CBSE Class 9 Constructions Assignment Set 03

Chapter - 11 (Constructions)

Key Concept

(1) Use only ruler and compass while drawing constructions.
(2) Protractor may be used for drawing non-standard angles.
(3) Constructions of a triangle given its base, a base angle and the difference of the other two sides.
(4) Constructions of a triangle given its perimeter and its two base angles.

Section - A

Question 1. With a ruler and compass which of the following angles cannot be constructed?
(a) \(60^\circ\)
(b) \(80^\circ\)
(c) \(90^\circ\)
(d) \(105^\circ\)
Answer: (b) \(80^\circ\)
Angles that can be constructed using a ruler and compass are multiples of \(15^\circ\) (i.e., \(15^\circ, 30^\circ, 45^\circ, 60^\circ, 75^\circ, 90^\circ, 105^\circ, 120^\circ\), etc.). Since \(80^\circ\) cannot be divided into standard constructible increments of \(15^\circ\) or \(7.5^\circ\) using basic angle bisectors from standard angles, it cannot be constructed with only a ruler and compass.
In simple words: Angles like \(80^\circ\) require a protractor because they are not multiples of standard constructible angles.

Exam Tip: Remember that standard constructible angles are obtained by bisecting \(60^\circ\) and \(90^\circ\).

Question 2. With a ruler and compass which of the following angles can be constructed?
(a) \(80^\circ\)
(b) \(90^\circ\)
(c) \(100^\circ\)
(d) \(110^\circ\)
Answer: (b) \(90^\circ\)
A \(90^\circ\) angle is constructed by drawing a perpendicular bisector or by intersecting arcs from a \(60^\circ\) and \(120^\circ\) angle on a line, which is a standard fundamental construction using only a ruler and compass.
In simple words: A \(90^\circ\) angle is a basic geometric construction learned in school.

Exam Tip: Always practice perpendicular bisectors and \(60^\circ\), \(90^\circ\), \(120^\circ\) angles as they form the basis for all other constructions.

Section - B

Question 3. Construct an angle of \(45^\circ\) at the initial point of a given ray and justify the construction.
Answer: Steps of construction:
1. Let \(OA\) be the given ray with initial point \(O\).
2. With \(O\) as center and any radius, draw an arc intersecting \(OA\) at point \(X\).
3. With \(X\) as center and the same radius, cut the previous arc at point \(Y\), and then with \(Y\) as center, cut the arc at point \(Z\).
4. With \(Y\) and \(Z\) as centers, draw two arcs intersecting each other at point \(P\). Join \(OP\) to form a ray making a \(90^\circ\) angle with \(OA\). Let this ray intersect the arc at \(T\).
5. With \(X\) and \(T\) as centers, draw two arcs of radius greater than half of \(XT\) intersecting at \(Q\).
6. Join \(OQ\). The angle \(AOQ\) is \(45^\circ\).
Justification: Since \(\angle AO\`P = 90^\circ\) and \(OQ\) is the bisector of \(\angle AOT\), \(\angle AOQ = \frac{1}{2} \times 90^\circ = 45^\circ\).
In simple words: To make \(45^\circ\), first construct \(90^\circ\) and then bisect it.

Exam Tip: Always write both the construction steps and the justification clearly to get full marks.

Question 4. Construct the following angles and verify by measuring them by a protractor. (i) \(75^\circ\) (ii) \(135^\circ\)
Answer: (i) Construction of \(75^\circ\):
1. Draw a ray \(OA\).
2. Construct a \(60^\circ\) angle and a \(120^\circ\) angle.
3. Bisect the angle between \(60^\circ\) and \(120^\circ\) to get \(90^\circ\).
4. Bisect the angle between \(60^\circ\) and \(90^\circ\) to obtain \(75^\circ\).
Verification: Measure with a protractor to confirm it reads exactly \(75^\circ\).
(ii) Construction of \(135^\circ\):
1. Draw a ray \(OA\) and produce it backwards to form a straight line. Alternatively, construct a \(90^\circ\) angle and a \(180^\circ\) angle.
2. Bisect the angle between \(90^\circ\) and \(180^\circ\) to get \(90^\circ + 45^\circ = 135^\circ\).
Verification: Measure with a protractor to confirm it reads exactly \(135^\circ\).
In simple words: \(75^\circ\) is halfway between \(60^\circ\) and \(90^\circ\), while \(135^\circ\) is halfway between \(90^\circ\) and \(180^\circ\).

Exam Tip: Leave all construction arcs clearly visible; do not rub them out.

Section - C

Question 5. Construct a \(\Delta PQR\) with base \(QR = 3.8\text{cm}\), \(\angle Q = 75^\circ\) and \(PQ + PR = 7.9\text{cm}\).
Answer: Steps of construction:
1. Draw base \(QR = 3.8\text{cm}\).
2. At point \(Q\), construct an angle of \(75^\circ\), say \(\angle XQR = 75^\circ\).
3. Cut a line segment \(QD = 7.9\text{cm}\) from ray \(QX\).
4. Join \(DR\).
5. Draw the perpendicular bisector of \(DR\), intersecting \(QD\) at point \(P\).
6. Join \(PR\). Then \(\Delta PQR\) is the required triangle.
In simple words: Form the base angle, mark the sum along the ray, connect to the endpoint, and draw a perpendicular bisector to locate the third vertex.

Exam Tip: Ensure your compass is locked tightly so the radius does not slip when drawing arcs.

Question 6. Construct a \(\Delta PQR\) with base \(QR = 3.4\text{cm}\), \(\angle R = 75^\circ\) and \(PR - PQ = 1.2\text{cm}\).
Answer: Steps of construction:
1. Draw base \(QR = 3.4\text{cm}\).
2. At point \(R\), construct an angle of \(75^\circ\), say \(\angle YRS = 75^\circ\).
3. Cut a line segment \(RD = 1.2\text{cm}\) from ray \(RY\) (since \(PR > PQ\), cut on the side opposite to the angle or along the ray depending on convention; here cut along \(RY\)).
4. Join \(DQ\).
5. Draw the perpendicular bisector of \(DQ\), intersecting \(RY\) at point \(P\).
6. Join \(PQ\). Then \(\Delta PQR\) is the required triangle.
In simple words: When the difference of sides is given, mark the difference along the base angle ray, join to the other base vertex, and bisect that segment.

Exam Tip: Pay close attention to whether \(PR - PQ\) or \(PQ - PR\) is given to determine which side of the ray to mark the difference.

Question 7. Construct an equilateral triangle with sides \(4\text{cm}\).
Answer: Steps of construction:
1. Draw a line segment \(AB = 4\text{cm}\).
2. With \(A\) as center and radius \(4\text{cm}\), draw an arc.
3. With \(B\) as center and radius \(4\text{cm}\), draw another arc intersecting the previous arc at point \(C\).
4. Join \(AC\) and \(BC\).
\(\Delta ABC\) is the required equilateral triangle with each side equal to \(4\text{cm}\).
In simple words: Open your compass to \(4\text{cm}\), draw intersecting arcs from both ends of the base line, and connect the intersection point.

Exam Tip: All three angles in this triangle will automatically measure \(60^\circ\)

Section - D

Question 8. Construct a triangle ABC in which \(\angle B = 60^\circ\), \(\angle C = 45^\circ\) and \(AB + BC + CA = 13\text{cm}\).
Answer: Steps of construction:
1. Draw a line segment \(XY = 13\text{cm}\) (which is equal to \(AB + BC + CA\)).
2. At point \(X\), make an angle equal to \(\angle B\), i.e., \(60^\circ\), and at point \(Y\), make an angle equal to \(\angle C\), i.e., \(45^\circ\).
3. Bisect these 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 \(AC\). \(\Delta ABC\) is the required triangle.
In simple words: Draw the total perimeter as a base, construct half of each base angle at the ends, find their intersection to get the top vertex, and use perpendicular bisectors to find the base points.

Exam Tip: Remember to bisect the base angles (taking half of \(60^\circ\) and \(45^\circ\)) at the perimeter line endpoints.

Question 9. Construct a right triangle whose base is \(12\text{cm}\) and sum of its hypotenuse and other side is \(18\text{cm}\).
Answer: Steps of construction:
1. Draw base \(BC = 12\text{cm}\).
2. At point \(B\), construct a right angle, i.e., \(\angle XBC = 90^\circ\).
3. Cut a line segment \(BD = 18\text{cm}\) from ray \(BX\).
4. Join \(CD\).
5. Draw the perpendicular bisector of \(CD\), intersecting \(BD\) at point \(A\).
6. Join \(AC\). Then \(\Delta ABC\) is the required right triangle.
In simple words: Draw the \(12\text{cm}\) base, make a \(90^\circ\) angle, cut the \(18\text{cm}\) sum, and use a perpendicular bisector to complete the triangle.

Exam Tip: Verify using Pythagoras theorem that the constructed sides satisfy the given conditions.

Question 10. Construct a \(\Delta PQR\) with its perimeter = \(11\text{cm}\) and the base angles of \(75^\circ\) and \(30^\circ\).
Answer: Steps of construction:
1. Draw a line segment \(XY = 11\text{cm}\) representing the perimeter.
2. At \(X\), construct an angle of \(75^\circ\), and at \(Y\), construct an angle of \(30^\circ\).
3. Bisect both angles. Let the bisectors meet at point \(P\).
4. Draw perpendicular bisectors of segment \(PX\) (intersecting \(XY\) at \(Q\)) and segment \(PY\) (intersecting \(XY\) at \(R\)).
5. Join \(PQ\) and \(PR\). Then \(\Delta PQR\) is the required triangle.
In simple words: Just like Question 8, draw the total perimeter, draw half of each given base angle at the ends, and bisect the sides to locate the triangle vertices.

Exam Tip: Double-check that you bisect the angles before finding the perpendicular bisectors.