ray \(PX\).

Exam Tip: Since \(PR > RQ\), the difference is cut directly along the positive ray \(PX\).

Question 11. Construct a triangle PQR, in which \(PQ = 7\) cm, \(\angle P = 60^\circ\) and \(RQ - PR = 2.5\) cm.
Answer: 1. Draw base \(PQ = 7\) cm.
2. At point \(P\), construct a \(60^\circ\) angle \(\angle QPX\). Since \(RQ > PR\), produce ray \(PX\) in the opposite direction below/behind \(P\) to form line \(X'PX\).
3. Cut off \(PD = 2.5\) cm along the opposite ray \(PX'\).
4. Join \(DQ\).
5. Draw the perpendicular bisector of \(DQ\) intersecting ray \(PX\) at point \(R\).
6. Join \(RQ\). Triangle \(PQR\) is constructed.
In simple words: When the second side is larger (\(RQ > PR\)), the difference is marked on the opposite extension of the angle ray.

Exam Tip: Pay careful attention to whether \(PR - RQ\) or \(RQ - PR\) is given to determine which side of \(P\) to mark the difference.

Question 12. Construct a triangle PQR, in which \(PQ = 7\) cm, \(\angle P = 30^\circ\) and \(PR - RQ = 2\) cm.
Answer: 1. Draw base \(PQ = 7\) cm.
2. At point \(P\), construct a \(30^\circ\) angle \(\angle QPX\).
3. Cut off \(PD = 2\) cm along ray \(PX\).
4. Join \(DQ\).
5. Draw the perpendicular bisector of \(DQ\) intersecting ray \(PX\) at point \(R\).
6. Join \(RQ\). Triangle \(PQR\) is completed.
In simple words: Construct \(30^\circ\) at \(P\), cut 2 cm, and bisect \(DQ\) to locate \(R\).

Exam Tip: Ensure the \(30^\circ\) angle is accurately bisected from an initial \(60^\circ\) construct.

Question 13. Construct a similar triangle PQR, in which \(\angle P = 30^\circ\), \(\angle Q = 60^\circ\) and \(PR + RQ + QP = 12\) cm.
Answer: 1. Draw a line segment \(XY = 12\) cm representing the perimeter.
2. At \(X\), construct an angle equal to \(\angle P = 30^\circ\) and bisect it.
3. At \(Y\), construct an angle equal to \(\angle Q = 60^\circ\) and bisect it.
4. Let the bisectors intersect at point \(R\).
5. Draw the perpendicular bisectors of \(RX\) and \(RY\) to intersect \(XY\) at \(P\) and \(Q\) respectively.
6. Join \(RP\) and \(RQ\). Triangle \(PQR\) is constructed.
In simple words: Use the total perimeter method with half of the given base angles at the endpoints.

Exam Tip: This is a standard perimeter-based triangle construction problem.

Question 14. Construct a triangle PQR, in which \(\angle P = 45^\circ\), \(\angle Q = 60^\circ\) and \(PR + RQ + QP = 9\) cm.
Answer: 1. Draw a line segment \(XY = 9\) cm representing the total perimeter.
2. At \(X\), construct an angle of \(45^\circ\) and bisect it (\(22.5^\circ\)).
3. At \(Y\), construct an angle of \(60^\circ\) and bisect it (\(30^\circ\)).
4. Let the bisectors intersect at vertex \(R\).
5. Draw perpendicular bisectors of \(RX\) and \(RY\) to intersect \(XY\) at \(P\) and \(Q\).
6. Join \(RP\) and \(RQ\). Triangle \(PQR\) is completed.
In simple words: Draw a 9 cm base line, make half-angles at each end, and use perpendicular bisectors to find the vertices.

Exam Tip: Verify that the sum of segments \(XP + PQ + QY\) equals the 9 cm perimeter line.