9th Class Mathematics Notes 2026 – Chapter 11

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Class 9 Mathematics Notes – Chapter 11: Loci and Construction

Class: 9th
Subject: Mathematics
Board: Punjab Board
Chapter: 11 – Loci and Construction
Website: QuranEducationPoint.com

 

Exercise 11.1

Question 1

Construct \(\triangle ABC\) with the given measurements and verify that the perpendicular bisectors of the triangle are concurrent.

(i) \(AB=5\text{ cm},\ BC=6\text{ cm},\ AC=7\text{ cm}\)

Steps of Construction

  1. Draw the line segment \(AC=7\text{ cm}\).

  2. With \(A\) as centre and radius \(5\text{ cm}\), draw an arc.

  3. With \(C\) as centre and radius \(6\text{ cm}\), draw another arc intersecting the first arc at \(B\).

  4. Join \(AB\) and \(BC\) to complete \(\triangle ABC\).

  5. Construct the perpendicular bisectors of \(AB\), \(BC\), and \(AC\) by drawing intersecting arcs from their respective endpoints. Join the arc intersections to form the perpendicular bisectors \(L_1,L_2,L_3\).

Verification / Result

  • All three perpendicular bisectors \(L_1,L_2,L_3\) pass through a single point \(O\), called the circumcentre.

  • Hence, the perpendicular bisectors of \(\triangle ABC\) are concurrent.

(ii) \(AB=7.1\text{ cm},\ \angle B=135^\circ,\ BC=6.5\text{ cm}\)

Steps of Construction

  1. Draw the line segment \(BC=6.5\text{ cm}\).

  2. At vertex \(B\), construct an angle of \(135^\circ\).

  3. On the terminal ray of the angle, mark point \(A\) such that \(BA=7.1\text{ cm}\).

  4. Join \(A\) to \(C\) to complete \(\triangle ABC\).

  5. Construct the perpendicular bisectors of sides \(AB\), \(BC\), and \(AC\).

Verification / Result

  • All three perpendicular bisectors meet at a point \(O\) outside the triangle.

  • Hence, the perpendicular bisectors are concurrent.

 

Question 2

Construct \(\triangle LMN\) with the following measurements and verify that the medians of the triangle are concurrent.

(i) \(LM=4.9\text{ cm},\ \angle L=51^\circ,\ \angle M=38^\circ\)

Steps of Construction

  1. Draw the line segment \(LM=4.9\text{ cm}\).

  2. At vertex \(L\), construct an angle of \(51^\circ\).

  3. At vertex \(M\), construct an angle of \(38^\circ\).

  4. Extend the two rays until they intersect at \(N\), forming \(\triangle LMN\).

  5. Find the midpoints \(P,Q,R\) of sides \(LM,MN,NL\), respectively.

  6. Draw the medians \(NP\), \(LQ\), and \(MR\).

Verification / Result

  • The medians \(NP,LQ,\) and \(MR\) intersect at a single point \(G\), called the centroid.

  • Hence, the medians of \(\triangle LMN\) are concurrent.

(ii) \(MN=4.8\text{ cm},\ \angle N=30^\circ,\ LM=8.1\text{ cm}\)

Steps of Construction

  1. Draw the line segment \(MN=4.8\text{ cm}\).

  2. At point \(N\), construct an angle of \(30^\circ\).

  3. With \(M\) as centre and radius \(8.1\text{ cm}\), draw an arc intersecting the ray from \(N\) at point \(L\).

  4. Join \(L\) to \(M\) to complete \(\triangle LMN\).

  5. Find the midpoints \(D,E,F\) of sides \(MN,NL,LM\), respectively.

  6. Draw the medians \(LD\), \(ME\), and \(NF\).

Verification / Result

  • All three medians intersect at a single point \(G\).

  • Hence, the medians are concurrent.

 

Question 3

Verify that the angle bisectors of \(\triangle ABC\) are concurrent with the following measurements.

(i) \(AB=4.5\text{ cm},\ \angle A=45^\circ,\ AC=5.3\text{ cm}\)

Steps of Construction

  1. Draw the line segment \(AC=5.3\text{ cm}\).

  2. At vertex \(A\), construct an angle of \(45^\circ\).

  3. On the ray of the angle, mark point \(B\) such that \(AB=4.5\text{ cm}\).

  4. Join \(B\) to \(C\) to form \(\triangle ABC\).

  5. Bisect \(\angle A\) using a compass to construct ray \(AX\).

  6. Repeat the process to bisect \(\angle B\) and \(\angle C\), obtaining rays \(BY\) and \(CZ\).

Verification / Result

  • The three angle bisectors \(AX,BY,\) and \(CZ\) intersect at a single point \(I\), called the incentre.

  • Hence, the angle bisectors of \(\triangle ABC\) are concurrent.

(ii) \(AB=6\text{ cm},\ \angle A=150^\circ,\ \angle B=60^\circ\)

Geometric Calculation

The sum of the interior angles of a triangle is \(180^\circ\).

\[ \angle A+\angle B+\angle C=180^\circ \]
\[ 150^\circ+60^\circ+\angle C=180^\circ \]
\[ 210^\circ+\angle C=180^\circ \]

This would require \(\angle C=-30^\circ\), which is impossible for a triangle.

Result: Construction is not possible because the sum of the given angles exceeds \(180^\circ\).

 

Question 4

Given \(DE=4.8\text{ cm}\), \(EF=4\text{ cm}\), and \(\angle E=45^\circ\), draw the altitudes of \(\triangle DEF\) and find the orthocentre.

Steps of Construction

  1. Draw the line segment \(DE=4.8\text{ cm}\).

  2. At vertex \(E\), construct an angle of \(45^\circ\).

  3. On the ray of the angle, mark point \(F\) such that \(EF=4\text{ cm}\).

  4. Join \(F\) to \(D\) to complete \(\triangle DEF\).

  5. From vertex \(F\), draw a perpendicular to \(DE\).

  6. From vertex \(D\), draw a perpendicular to \(EF\), extending the side if necessary.

  7. From vertex \(E\), draw a perpendicular to \(DF\).

Result

  • The three altitudes intersect at a single point \(H\).

  • The point \(H\) is the orthocentre of \(\triangle DEF\).

 

Question 5

Construct the following triangles and determine whether an ambiguous case exists.

(i) \(\triangle BCD\), where \(BC=5\text{ cm},\ \angle B=62^\circ,\ CD=4.7\text{ cm}\)

Mathematical Analysis

  • Base side: \(BC=5\text{ cm}\).

  • Given angle: \(\angle B=62^\circ\).

  • Opposite side: \(CD=4.7\text{ cm}\).

Calculate the perpendicular height:

\[ h=BC\sin B \]
\[ h=5\sin62^\circ \]
\[ h\approx5(0.8829)=4.41\text{ cm} \]

Since

\[ 4.41<4.7<5 \]

the given side is greater than the height but less than the base.

Result

  • Yes, this is an ambiguous case.

  • Two valid triangles, \(\triangle BCD_1\) and \(\triangle BCD_2\), can be constructed.

(ii) \(\triangle KLM\), where \(LM=6\text{ cm},\ \angle M=42^\circ,\ LK=5\text{ cm}\)

Note: The supplied question mentions \(LN=5\text{ cm}\). The following solution assumes this means \(LK=5\text{ cm}\).

Mathematical Analysis

  • Base side: \(LM=6\text{ cm}\).

  • Given angle: \(\angle M=42^\circ\).

  • Opposite side: \(LK=5\text{ cm}\).

Calculate the perpendicular height:

\[ h=LM\sin42^\circ \]
\[ h=6(0.6691)\approx4.01\text{ cm} \]

Since

\[ 4.01<5<6 \]

the given side is greater than the height but less than the base.

Result

  • Yes, this is an ambiguous case.

  • Two distinct triangles can be constructed.

 

Exercise 11.2

Question 1

Two points \(A\) and \(B\) are \(8.2\text{ cm}\) apart. Construct the locus of points \(5\text{ cm}\) from point \(A\).

Steps of Construction

  1. Mark point \(A\).

  2. Open the compass to a radius of \(5\text{ cm}\).

  3. With \(A\) as centre, draw a complete circle.

Locus Definition

The locus of points at a fixed distance of \(5\text{ cm}\) from a fixed point \(A\) is a circle of radius \(5\text{ cm}\) centred at \(A\).

Question 2

Construct the locus of points \(2.2\text{ cm}\) from line segment \(CD\), where \(CD=5.7\text{ cm}\).

Steps of Construction

  1. Draw the line segment \(CD=5.7\text{ cm}\).

  2. Draw two parallel line segments at a perpendicular distance of \(2.2\text{ cm}\) on both sides of \(CD\).

  3. With \(C\) and \(D\) as centres, draw semicircles of radius \(2.2\text{ cm}\) to connect the ends of the parallel lines.

Locus Definition

The locus consists of two parallel lines joined by semicircular ends, forming a capsule-shaped boundary around the segment \(CD\).

Question 3

Construct \(\angle ABC=105^\circ\). Construct the locus of a point \(P\) that moves so that it is equidistant from rays \(BA\) and \(BC\).

Steps of Construction

  1. Construct \(\angle ABC=105^\circ\) using a ruler and protractor or compass construction.

  2. Bisect \(\angle ABC\) using a compass.

Locus Definition

The locus of points equidistant from two intersecting rays is the angle bisector of the angle between them.

Question 4

Two points \(E\) and \(F\) are \(5.4\text{ cm}\) apart. Construct the locus of a point \(P\) that is equidistant from \(E\) and \(F\).

Steps of Construction

  1. Draw the line segment \(EF=5.4\text{ cm}\).

  2. Construct the perpendicular bisector of \(EF\).

Locus Definition

The locus of points equidistant from two fixed points \(E\) and \(F\) is the perpendicular bisector of the line segment joining them.

Question 5

Two cities \(A\) and \(B\) on an island are \(8\text{ km}\) apart. Kashif lives exactly \(6.8\text{ km}\) from city \(A\) and \(7.3\text{ km}\) from city \(B\). Mark the possible locations where Kashif could live.

Steps of Construction

  1. Use the scale \(1\text{ km}=1\text{ cm}\). Draw \(AB=8\text{ cm}\).

  2. With \(A\) as centre, draw a circle or arc of radius \(6.8\text{ cm}\).

  3. With \(B\) as centre, draw a circle or arc of radius \(7.3\text{ cm}\).

  4. Mark both intersection points with a cross \((\times)\).

Result

There are two possible locations where Kashif could live.

Question 6

Construct \(\triangle CDE\), where \(CD=7.6\text{ cm}\), \(\angle D=45^\circ\), and \(DE=5.9\text{ cm}\). Draw the loci of points that are:

(a) Equidistant from \(C\) and \(D\).

(b) Equidistant from lines \(CD\) and \(DE\).

Mark the intersection point \(X\).

Steps of Construction

  1. Construct \(\triangle CDE\) using \(CD=7.6\text{ cm}\), \(\angle D=45^\circ\), and \(DE=5.9\text{ cm}\).

  2. Locus (a): Draw the perpendicular bisector of \(CD\).

  3. Locus (b): Draw the angle bisector of \(\angle CDE\), formed by \(DC\) and \(DE\).

  4. Label the intersection of the two loci as \(X\).

Question 7

Construct \(\triangle LMN\), where \(LM=7\text{ cm}\), \(\angle L=70^\circ\), and \(\angle M=45^\circ\). Find a point inside the triangle that is equidistant from \(L\) and \(M\) and \(3\text{ cm}\) from \(L\).

Steps of Construction

  1. Construct \(\triangle LMN\) with \(LM=7\text{ cm}\), \(\angle L=70^\circ\), and \(\angle M=45^\circ\).

  2. Construct the perpendicular bisector of \(LM\). This is the locus of points equidistant from \(L\) and \(M\).

  3. With \(L\) as centre and radius \(3\text{ cm}\), draw an arc inside the triangle.

  4. Mark the intersection of the arc and the perpendicular bisector inside the triangle.

Result

The intersection point satisfies both conditions.

Question 8

Construct a right-angled triangle \(RST\), where \(RS=6.8\text{ cm}\), \(\angle S=90^\circ\), and \(ST=7.5\text{ cm}\). Find a point inside the triangle that is equidistant from lines \(RS\) and \(RT\) and \(4.5\text{ cm}\) from \(R\).

Steps of Construction

  1. Construct right-angled \(\triangle RST\) using \(RS=6.8\text{ cm}\), \(\angle S=90^\circ\), and \(ST=7.5\text{ cm}\).

  2. Construct the angle bisector of \(\angle SRT\).

  3. With \(R\) as centre, draw an arc of radius \(4.5\text{ cm}\).

  4. Mark the intersection of the angle bisector and the arc inside the triangle.

Result

The intersection point satisfies both conditions.

Question 9

Construct rectangle \(UVWX\), where \(UV=7.2\text{ cm}\) and \(VW=5.6\text{ cm}\). Draw the loci of points that are \(2\text{ cm}\) from \(UV\) and \(3.5\text{ cm}\) from \(W\).

Steps of Construction

  1. Construct rectangle \(UVWX\) with \(UV=7.2\text{ cm}\) and \(VW=5.6\text{ cm}\).

  2. Draw a line parallel to \(UV\), at a perpendicular distance of \(2\text{ cm}\) inside the rectangle.

  3. With \(W\) as centre, draw a circle or arc of radius \(3.5\text{ cm}\).

  4. Mark the intersection points of the parallel line and the circle that lie within the rectangle.

Result

The required points are the intersections that satisfy both distance conditions.

Question 10

Two cell towers are located at points \(A\) and \(B\). Draw the locus of points equidistant from the two towers.

Answer

The locus of points equidistant from two fixed points \(A\) and \(B\) is the perpendicular bisector of the line segment \(AB\).

Question 11

Draw the locus of all points \(10\text{ km}\) from a source defining a quarantine area.

Steps of Construction

  1. Choose a suitable scale, for example, \(10\text{ km}=1\text{ cm}\).

  2. Mark the source of infection as point \(O\).

  3. With \(O\) as centre, draw a circle of radius representing \(10\text{ km}\) on the chosen scale.

Answer

The locus is a circle centred at the source, with radius \(10\text{ km}\) in actual distance.

Question 12

A treasure is buried somewhere on an island. It is \(24\text{ km}\) from \(A\) and equidistant from \(B\) and \(C\). Use a scale of \(1\text{ cm}\) to represent \(10\text{ km}\) to find the possible location.

Steps of Construction

1. Convert the distance to the given scale:

\[ 24\text{ km}=\frac{24}{10}=2.4\text{ cm} \]

  1. Construct the perpendicular bisector of \(BC\). This is the locus of points equidistant from \(B\) and \(C\).

  2. With \(A\) as centre, draw a circle of radius \(2.4\text{ cm}\).

  3. Mark the intersection point or points of the circle and the perpendicular bisector.

Result

The treasure could be located at the intersection point or points of the two loci.

Question 13

Using a scale of \(1\text{ cm}\) to represent \(10\text{ m}\), find the possible points where a mango tree should be planted.

Given data and scale

  • Distance between the apple tree \(A\) and banana tree \(B\): \(90\text{ m}=9\text{ cm}\).

  • Distance of mango tree \(M\) from \(A\): \(64\text{ m}=6.4\text{ cm}\).

  • Distance of mango tree \(M\) from \(B\): between \(54\text{ m}=5.4\text{ cm}\) and \(82\text{ m}=8.2\text{ cm}\).

Steps of Construction

  1. Draw the line segment \(AB=9\text{ cm}\).

  2. With \(A\) as centre, draw a circle of radius \(6.4\text{ cm}\).

  3. With \(B\) as centre, draw two concentric arcs of radii \(5.4\text{ cm}\) and \(8.2\text{ cm}\).

  4. Identify the portion of the circle centred at \(A\) that lies between the two arcs centred at \(B\).

Result

The mango tree should be planted at any point on the \(6.4\text{ cm}\) radius circle centred at \(A\) that lies between the two distance boundaries centred at \(B\).

 

Review Exercise 11

Question 1: Multiple-choice questions (MCQs)

(i) A triangle can be constructed if the sum of any two sides is ______ the third side.

Correct Answer: (b) Greater than.

(ii) An equilateral triangle ______.

Correct Answer: (a) Can be isosceles.

Every equilateral triangle also satisfies the conditions for an isosceles triangle.

(iii) If the sum of two angles is less than \(90^\circ\), then the triangle is ______.

Correct Answer: (c) Obtuse-angled.

(iv) The line segment joining the midpoint of a side to its opposite vertex in a triangle is called ______.

Correct Answer: (a) Median.

(v) The angle bisectors of a triangle intersect at ______.

Correct Answer: (a) One point.

(vi) The locus of all points equidistant from a fixed point is ______.

Correct Answer: (a) Circle.

(vii) The locus of points equidistant from two fixed points is ______.

Correct Answer: (b) Perpendicular bisector.

(viii) The locus of points equidistant from a fixed line is ______.

Correct Answer: (d) Parallel lines.

(ix) The locus of points equidistant from two intersecting lines is ______.

Correct Answer: (c) Angle bisector.

(x) The set of all points farther than \(2\text{ km}\) from a fixed point \(B\) is a region outside a circle of radius ______ and centre \(B\).

Correct Answer: (c) \(2\text{ km}\).

Question 2

Construct a right-angled triangle with sides \(6\text{ cm}\), \(8\text{ cm}\), and \(10\text{ cm}\).

Steps of Construction

  1. Draw the base segment \(AB=8\text{ cm}\).

  2. At vertex \(A\), construct an angle of \(90^\circ\).

  3. With \(A\) as centre and radius \(6\text{ cm}\), draw an arc intersecting the perpendicular ray at \(C\).

  4. Join \(C\) to \(B\).

  5. Verify that \(BC=10\text{ cm}\) using the Pythagorean theorem:

\[ 6^2+8^2=36+64=100=10^2 \]

Result

The required right-angled triangle has sides \(6\text{ cm}\), \(8\text{ cm}\), and \(10\text{ cm}\).

Question 3

Construct \(\triangle ABC\), where \(AB=5.3\text{ cm}\), \(\angle A=30^\circ\), and \(\angle B=120^\circ\). Draw the locus of points equidistant from \(A\) and \(B\).

Steps of Construction

  1. Draw the base segment \(AB=5.3\text{ cm}\).

  2. At \(A\), construct an angle of \(30^\circ\).

  3. At \(B\), construct an angle of \(120^\circ\) on the same side of \(AB\).

  4. Let the two rays intersect at \(C\), forming \(\triangle ABC\).

  5. Construct the perpendicular bisector of \(AB\).

Result

The perpendicular bisector of \(AB\) is the required locus of points equidistant from \(A\) and \(B\).

Question 4

Construct a triangle with \(DE=7.3\text{ cm}\), \(\angle D=42^\circ\), and \(EF=5.4\text{ cm}\).

Steps of Construction

  1. Draw the line segment \(DE=7.3\text{ cm}\).

  2. At \(D\), construct an angle of \(42^\circ\).

  3. With \(E\) as centre and radius \(5.4\text{ cm}\), draw an arc intersecting the ray from \(D\) at \(F\).

  4. Join \(F\) to \(E\) to complete \(\triangle DEF\).

Result

The required triangle is constructed.

Question 5

Construct \(\triangle XYZ\), where \(YX=8\text{ cm}\), \(YZ=7\text{ cm}\), and \(XZ=6.5\text{ cm}\). Draw the locus of all points equidistant from lines \(XY\) and \(XZ\).

Steps of Construction

  1. Draw the base \(YX=8\text{ cm}\).

  2. With \(X\) as centre and radius \(6.5\text{ cm}\), draw an arc.

  3. With \(Y\) as centre and radius \(7\text{ cm}\), draw an arc intersecting the first arc at \(Z\).

  4. Join \(ZX\) and \(ZY\) to complete \(\triangle XYZ\).

  5. Construct the angle bisector of \(\angle YXZ\).

Result

The angle bisector of \(\angle YXZ\) is the required locus of points equidistant from lines \(XY\) and \(XZ\).

Question 6

Construct \(\triangle FGH\), where \(FG=GH=6.4\text{ cm}\) and \(\angle G=122^\circ\). Draw the loci of points that are:

(a) Equidistant from \(F\) and \(G\).

(b) Equidistant from lines \(FG\) and \(GH\).

(c) Mark the point where the two loci intersect.

Steps of Construction

  1. Construct \(\triangle FGH\) with \(FG=6.4\text{ cm}\), \(GH=6.4\text{ cm}\), and \(\angle G=122^\circ\).

  2. Locus (a): Construct the perpendicular bisector of \(FG\).

  3. Locus (b): Construct the angle bisector of \(\angle FGH\).

  4. Mark the intersection of the two loci as \(P\).

Result

Point \(P\) satisfies both locus conditions.

Question 7

Two houses \(Q\) and \(R\) are \(73\text{ m}\) apart. Using a scale of \(1\text{ cm}\) to represent \(10\text{ m}\), construct the loci of a point \(P\) that is:

(i) \(32\text{ m}\) from \(Q\).

(ii) \(48\text{ m}\) from the line joining \(Q\) and \(R\).

Scale calculations

  • \(QR=73\text{ m}=7.3\text{ cm}\).

  • Distance from \(Q\): \(32\text{ m}=3.2\text{ cm}\).

  • Distance from line \(QR\): \(48\text{ m}=4.8\text{ cm}\).

Steps of Construction

  1. Draw \(QR=7.3\text{ cm}\).

  2. Locus (i): With \(Q\) as centre and radius \(3.2\text{ cm}\), draw a circle.

  3. Locus (ii): Draw two parallel lines at a perpendicular distance of \(4.8\text{ cm}\) on both sides of \(QR\).

Result

The required point or points are the intersections of the circle and the parallel lines.

Question 8

A field is in the form of rectangle \(ABCD\), where \(AB=70\text{ m}\) and \(BC=60\text{ m}\). Construct the rectangle using a scale of \(1\text{ cm}\) to represent \(10\text{ m}\). Show the region inside the field that is less than \(30\text{ m}\) from \(C\) and farther than \(25\text{ m}\) from \(AB\).

Scale calculations

  • \(AB=70\text{ m}=7\text{ cm}\).

  • \(BC=60\text{ m}=6\text{ cm}\).

  • Distance from \(C\): \(30\text{ m}=3\text{ cm}\).

  • Distance from \(AB\): \(25\text{ m}=2.5\text{ cm}\).

Steps of Construction

  1. Construct rectangle \(ABCD\) with \(AB=7\text{ cm}\) and \(BC=6\text{ cm}\).

  2. With \(C\) as centre, draw an arc of radius \(3\text{ cm}\) inside the rectangle. The region less than \(30\text{ m}\) from \(C\) lies inside this circular boundary.

  3. Draw a line parallel to \(AB\), at a perpendicular distance of \(2.5\text{ cm}\) inside the rectangle.

  4. Shade the region that lies inside the rectangle, within the \(3\text{ cm}\) boundary around \(C\), and farther than \(2.5\text{ cm}\) from \(AB\).

Result

The required region is the part of the field satisfying both distance conditions.

 

Notes by Sir Dilawar

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