Unit 11 Class 9 Math Solutions – Loci and Construction
Unit 11 Class 9 Math Solutions contain complete exercise-wise PDF solutions of the chapter Loci and Construction. Students can find the solutions of Exercise 11.1, Exercise 11.2, and Review Exercise 11 on this page.
This unit explains triangle constructions, concurrency of special lines, circumcentre, centroid, incentre, orthocentre, ambiguous triangle constructions, loci, scale drawings, and practical construction problems.
The solution PDFs include complete compass-and-ruler steps, labelled diagrams, construction checks, locus intersections, possible and impossible cases, printing notes, and clearly stated final answers for the Punjab Board Class 9 Mathematics book.
Students can view the PDFs online and use the save or download option in the PDF viewer to keep the solutions for offline study.
These Unit 11 solutions are useful for homework, revision, practical geometry, class tests, annual examinations, and board exam preparation.
Solutions of Exercise 11.1 Unit 11 Class 9 Math Notes
Exercise 11.1 focuses on triangle construction and concurrency. Students construct triangles from given sides and angles, draw special lines, identify their common points, and investigate ambiguous or impossible constructions.
Basic Tools Used in Geometrical Construction
The main tools are a ruler, compass, protractor, and a sharp pencil.
Construction arcs should remain visible because they show how the required points and lines were obtained.
Measurements should be taken carefully, and the final triangle should be labelled clearly.
Constructing a Triangle by SSS
The SSS condition is used when all three side lengths are given.
Draw one side as the base.
From one endpoint, draw an arc with radius equal to the second side.
From the other endpoint, draw another arc with radius equal to the third side.
The intersection of the arcs gives the third vertex.
Constructing a Triangle by SAS
The SAS condition is used when two sides and the included angle are given.
Draw one given side, construct the included angle at the correct endpoint, and mark the second side on the constructed ray.
Join the remaining vertices to complete the triangle.
Constructing a Triangle from One Side and Two Angles
When one side and two angles are given, draw the known side first.
Construct the given angle at each endpoint on the same side of the base.
The two rays meet at the third vertex.
The third angle may be checked from:
\(\angle A+\angle B+\angle C=180^\circ\)
Perpendicular Bisectors of a Triangle
A perpendicular bisector passes through the midpoint of a side and forms a right angle with that side.
To construct it, draw equal arcs from both endpoints of the side using a compass radius greater than half the side.
Join the two arc-intersection points.
Circumcentre
The three perpendicular bisectors of a triangle are concurrent at the circumcentre.
If the circumcentre is represented by \(O\), then:
\(OA=OB=OC\)
This point is equally distant from all three vertices and is the centre of the circumcircle.
Position of the Circumcentre
For an acute triangle, the circumcentre lies inside the triangle.
For a right triangle, it lies at the midpoint of the hypotenuse.
For an obtuse triangle, it lies outside the triangle.
Exercise 11.1 includes an obtuse triangle whose circumcentre is outside the triangle.
Medians of a Triangle
A median joins a vertex to the midpoint of the opposite side.
To construct the three medians, first locate the midpoint of every side.
Then join each vertex to the midpoint of the opposite side.
Centroid
The three medians of a triangle are concurrent at the centroid.
The centroid is usually represented by \(G\).
It divides every median in the ratio:
\(2:1\)
The longer part lies between the vertex and the centroid.
For example:
\(LG:GP=2:1\)
Angle Bisectors of a Triangle
An angle bisector divides an angle into two equal parts.
To construct it, draw an arc from the vertex that cuts both arms of the angle.
From those two points, draw equal arcs and join their intersection to the vertex.
Incentre
The three internal angle bisectors are concurrent at the incentre.
The incentre is usually represented by \(I\).
It is equally distant from the three sides of the triangle.
\(d(I,AB)=d(I,BC)=d(I,CA)\)
A circle centred at the incentre touches all three sides of the triangle.
Altitudes of a Triangle
An altitude is a perpendicular line segment drawn from a vertex to the opposite side or its extension.
To draw an altitude, construct a perpendicular from the chosen vertex to the opposite side.
Orthocentre
The three altitudes of a triangle are concurrent at the orthocentre.
The orthocentre is usually represented by \(H\).
For an acute triangle, the orthocentre lies inside the triangle.
For a right triangle, it lies at the right-angle vertex.
For an obtuse triangle, it lies outside the triangle.
Checking Whether a Triangle Is Possible
The interior angles of every triangle must add to 180 degrees.
\(\angle A+\angle B+\angle C=180^\circ\)
If two stated angles already have a sum greater than 180 degrees, the triangle cannot be constructed.
Exercise 11.1 identifies such a case and explains that the printed measurements are impossible.
Ambiguous SSA Construction
An SSA construction gives two sides and an angle that is not included between them.
Depending on the measurements, the construction may produce no triangle, one triangle, or two triangles.
Two triangles are possible when a compass arc cuts the angle ray at two points.
Height Test for the Ambiguous Case
For an acute given angle \(A\), adjacent side \(b\), and opposite side \(a\), first calculate:
\(h=b\sin A\)
Two different triangles are possible when:
\(h<a<b\)
The PDF applies this test to confirm that the construction arc meets the ray twice.
One-Triangle and No-Triangle Cases
If the construction arc touches the ray at one point, only one triangle is possible.
If the arc does not meet the ray, no triangle can be constructed.
The geometrical arc test and the numerical height test should agree.
Printing Notes in Exercise 11.1
One question gives a segment name that does not belong to the stated triangle.
The solution explains the likely intended side and completes the ambiguous-case construction using that corrected interpretation.
Another question gives two angles whose sum exceeds 180 degrees, so no triangle is mathematically possible.
Solutions of Exercise 11.2 Unit 11 Class 9 Math Notes
Exercise 11.2 focuses on loci and practical constructions. Students identify the path or region satisfying one or more distance conditions and construct intersections of circles, perpendicular bisectors, angle bisectors, parallel lines, and bounded regions.
What Is a Locus?
A locus is the complete set of points satisfying a stated condition.
A correct locus must include every point that satisfies the condition and exclude every point that does not satisfy it.
Fixed Distance from a Fixed Point
The locus of points at a fixed distance \(r\) from a fixed point is a circle.
The fixed point is the centre, and the fixed distance is the radius.
\(AP=r\)
When the condition says within a distance, the required set is the circular region inside the boundary.
Points Equidistant from Two Fixed Points
The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.
For fixed points \(A\) and \(B\), every point \(P\) on this locus satisfies:
\(PA=PB\)
Points Equidistant from Two Intersecting Lines
The complete locus consists of the internal and external angle bisectors.
When the required point must lie inside the given angle or triangle, the internal angle bisector is used.
Every point on an angle bisector has equal perpendicular distances from the two lines.
Fixed Distance from an Infinite Line
The locus of points at a fixed perpendicular distance from an infinite straight line consists of two lines parallel to the given line.
One parallel lies on each side of the original line.
Fixed Distance from a Finite Line Segment
The locus around a finite segment is not made only of two infinite parallel lines.
It consists of two parallel portions joined by semicircular ends.
The semicircles are centred at the endpoints of the segment.
This closed boundary is sometimes described as a stadium-shaped curve.
Intersection of Two Loci
When a point must satisfy two conditions, construct both loci.
Any common intersection point satisfies both conditions.
There may be two intersections, one tangent intersection, one valid interior intersection, or no intersection.
Two-Circle Location Problems
A point at one distance from \(A\) and another distance from \(B\) lies at the intersection of two circles.
Two intersection points represent two possible locations.
The circles intersect at two points when:
\(|r_1-r_2|<AB<r_1+r_2\)
Circle and Perpendicular-Bisector Problems
A point at a fixed distance from one point and equidistant from two other points is found by intersecting a circle with a perpendicular bisector.
Only intersections lying in the permitted region or on the supplied map are valid.
Circle and Angle-Bisector Problems
A point at a fixed distance from a vertex and equidistant from two sides is found by intersecting a circle centred at the vertex with the angle bisector.
When the required point must be inside a triangle, use the internal angle-bisector ray and choose the interior intersection.
Perpendicular-Bisector and Angle-Bisector Intersection
Some practical constructions combine two different equal-distance conditions.
The perpendicular bisector gives points equidistant from two vertices.
The angle bisector gives points equidistant from two intersecting sides.
Their intersection satisfies both requirements.
When Two Loci Do Not Intersect
Not every pair of conditions has a common solution.
After drawing both loci, compare the relevant distances.
If one locus cannot reach the other, no point satisfies both conditions.
Exercise 11.2 includes examples where a circle is slightly too small to meet another locus.
Scale Drawings
Practical locus problems often use a scale to represent large real distances.
Convert every measurement before drawing.
For example, under a scale of 1 cm representing 10 km:
\(24\text{ km}=2.4\text{ cm}\)
Use the scaled measurements in the construction and interpret the result in the original units.
Quarantine-Zone Locus
A region within a fixed distance of an infection source is represented by the interior of a circle.
The circle is centred at the source, and its radius represents the maximum allowed distance.
The circumference gives points exactly at the stated distance.
Navigation Between Two Towers
Positions equidistant from two towers lie on the perpendicular bisector of the segment joining the towers.
Points on one side of the bisector are nearer to one tower, while points on the other side are nearer to the other tower.
Treasure-Map Construction
A treasure at a fixed distance from one point and equidistant from two other points lies where a circle intersects a perpendicular bisector.
The actual positions must be taken from the supplied map.
Only intersections that lie on the valid land or map region are accepted.
Allowed Distance Between Two Limits
A condition such as between 54 m and 82 m from one point describes an annular region between two circles.
When it is combined with a fixed distance from another point, only the portions of the fixed-distance circle lying inside the outer boundary and outside the inner boundary are valid.
The resulting locus may consist of arcs rather than isolated points.
Strict and Non-Strict Inequalities
Words such as less than and farther than describe open regions when interpreted strictly.
The exact boundary line or circle is then excluded.
Words such as at most, not more than, or between with inclusive limits may include the boundary.
Regions Inside a Rectangle
A practical question may require points inside a field that also satisfy distance conditions.
First construct the complete geometric loci.
Then keep only the overlapping part that lies inside the field.
The required answer may be a quarter-disc, strip, arc, or another bounded region.
Solutions of Review Exercise 11 Unit 11 Class 9 Math Notes
Review Exercise 11 revises triangle construction, concurrency, loci, ambiguous cases, scale drawings, and regions satisfying two distance conditions.
Multiple-Choice Questions
The multiple-choice section checks the triangle inequality, types of triangles, medians, concurrency, circles, perpendicular bisectors, angle bisectors, parallel loci, and inside or outside regions.
Triangle Inequality
A triangle can be constructed only when the sum of any two side lengths is greater than the third side.
\(a+b>c,\qquad b+c>a,\qquad c+a>b\)
If any one of these conditions fails, no triangle exists.
Constructing a 6-8-10 Right Triangle
The review constructs a triangle with side lengths 6 cm, 8 cm, and 10 cm.
Its right angle is verified by the converse of Pythagoras’ theorem:
\(6^2+8^2=10^2\)
Equidistant from Two Vertices
After constructing the triangle, the locus of points equidistant from two named vertices is the perpendicular bisector of the segment joining them.
Ambiguous SSA Revision
The review includes an SSA construction in which the compass arc meets the ray at two points.
Therefore, two different triangles satisfy the same given measurements.
Equidistant from Two Sides
Points equidistant from two intersecting sides lie on their angle bisectors.
Inside a triangle, the required locus is the internal angle bisector.
Intersection of Two Different Loci
A point may be required to lie on both a perpendicular bisector and an angle bisector.
Construct both accurately and extend them when necessary.
Their common point satisfies both equal-distance conditions.
No-Solution Scale Problem
One review question requires a point 32 m from a house and 48 m from a line passing through that house.
The maximum perpendicular distance from a point on the 32 m circle to that line cannot exceed 32 m.
Therefore, the two conditions are incompatible and the loci do not intersect.
Required Region in a Rectangular Field
The final review question combines a distance from one corner with a minimum distance from one side.
A circular region and a parallel boundary are constructed inside the rectangle.
The required answer is the part of the quarter-disc that satisfies both conditions.
Important Rules and Facts of Unit 11 Class 9 Math
Angle Sum of a Triangle
\(\angle A+\angle B+\angle C=180^\circ\)
Triangle Inequality
\(a+b>c,\qquad b+c>a,\qquad c+a>b\)
Circumcentre Property
\(OA=OB=OC\)
Centroid Ratio
\(\text{vertex to centroid}:\text{centroid to midpoint}=2:1\)
Incentre Property
\(d(I,AB)=d(I,BC)=d(I,CA)\)
Ambiguous SSA Height
\(h=b\sin A\)
Two-Triangle SSA Condition
\(h<a<b\)
Equidistant from Two Points
\(PA=PB\)
Equidistant from Two Lines
\(d(P,l_1)=d(P,l_2)\)
Two-Circle Intersection Condition
\(|r_1-r_2|<d<r_1+r_2\)
Common Mistakes in Unit 11 Class 9 Math Notes
Students may erase the construction arcs that are required to justify the method.
A compass radius smaller than half the segment may fail to produce intersecting arcs for a perpendicular bisector.
The wrong side may be chosen as the base, making the construction unnecessarily difficult.
Angles may be constructed on opposite sides of the base when both should be on the same side.
A median is sometimes confused with a perpendicular bisector.
An altitude is sometimes drawn perpendicular to the wrong side.
Students may name the circumcentre, centroid, incentre, and orthocentre incorrectly.
The centroid ratio should be measured from the vertex as 2 to 1.
The circumcentre is not always inside the triangle.
The complete locus of points equidistant from two intersecting lines includes both angle bisectors, although an interior problem usually requires only the internal bisector.
For a finite segment, the fixed-distance locus needs semicircular ends.
Scale measurements should be converted before construction.
Students may mark every intersection even when the problem restricts the point to a triangle, field, island, or another region.
A construction should not be forced when the loci do not intersect.
Strict inequalities may exclude the boundary.
In an SSA problem, students may overlook the second possible triangle.
Printing mistakes should be identified from the geometry instead of copied without checking.
Exam Preparation Tips for Unit 11 Class 9 Math Notes
Practise constructing perpendicular bisectors, angle bisectors, parallels, and perpendiculars separately.
Memorize the four concurrency points and the special lines that meet at each point.
Keep construction arcs light but visible.
Use a sharp pencil and label every point clearly.
Write the construction steps in the same order as the diagram.
Check triangle angles and side conditions before starting.
Learn the ambiguous SSA height test.
Translate every locus statement into a circle, perpendicular bisector, angle bisector, parallel line, or bounded region.
When two conditions are given, construct both loci and inspect their intersections.
Convert map and field measurements according to the stated scale.
State when no common point exists and explain why.
Attempt Review Exercise 11 independently before checking the solution PDF.
Use the save or download option in the PDF viewer to keep the solutions for offline revision.
Why Unit 11 Class 9 Math Solutions Are Important
Unit 11 Class 9 Math Solutions are important because geometrical construction develops accuracy, reasoning, and visual problem-solving.
Concurrency explains important centres of a triangle, while loci convert distance conditions into clear geometric paths and regions.
These ideas are used in surveying, navigation, maps, architecture, engineering, design, location planning, and computer graphics.
The chapter also prepares students for advanced geometry, coordinate geometry, trigonometry, and practical measurement.
FAQs About Unit 11 Class 9 Math Solutions
What is the topic of Unit 11 Class 9 Math?
The topic of Unit 11 Class 9 Math is Loci and Construction.
How many exercises are included in Unit 11?
Unit 11 includes Exercise 11.1, Exercise 11.2, and Review Exercise 11.
What is covered in Exercise 11.1?
Exercise 11.1 covers triangle constructions, perpendicular bisectors, medians, angle bisectors, altitudes, concurrency points, impossible triangles, and ambiguous SSA constructions.
What is covered in Exercise 11.2?
Exercise 11.2 covers circle loci, perpendicular-bisector loci, angle-bisector loci, parallel loci, fixed distance from a segment, scale drawings, locus intersections, bounded regions, and practical applications.
What is covered in Review Exercise 11?
Review Exercise 11 revises triangle inequality, right-triangle construction, ambiguous cases, basic loci, locus intersections, no-solution cases, and field-region problems.
Where do the perpendicular bisectors of a triangle meet?
They meet at the circumcentre, which is equally distant from all three vertices.
Where do the medians of a triangle meet?
They meet at the centroid, which divides every median in the ratio 2 to 1 from the vertex.
Where do the angle bisectors of a triangle meet?
They meet at the incentre, which is equally distant from all three sides.
Where do the altitudes of a triangle meet?
They meet at the orthocentre.
What is the locus of points at a fixed distance from a point?
It is a circle centred at the fixed point.
What is the locus of points equidistant from two fixed points?
It is the perpendicular bisector of the segment joining the two points.
What is the locus of points equidistant from two intersecting lines?
The complete locus consists of the internal and external angle bisectors.
What is an ambiguous construction?
An ambiguous SSA construction is one in which the same measurements may produce two different triangles.
Can students download the Unit 11 solutions PDFs?
Yes. Students can view the exercise-wise PDFs on this page and use the save or download option provided by the PDF viewer.
Are these Unit 11 Class 9 Math Solutions useful for exam preparation?
Yes. The solutions are useful for practical construction work, homework, revision, class tests, annual examinations, and board exam preparation.
Disclaimer
These Unit 11 Class 9 Math Solutions are provided for educational help. Students should use them to understand construction methods, check their work, and study alongside the official textbook and their teacher’s instructions.
Where a printed measurement, segment name, or diagram appears inconsistent, the solution explains the mathematical issue and follows the most reasonable interpretation.
Final Words
Unit 11 Class 9 Math Solutions help students understand Loci and Construction in a clear and organized way.
Exercise 11.1 develops triangle-construction and concurrency skills, Exercise 11.2 applies loci to practical situations, and Review Exercise 11 revises the complete chapter.
Keep the construction arcs visible, verify every special point, and use the intersection of loci to decide whether a problem has one solution, more than one solution, a region of solutions, or no solution.
