ML Aggarwal Class 10 Maths Solutions Chapter 14 Locus

Exercise 14

Question 1. A point moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path travelled by P?
Answer: Suppose point P shifts in a way that keeps it at a fixed distance from line AB. In that case, the moving point traces a set of two lines parallel to AB, positioned on either side of it at equal distance from it. Therefore, the path consists of a pair of straight lines parallel to AB.
In simple words: When a point stays the same distance away from a line, it traces two straight lines - one on each side of the original line, both running parallel to it.

Exam Tip: Always sketch the two parallel lines on opposite sides of the given line to clearly show the complete locus - examiners look for both lines, not just one.

Question 2. A point P moves so that its perpendicular distances from two given lines AB and CD are equal. State the locus of the point P.
Answer: Two different situations can happen here. When lines AB and CD are parallel, the locus of point P, which maintains equal distance from both lines, is a single line running midway between AB and CD, also parallel to them. When AB and CD are intersecting lines, the locus of point P forms a pair of straight lines l and m that bisect the angles formed by the given lines AB and CD.
In simple words: If the two lines are parallel, the locus is a line in the middle. If the two lines cross, the locus is two lines that split the angles equally.

Exam Tip: Distinguish between the two cases clearly - parallel lines produce one locus line at the midpoint, while intersecting lines produce two angle bisectors. Always state which case applies to earn full marks.

Question 3. P is a fixed point and a point Q moves such that the distance PQ is constant. What is the locus of the path traced out by the point Q?
Answer: Since P is a fixed point and Q is a moving point that always stays at a constant distance from P, point P becomes the center of the circular path that Q traces. The distance PQ acts as the radius of this circle. Therefore, the locus of point Q is a circle with P as its center.
In simple words: A point that stays the same distance from a fixed point traces a circle. The fixed point is the center, and the constant distance is the radius.

Exam Tip: Clearly identify the fixed point as the center and state the radius in terms of the given constant distance - this shows complete understanding of the locus definition.

Question 4(i). AB is a fixed line. State the locus of point P so that \( \angle APB = 90° \).
Answer: We know from circle theory that any angle inscribed in a semicircle equals 90°. When AB is the diameter of a circle, every point P on that circle (excluding points A and B) forms a right angle with A and B. Therefore, the locus of point P will be the circle whose diameter is AB.
In simple words: All points that make a 90-degree angle with two fixed points lie on a circle where those two points are at opposite ends (the diameter).

Exam Tip: State that AB is the diameter and draw the complete circle - the locus includes the entire circle except the endpoints A and B themselves, though some examiners may accept the full circle.

Question 4(ii). A, B are fixed points. State the locus of P so that \( \angle APB = 60° \).
Answer: The locus of P consists of an arc (or arcs) of a circle for which AB serves as a chord. Points on this arc all maintain the same angle of 60° when connected to the endpoints A and B.
In simple words: All points making a 60-degree angle with two fixed points lie on an arc of a circle having those two points as a chord.

Exam Tip: Remember that there are actually two arcs (one on each side of the chord AB) that satisfy this condition, so mention both when describing the complete locus.

Question 5(i). Draw and describe the locus of points at a distance 2.5 cm from a fixed line.
Answer: The locus of points that are 2.5 cm away from a fixed line forms a pair of straight lines, each running parallel to the fixed line and positioned at a perpendicular distance of 2.5 cm from it - one line on each side.
In simple words: Points that stay the same distance from a line make two parallel lines, one above and one below the original line.

Exam Tip: Draw both lines clearly at the specified distance, and mark the perpendicular distance on the diagram to show understanding of what "perpendicular distance" means.

Question 5(ii). Draw and describe the locus of vertices of all isosceles triangles having a common base.
Answer: Consider an isosceles triangle ABC where AB equals AC. Draw AD perpendicular to the base BC. In triangles ABD and ACD, we have AD in common, AB equals AC (isosceles property), and both angles ADB and ADC are right angles. By RHS congruence, triangle ABD is congruent to triangle ACD. This means BD equals DC, so AD bisects BC. Since AD is perpendicular to BC and bisects it, AD is the perpendicular bisector of the base. Every vertex of an isosceles triangle with base BC must lie on this perpendicular bisector. Therefore, the locus of vertices is the perpendicular bisector of the base.
In simple words: If you draw all possible isosceles triangles on the same base, all their top vertices lie on one line - the perpendicular bisector of the base.

Exam Tip: Use congruence of triangles to prove that the altitude from the apex to the base also bisects the base - this is the key step that establishes the locus as the perpendicular bisector.

Question 5(iii). Draw and describe the locus of points inside a circle and equidistant from two fixed points on the circle.
Answer: Let the two fixed points on the circle be A and B. Construct the perpendicular bisector of line segment AB. This perpendicular bisector passes through the center O and intersects the circle at points C and E. The line segment CE is the locus of all points that maintain equal distance from A and B. Therefore, the locus consists of the diameter of the circle that is perpendicular to the chord AB.
In simple words: Points inside a circle that are equally far from two points on the circle form a straight line - the diameter that cuts through the midpoint of the chord joining those two points.

Exam Tip: Clearly show that the perpendicular bisector of a chord passes through the center of the circle - this demonstrates why the locus is a diameter.

Question 5(iv). Draw and describe the locus of centres of all circles passing through two fixed points.
Answer: Let the two fixed points be A and B, and let C₁, C₂, C₃ be the centers of various circles that pass through both A and B. From observation, all these centers lie on a single line. The locus of centers of all circles passing through two fixed points is the perpendicular bisector of the line segment connecting those two fixed points.
In simple words: All centers of circles passing through two fixed points lie on the perpendicular bisector of the line joining those two points.

Exam Tip: This is a key result in circle geometry - remember that circles of different radii can pass through two fixed points, but all their centers must lie on one specific line: the perpendicular bisector.

Question 5(v). Draw and describe the locus of a point in rhombus ABCD which is equidistant from AB and AD.
Answer: In rhombus ABCD, AC is a diagonal. By the properties of a rhombus, diagonal AC bisects angle A. A key property states that any point lying on the angle bisector of an angle is equidistant from the two rays forming that angle. Since AC bisects angle DAB, every point on AC maintains equal perpendicular distance from sides AD and AB. Therefore, the locus of the point is the diagonal AC of rhombus ABCD.
In simple words: Points in a rhombus that are equally far from two sides lie on the diagonal connecting the angle between those sides.

Exam Tip: Use the angle bisector property clearly - points on an angle bisector are equidistant from the two arms of the angle, which directly gives the locus as diagonal AC.

Question 5(vi). Draw and describe the locus of a point in the rhombus ABCD which is equidistant from the points A and C.
Answer: In rhombus ABCD, let AC and BD be the two diagonals, intersecting at point O. Because the diagonals of a rhombus bisect each other, O is the midpoint of AC. Additionally, the diagonals of any rhombus intersect at right angles. Since BD is the perpendicular bisector of diagonal AC, any point lying on BD is equidistant from points A and C. Therefore, the locus is the diagonal BD of rhombus ABCD.
In simple words: Points in a rhombus that are equally far from two opposite vertices lie on the other diagonal.

Exam Tip: Remember that the perpendicular bisector of any line segment is the locus of all points equidistant from its endpoints - since BD is perpendicular to AC and bisects it, BD is the perpendicular bisector.

Question 6(i). Describe completely the locus of mid-point of radii of a circle.
Answer: Let the radius of the original circle be r. The midpoint of any radius lies at a distance of r/2 from the center of the circle. As we consider all possible radii of the circle, their midpoints form a set of points that are all at distance r/2 from the center. These points trace out another circle that shares the same center (concentric) with the original circle but has a radius of r/2. Therefore, the locus of midpoints of all radii is a concentric circle with half the radius of the original circle.
In simple words: The midpoints of all radii in a circle form another circle with the same center but half the radius.

Exam Tip: Make sure to state "concentric" clearly - this means both circles share the same center - and always specify that the new radius is exactly half the original radius.

Question 6(ii). Describe completely the locus of centre of a ball, rolling along a straight line on a level floor.
Answer: Suppose a ball with a fixed radius rolls from position A to position B on a level floor. When the ball is in contact with the floor, the center of the ball is always located at a height equal to the radius of the ball above the surface. As the ball moves from A to B, its center traces a path that remains parallel to the floor. Therefore, the locus of the center of the ball is a line parallel to the floor at a height equal to the radius of the ball.
In simple words: As a ball rolls along a flat floor, its center moves in a straight line that is parallel to the floor and sits at a height equal to the ball's radius.

Exam Tip: The key insight is that the height of the center above the floor equals the radius - this is constant throughout the motion, making the path a straight line parallel to the floor.

Question 6(iii). Describe completely the locus of point in a plane equidistant from a given line.
Answer: Let the given line be AB. Consider two points P and P' positioned on opposite sides of AB, each at an equal perpendicular distance from it. Through P and P', draw lines CD and EF respectively, both parallel to AB. All points on line CD maintain the same perpendicular distance from AB (on one side), and all points on line EF maintain that same distance (on the other side). Therefore, the locus consists of a pair of lines, each parallel to the given line AB and positioned at equal distances on either side.
In simple words: Points at the same distance from a line form two straight lines - one on each side of the original line, both running parallel to it.

Exam Tip: Always draw both parallel lines to show the complete locus - the answer is not just one line, but a pair of parallel lines on opposite sides of the given line.

Question 6(iv). Describe completely the locus of point in a plane, at a constant distance of 5 cm from a fixed point (in the plane).
Answer: Let O be the fixed point and P be any point such that OP equals 5 cm. By using O as the center and OP as the radius, a circle can be drawn. All points on this circle are exactly 5 cm away from the fixed point O. Therefore, the locus is a circle with the fixed point as its center and a radius of 5 cm.
In simple words: All points at the same distance from a fixed point form a circle with that fixed point as the center and the constant distance as the radius.

Exam Tip: State both the center and the radius clearly when describing the locus circle - this provides a complete description of the geometric shape.

Question 6(v). Describe completely the locus of centre of a circle of varying radius and touching two arms of \( \angle ABC \).
Answer: Let two circles with centers O and O' be tangent to both arms AB and BC of angle ABC. Let BX be the angle bisector of angle ABC. Since a radius is perpendicular to a tangent, angles OEB and OFB are both 90°. Since BX bisects angle ABC, angles OBE and OBF are equal. By the AA similarity criterion, triangle OEB is similar to triangle OFB. From this similarity and the fact that O is the center with OE and OF as radii, we get OE equals OF. This means the circle with center O is equidistant from both arms of the angle. Since this holds for any circle touching both arms, the locus of all possible centers lies on the angle bisector BX. Therefore, the locus is the angle bisector of angle ABC.
In simple words: All centers of circles that touch both arms of an angle lie on the angle bisector of that angle.

Exam Tip: Use the perpendicularity of radius and tangent (90° angles) combined with the angle bisector property to establish that all such centers are equidistant from both arms - this proves the locus is the angle bisector.

Question 6(vi). Describe completely the locus of centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with centre O.
Answer: When a circle with radius 2 cm touches the given fixed circle externally, the locus of its centre forms a concentric circle with radius (3 + 2) = 5 cm. When the same circle touches the given fixed circle internally, the locus of its centre becomes a concentric circle with radius (3 - 2) = 1 cm.
In simple words: If a small circle rolls around the outside of a fixed circle, its centre traces a bigger concentric circle. If it rolls inside, the centre traces a smaller concentric circle.

Exam Tip: Always identify whether the touching is external or internal - each case produces a different concentric circle. Show both radii clearly in your answer.

Question 7. Using ruler and compasses, construct (i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm. (ii) the locus of points equidistant from A and C.
Answer:
(i) The constructed triangle ABC is shown in the figure with sides AB = 5.5 cm, BC = 3.4 cm, and CA = 4.9 cm, along with the perpendicular bisector of AC marked with equal-length tick marks on either side.
(ii) The set of all points that are the same distance from A as they are from C forms the perpendicular bisector of the line segment AC.
In simple words: Any point sitting on the perpendicular bisector of AC will be equally far from both A and C.

Exam Tip: The perpendicular bisector must pass through the midpoint of AC at exactly 90 degrees - check this with your compass and set-square before finalizing.

Question 8. Construct triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that: (i) P is equidistant from B and C and (ii) P is equidistant from AB and BC. (iii) Measure and record the length of PB.
Answer: First, build triangle ABC with the given measurements. Since P must be equidistant from B and C, it must lie on the perpendicular bisector of BC, marked as line YZ. Since P must also be equidistant from sides AB and BC, it must lie on the angle bisector at B, marked as line BX. Point P is where lines BX and YZ meet. When measured from the construction, PB = 4.6 cm.
In simple words: P sits at the spot where two important lines cross - the perpendicular bisector of BC and the angle bisector of angle B.

Exam Tip: Mark the intersection point clearly and verify that it satisfies both conditions before measuring - P must lie on both loci at the same spot.

Question 9. A line segment AB is 8 cm long. Locate by construction the locus of a point which is: (i) Equidistant from A and B. (ii) Always 4 cm from the line AB. (iii) Mark two points X and Y, which are 4 cm from AB and equidistant from A and B. Name the figure AXBY.
Answer:
(i) All points that are equally distant from A and B lie on the perpendicular bisector of AB.
(ii) Let the perpendicular bisector meet AB at point O. Mark two points X and Y on this bisector, each exactly 4 cm away from O on either side. Draw line PQ parallel to AB and passing through X. Draw line RS parallel to AB and passing through Y. The locus is the pair of straight lines PQ and RS, both parallel to AB and each 4 cm away from it.
(iii) Joining the four points A, B, X, and Y forms a quadrilateral. Since the diagonals AX and BY meet at right angles at O and are equal in length, AXBY is a square.
In simple words: Points equidistant from A and B sit on a perpendicular line. Points 4 cm from AB form two parallel lines. Where these two conditions meet, we get a square.

Exam Tip: Verify that AXBY has all four sides equal and all angles 90 degrees before naming it as a square - use your ruler and set-square to double-check.

Question 10. Use ruler and compasses only for this question. (i) Construct △ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°. (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC. (iii) Construct the locus of points inside the triangle which are equidistant from B and C. (iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Answer:
(i) The triangle ABC is constructed with the given measurements shown in the figure.
(ii) The locus of points lying at the same distance from two lines is the angle bisector between them. From the figure, BE is the angle bisector of ∠ABC, meeting side AC at point D. Therefore, BD is the locus of points inside the triangle that are equidistant from sides BA and BC.
(iii) The locus of points at the same distance from two points is the perpendicular bisector joining them. From the figure, XY is the perpendicular bisector of BC, meeting AC at H and BC at O. The segment OH is the locus of points inside the triangle equidistant from B and C.
(iv) Point P is located at the intersection of BD and OH. Since P lies on both loci, it is equidistant from sides AB and BC as well as from points B and C. Measuring from the construction gives PB = 3.4 cm.
In simple words: P is found where the angle bisector from B meets the perpendicular bisector of BC - this special spot is equally distant from both sides and both points.

Exam Tip: Ensure the angle bisector from B and the perpendicular bisector of BC are accurately drawn - their intersection point must satisfy all four distance conditions.

Question 11. Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence, (i) Construct the locus of points equidistant from BA and BC. (ii) Construct the locus of points equidistant from B and C. (iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Answer:
(i) The set of all points sitting at equal distance from two intersecting lines is the angle bisector between them. From the figure, the locus of points equidistant from BA and BC is the angle bisector BX of ∠ABC.
(ii) The set of all points at equal distance from two given points is the perpendicular bisector of the line joining them. From the figure, the locus of points equidistant from B and C is the perpendicular bisector YZ of BC.
(iii) Point P is where lines BX and YZ intersect. Since P lies on both loci, it satisfies both conditions. When measured from the construction, PC = 5.1 cm.
In simple words: P is the meeting point of two important lines - the angle bisector from B and the perpendicular bisector of BC.

Exam Tip: Mark the intersection point clearly and verify that P actually lies on both constructed lines before taking your measurement.

Question 12. Points A, B and C represent position of three towers such that AB = 60 m, BC = 73 m and CA = 52 m. Taking a scale of 10 m to 1 cm, make an accurate drawing of △ABC. Find by drawing, the location of a point which is equidistant from A, B and C, and its actual distance from any of the towers.
Answer: Build triangle ABC using the given side lengths and scale. Next, construct the perpendicular bisectors of all three sides of the triangle. Line FG is the perpendicular bisector of AB, line DE is the perpendicular bisector of AC, and line HI is the perpendicular bisector of BC. All three bisectors meet at a single point P. This point P is equidistant from all three vertices A, B, and C. By measuring the distance BP in the drawing and converting back using the scale, the actual distance from any tower to point P is approximately 37 m.
In simple words: The three perpendicular bisectors of a triangle always meet at one point, and this point is equally far from all three corners of the triangle.

Exam Tip: Use an accurate scale and sharp pencil for the perpendicular bisectors - they must meet at exactly one point. Always convert your measured distance back to actual distance using the given scale.

Question 13. Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such points exist?
Answer: Draw two intersecting lines AB and CD making angles of 30° and 150°, meeting at point O. The locus of points equidistant from two intersecting lines consists of the angle bisectors between them. From the figure, EF and GH are the two angle bisectors of the lines AB and CD. To find points that are also 2 cm away from O, construct a circle with O as centre and radius 2 cm. This circle meets the two angle bisectors at four different points - P, Q, R, and S. Therefore, there are exactly 4 points that are equidistant from the two lines and 2 cm away from their point of intersection.
In simple words: A circle drawn around the meeting point will cross both angle bisectors at four spots, giving four points that satisfy both conditions.

Exam Tip: Draw the angle bisectors accurately using compass and ruler - the two bisectors are perpendicular to each other and divide the four angles equally. Count carefully where the circle meets each bisector.

Question 14. Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm. (i) Measure ∠BCD. (ii) Locate the point P on BD which is equidistant from BC and CD.
Answer: Draw the base AB = 6 cm. At point A, construct a 90° angle and bisect it to get 45°. Cut off AD = 6 cm along this bisected line so that ∠BAD = 45°. From point D, cut an arc of 5 cm, and from point B, cut an arc of 3.6 cm. Mark their intersection point as C. Join all points to complete quadrilateral ABCD.
(i) Measuring angle BCD from the constructed figure gives ∠BCD = 65°.
(ii) The locus of points equidistant from two lines is the angle bisector of the angle between them. CE is the angle bisector of ∠BCD, so all points on CE are equidistant from lines BC and CD. Point P is where CE meets diagonal BD.
In simple words: The angle bisector from C is the path of all points equally far from sides BC and CD. Where this path crosses BD is the point P we need.

Exam Tip: To construct 45° without a protractor, first make 90° at A, then bisect it carefully. Verify your angle measurement by checking that the angle bisector actually splits ∠BCD into two equal parts.

Question 15. Without using set square or protractor, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure ∠ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
Answer: Draw AB = 4 cm as the base of the rhombus. From point A, cut an arc of 5 cm and from B, cut an arc of 4 cm, marking their intersection as C. From C, cut an arc of 4 cm and from A, cut another arc of 4 cm, marking their intersection as D. Join all points to form the rhombus ABCD and draw diagonal AC. When measured, ∠ABC = 78°. To find point R such that RB = RC, use the fact that all points equidistant from two points lie on their perpendicular bisector. PQ is the perpendicular bisector of BC and meets side AD at point R. When measured, AR = 1.2 cm. Therefore, ∠ABC = 78° and AR = 1.2 cm.
In simple words: A rhombus has all sides equal. The perpendicular bisector of BC finds the point on AD that is equally far from B and C.

Exam Tip: In a rhombus, opposite angles are equal and diagonals bisect each other at right angles. Use these properties to verify your construction before measuring.

Question 16.
Answer:

Exam Tip:

Question 16. Without using set square or protractor construct:
(i) Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
(ii) Draw the locus of a point which moves so that it is always 2.5 cm from B.
(iii) Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
(iv) Mark the point of intersection of the loci with the letter P and measure PC.
Answer:
(i) Steps of construction:
1. Draw BC = 3.2 cm as base.
2. From B cut an arc of 5.5 cm and from C cut an arc of 4.8 cm. Take their intersection as point A.
3. Join the points to form triangle ABC.

(ii) The locus of a point that stays 2.5 cm away from point B is a circle with centre B and radius 2.5 cm.

(iii) We know that the locus of points equidistant from two lines is the angle bisector of the angle between them. From the figure, CD is the angle bisector of ∠ACB, hence it is equidistant from CA and BC.

(iv) There are two points P₁ and P₂ which intersect the circle. On measuring we get, P₁C = 1.1 cm and P₂C = 3.6 cm.
In simple words: To find points that are a certain distance from B, draw a circle. To find points that are equally far from two sides, draw the angle bisector. Where these two lines meet is the point P.

Exam Tip: Always draw construction lines lightly and keep them visible - examiners need to see your angle bisector and the circle clearly to award marks. Mark the intersection points with precision.

Question 17. By using ruler and compasses only, construct an isosceles triangle ABC in which BC = 5 cm, AB = AC and ∠BAC = 90°. Locate the point P such that (i) P is equidistant from the sides BC and AC. (ii) P is equidistant from the points B and C.
Answer:
Steps of construction:
1. Make BC = 5 cm as base.
2. Create a semicircle with BC as diameter.
3. Make the right bisector of BC and mark it as D.
4. Make the angle bisector of ∠ACB. From the figure, CE is the angle bisector.
5. Mark point P where the angle bisector ∠ACB (i.e. CE) meets the perpendicular bisector of BC (i.e. AD).
6. Draw a perpendicular from point P to BC and let it meet the semicircle at point A.

(i) We know that the locus of points equidistant from two lines is the angle bisector of the angle between them. From the figure, CE is the angle bisector of ∠ACB, hence it is equidistant from AC and BC.

(ii) We know that the locus of points equidistant from two points is the perpendicular bisector of the line segment joining them. From the figure, AD is the perpendicular bisector of BC, hence it is equidistant from B and C.

Both AD and CE meet at point P, so P is equidistant from BC and CA and also from B and C.
In simple words: Draw the perpendicular bisector of BC to find points equally far from B and C. Draw the angle bisector of ∠ACB to find points equally far from the two sides. Where they cross is point P.

Exam Tip: Show all construction lines clearly - the perpendicular bisector, the angle bisector, and their intersection point. Verify by measuring distances from P to confirm equal distances.

Question 18. Using ruler and compass only, construct a semicircle with diameter BC = 7 cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD such that D is equidistant from AB and BC. Measure ∠ADC and write it down.
Answer:
Steps of construction:
1. Draw a line segment BC = 7 cm.
2. Create a semicircle with BC as diameter.
3. Make the right bisector of BC and construct a perpendicular from it such that it meets the semicircle at A as shown in the figure.
4. Construct the angle bisector of ∠ABC, and let it meet the semicircle at point D.
5. Join the points to form quadrilateral ABCD.

On measuring, we get ∠ADC = 135°.
In simple words: The perpendicular bisector of BC passes through any point that is equally far from B and C. Point D sits on the angle bisector of ∠ABC. These two loci meet at P, and the angle ∠ADC turns out to be 135 degrees.

Exam Tip: Keep all construction lines visible and ensure the quadrilateral is completed correctly by joining points in order. The angle bisector and perpendicular bisector are key constructions - they must be accurate for the correct angle measurement.

Question 19. Using ruler and compasses only, construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, ∠B = 60°, AD = 5 cm and D is equidistant from AB and BC. Measure CD.
Answer:
Steps of construction:
1. Draw BC = 5 cm as base.
2. Make an angle of 60° at B.
3. Cut off an arc of 6 cm from B at the angle and mark it A as in the figure.
4. Since D is equidistant from AB and BC, it will lie on the angle bisector of ∠ABC.
5. Make an arc of 5 cm from point A and take point D where the arc cuts the angle bisector BE.
6. Join the points A, B, C and D to form quadrilateral ABCD.

On measuring we get, CD = 5.25 cm approximately.
In simple words: Draw the base BC and the angle at B. Mark A at 6 cm from B. Since D must be equally far from the two sides at B, place it on the angle bisector. The distance from A to D is 5 cm, and the angle bisector helps locate D exactly.

Exam Tip: The angle bisector of ∠ABC is crucial for positioning D correctly. Measure CD carefully at the end - this is the final answer being asked for. Keep construction lines clear throughout.

Question 20. Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Answer:
Steps of construction:
1. Draw AB = 6 cm as base.
2. Make an arc of 4 cm from A and B. Take the point of intersection as C.
3. Join the points A, B and C to form the triangle.
4. Make the angle bisector of C as shown in the figure.
5. Cut an arc from CZ of 5 cm and mark point P such that CP = 5 cm.
6. Make a line XY parallel to AB at a distance of 5 cm.
7. From point P make an arc of 5 cm cutting XY at Q and R.
In simple words: The two points Q and R must satisfy two conditions: they are 5 cm away from P (draw a circle around P), and they are 5 cm away from the line AB (draw two parallel lines, one above and one below AB). Q and R are where the circle meets these parallel lines.

Exam Tip: The two loci here are a circle (5 cm from P) and a pair of parallel lines (5 cm from AB). Their intersections give two possible points Q and R. Both must be marked and labeled clearly. Make sure the parallel lines are drawn accurately using construction methods.

Question 21. Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of length 6 cm and 5 cm respectively. (i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction. (ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Answer:
Steps of construction:
1. Construct a circle with centre O and radius 4 cm.
2. Take a point A on the circle. From A make arcs of radius 6 cm and 5 cm and where they intersect the circle mark those points as B and C respectively.

(i) We know that the locus of points that are equidistant from two points is the perpendicular bisector of the line segment joining those points.

From the figure, IH is the locus of points inside the circle that are equidistant from A and C. Hence, the locus is the diameter of the circle which is perpendicular to the chord AC.

Proof:
Consider △GPA and △GPC.
∠PGC = ∠PGA = 90°
PG = PG (Common side)
CG = AG (as GH bisects AC)
Hence, by SAS congruence, △GPA ≅ △GPC.
Therefore, AP = PC [By C.P.C.T.C.]
Hence, every point on the perpendicular bisector of AC is equidistant from A and C.

(ii) We know that the locus of points that are equidistant from two lines is the angle bisector of the lines.

From the figure, AZ is the angle bisector of the angle between AB and AC. Hence, the locus is the chord of the circle bisecting ∠BAC.
In simple words: For part (i), draw a line through the centre O that is perpendicular to chord AC - any point on this line is equally far from A and C. For part (ii), the angle bisector of ∠BAC gives all points equally far from the two chords AB and AC.

Exam Tip: The proof for part (i) uses SAS congruence - make sure to clearly state the three conditions for congruence. For part (ii), recognize that an angle bisector is the standard locus for points equidistant from two lines. Both constructions must be shown clearly with all working visible.

Question 22. Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit the assessment. (i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm, and ∠ABC = 60°. (ii) Construct the locus of all points, inside △ABC, which are equidistant from B and C. (iii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to △ABC. (iv) Mark the point Q, in your construction, which would make △QBC equal in area to △ABC, and isosceles. (v) Measure and record the length of CQ.
Answer:
(i) Steps of construction:
1. Draw BC = 6 cm as base.
2. Construct ∠ABC = 60° and make an arc from B such that AB = 9 cm.
3. Join points A, B and C such that ABC forms a triangle.

(ii) We know that the locus of points equidistant from two points is the perpendicular bisector of the line segment joining the two points.

From the figure, the right bisector of BC (i.e. DE in the figure) lies inside △ABC.

(iii) The locus of the vertices of the triangles with BC as base, which are equal in area to △ABC is a straight line through A and parallel to BC. The area will be the same since the triangles will be on the same base and between the same parallel lines.

(iv) Q is the point of intersection of the right bisector of BC and the straight line through A parallel to BC.

(v) On measuring, CQ = 8.2 cm approximately.
In simple words: The perpendicular bisector of BC gives all points equally far from B and C. A line through A parallel to BC contains all possible positions for a third vertex that keeps the same area. Where these two lines meet is point Q, which is both equidistant from B and C, and gives a triangle of equal area.

Exam Tip: All construction lines must remain visible - the perpendicular bisector and the parallel line are the two key loci here. Show the intersections clearly and measure CQ accurately at the end. The parallel line through A ensures equal area by the property of triangles on the same base and between same parallels.

Chapter Test

Question 1. Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Answer:
We know that the locus of a point equidistant from two points is the perpendicular bisector of the line segment joining them.

From the figure, CD is the locus of all points which are equidistant from A and B.

Proof:
Consider △GOA and △GOB.
∠GOA = ∠GOB (Both are equal to 90°)
OG = OG (Common side)
AO = OB (They are equal as CD bisects AB at O).
Hence, by SAS congruence, △GOA ≅ △GOB.
Therefore, AG = BG.
Thus, any point on CD is equidistant from A and B.
In simple words: A perpendicular bisector is a line that cuts a segment exactly in the middle and stands at a right angle to it. Every point on this line is the same distance from both endpoints of the segment.

Exam Tip: The SAS congruence criterion is the key to this proof - state all three equal parts clearly. The conclusion AG = BG follows directly from congruent triangles, proving that every point on CD is equidistant from A and B.

Question 2. A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Answer:
Since the point P is allowed to travel in space, the point can exist in any dimension.

Hence, the locus will be a sphere with C as centre and the constant distance as radius.
In simple words: If you move a point in all directions while keeping it the same distance away from C, you trace out a sphere - like an imaginary ball centred at C.

Exam Tip: This question moves into three-dimensional space. The 2D equivalent is a circle; the 3D equivalent is a sphere. Always state clearly that the constant distance becomes the radius of the sphere, and C is its centre.

Question 3. Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².
Answer:
Let the height of triangle PAB = h cm.

So, area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \)

\[ 14 = \frac{1}{2} \times 7 \times h \]

\[ \frac{14 \times 2}{7} = h \]

\[ h = \frac{28}{7} \]

\[ h = 4 \text{ cm} \]

The locus of point P is a pair of straight lines, each parallel to AB and at a distance of 4 cm above and below AB. Any point on these two parallel lines will form a triangle with base AB and area 14 cm².
In simple words: First, find how far P must be from the line AB using the area formula. Then draw two parallel lines - one on each side of AB - at that distance. Any point on either line will make a triangle with area 14 cm².

Exam Tip: The key step is using the area formula to find the perpendicular height h. Once h is found, drawing the two parallel lines (one above and one below AB) completes the locus. Both lines together form the complete answer.

Question 4. Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is (i) at a distance of 3 cm from AB. (ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS? Compute the area of quadrilateral PQRS.
Answer: To find the locus satisfying both conditions, we first construct two parallel lines, CD and EF, each 3 cm away from line AB on opposite sides. Next, we draw a circle with center M and radius 5 cm. The four points where this circle meets the parallel lines are P, Q, R, and S. When these points are joined together, they form a rectangle. By measurement, PQ = 8 cm and PR = 6 cm. The area of rectangle PQRS is found using the formula: Area = Length × Breadth = 8 × 6 = 48 cm².
In simple words: Draw two lines parallel to AB at 3 cm distance. Draw a circle from point M with radius 5 cm. The circle meets these lines at four points P, Q, R, S, which form a rectangle with area 48 cm².

Exam Tip: Always label the locus lines and circle clearly, and measure the rectangle's dimensions carefully to ensure accuracy in area calculation.

Question 5. AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
Answer: To locate the required point, we construct a line EF parallel to AB at a distance of 2 cm. Similarly, we draw another line GH parallel to CD at a distance of 1.6 cm. The point where these two lines intersect is the point M. This point M satisfies both conditions - it is 2 cm away from line AB and 1.6 cm away from line CD.
In simple words: Draw a line parallel to AB at 2 cm distance, and another line parallel to CD at 1.6 cm distance. Where these two lines cross is your answer point.

Exam Tip: Ensure that the parallel lines are drawn on the correct side of the original lines to match the distance requirements.

Question 6. Two straight roads PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flag staff X, which is equidistant from P and S, and is also equidistant from the roads.
Answer: Using the scale 1 cm = 100 m, 800 m becomes 8 cm on the drawing. We draw two roads PK and PQ at an angle of 75° to each other. The locus of points equidistant from two intersecting lines is the angle bisector between them. Therefore, PR is the angle bisector of angle KPQ. Additionally, the locus of points equidistant from two points is the perpendicular bisector of the line segment joining them. Thus, AB is the perpendicular bisector of segment PS. The point X, where the angle bisector PR and the perpendicular bisector AB intersect, is equidistant from both the roads and from points P and S.
In simple words: Draw the angle bisector of the angle between the two roads, and also draw a line that cuts segment PS in half at a right angle. Where these two lines meet is the position of the flag staff.

Exam Tip: Draw the angle bisector accurately using a compass, and ensure the perpendicular bisector passes through the midpoint of PS at right angles.

Question 7. Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Answer: Steps of construction: First, draw the diagonal QS = 6 cm and the diagonal PR = 8 cm, intersecting each other. Connect the endpoints of these diagonals to form the rhombus PQRS. Since the diagonals of a rhombus bisect each other at right angles, PR acts as the angle bisector of angle SPQ. Next, draw CD as the perpendicular bisector of side RS. The point X, where CD and PR intersect, satisfies both conditions - it is equidistant from sides PQ and PS (lying on the angle bisector), and it is equidistant from points R and S (lying on the perpendicular bisector of RS). By measurement, XR = 3.15 cm.
In simple words: Draw the two diagonals to make the rhombus. Draw the angle bisector of angle P and the perpendicular bisector of side RS. These lines meet at point X.

Exam Tip: Measure diagonals precisely at the start, and use both an angle bisector and a perpendicular bisector to locate point X accurately.

Question 8. Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm, the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Answer: Steps of construction: Draw AB = 5.1 cm as the base. At point A, draw an arc with radius 2.8 cm (half of diagonal AC), and at point B, draw an arc with radius 3.5 cm (half of diagonal BD). These arcs meet at point O. Now extend line AO beyond O to mark point C such that OC = AO = 2.8 cm. Similarly, extend line BO beyond O to mark point D such that OD = BO = 3.5 cm. Connect A, B, C, and D to complete the parallelogram. To find point P on DC that is equidistant from sides AB and BC, we use the angle bisector of angle ABC. This angle bisector meets side DC at point P, which satisfies the given condition.
In simple words: Use the diagonals' halves to find point O, then extend to get C and D. The angle bisector of corner B meets side DC at the required point P.

Exam Tip: Ensure the arcs drawn from A and B intersect precisely at O, and that you extend AO and BO correctly to get C and D on opposite sides.

Question 9. By using ruler and compasses only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4 cm and angle DAB = 75°. C is equidistant from the sides AB and AD, also C is equidistant from the points A and B.
Answer: Steps of construction: Draw AB = 6.5 cm as the base line. At point A, construct an angle of 75° and mark point D on this angle such that AD = 4 cm. Since point C must be equidistant from sides AB and AD, it lies on the angle bisector of angle DAB. Since C is also equidistant from points A and B, it lies on the perpendicular bisector of AB. Point C is located at the intersection of these two loci - the angle bisector AE and the perpendicular bisector FG of AB. Finally, join A, B, C, and D to form the required quadrilateral ABCD.
In simple words: Draw the base and the 75° angle with D marked on it. Find C by drawing the angle bisector of angle A and the perpendicular bisector of AB, then joining where they meet.

Exam Tip: Use a protractor's construction method to divide 75° into two 37.5° angles for accuracy, and mark the midpoint of AB carefully for the perpendicular bisector.

Question 10. Use ruler and compass to answer this question. Construct angle ABC = 90°, where AB = 6 cm, BC = 8 cm. (a) Construct the locus of points equidistant from B and C. (b) Construct the locus of points equidistant from A and B. (c) Mark the point which satisfies both the conditions (a) and (b) as O. Construct the locus of points keeping a fixed distance OA from the fixed point O. (d) Construct the locus of points which are equidistant from BA and BC.
Answer: Steps of construction: Draw line segment BC = 8 cm. Construct angle ABC = 90° with AB = 6 cm meeting at B. Draw XY as the perpendicular bisector of BC - this is the locus of all points equidistant from B and C. Draw PQ as the perpendicular bisector of AB - this is the locus of all points equidistant from A and B. Mark O as the intersection point of XY and PQ. The required locus for part (c) is a circle with center O and radius OA. Draw BZ as the angle bisector of angle ABC - this is the locus of points equidistant from sides BA and BC. Summary: (a) Locus is XY (perpendicular bisector of BC). (b) Locus is PQ (perpendicular bisector of AB). (c) Locus is a circle with center O and radius OA. (d) Locus is BZ (angle bisector of angle ABC).
In simple words: Perpendicular bisectors of two sides give you point O. From O, draw a circle through A. The angle bisector from B splits the right angle equally.

Exam Tip: Mark all perpendicular bisectors and the angle bisector clearly with different line styles, and ensure the circle in part (c) passes exactly through point A.