OP Malhotra Class 10 Maths Solutions Chapter 13 Loci Exercise 13 (C)

S Chand Class 10 ICSE Maths Solutions Chapter 13 Loci Ex 13(c)

Question 1. Without using set-square or protractor, draw
(i) the perpendicular CN from C to AB;
(ii) the point P on CN equidistant from AB and AC.
Answer:
(i) Steps of construction:
(a) Take a line segment AB and a point C outside it.
(b) From C, draw an arc with a suitable radius that intersects AB at E and F.
(c) With centers E and F, draw arcs that intersect each other at G.
(d) Join CG, making it meet AB at N. Then CN is perpendicular to AB.
(e) Join CA. This completes the first part of the construction.
(ii) To find point P:
Draw the angle bisector `l` of angle CAB. This bisector `l` will intersect the perpendicular line CN at point P. This point P is the desired location, as it is equidistant from both lines AB and AC by definition of an angle bisector.
In simple words: First, draw a line and mark a point outside it. Then, draw a line from the outside point that crosses the first line at a right angle (perpendicular). Next, draw a line that cuts the angle at A exactly in half (angle bisector). The spot where this bisector crosses the perpendicular line is point P. This point P is special because it's the same distance from both lines AB and AC.

🎯 Exam Tip: Remember that a point equidistant from two intersecting lines lies on the angle bisector of the angle formed by those lines. For perpendiculars, use arcs from the external point to create two points on the line, then bisect those two points from the external point.

Question 2. Without using set-squares or protractor, draw a triangle FGH in which GH = 2.5 units, FG = FH, and the altitude FX = 3 units. Measure FG.
Answer:
Steps of construction:
(i) Draw a line segment GH = 2.5 units (cm).
(ii) Draw its perpendicular bisector (a line that cuts GH exactly in half at a right angle) and mark off XF = 3 units (cm) along this bisector from the midpoint X of GH. This creates the altitude.
(iii) Join points FG and FH. Since FG = FH, this makes triangle FGH an isosceles triangle.
On measuring FG, it is 3.4 units (cm).
In simple words: First, draw the base line GH. Then, find the middle of GH and draw a straight line upwards from there at a right angle. Mark a point F on this line, 3 units up from GH. Finally, connect F to G and F to H. This creates a triangle where the sides FG and FH are equal, and FX is its height. When you measure FG, you will find it is 3.4 units long.

🎯 Exam Tip: When constructing an isosceles triangle with a given base and altitude, always draw the perpendicular bisector of the base first to locate the apex of the triangle.

Question 3. Using ruler and compasses only, construct a quadrilateral ABCD in which AB = 2 units, BC = 2.5 units, ∠ABC = 60°, ∠BCD = 90° and AD = DC. Measure the length of BD.
Answer:
Steps of Construction:
(i) Draw a line segment BC = 2.5 units.
(ii) At point B, draw a ray (BX) making an angle of 60° with BC. At point C, draw a ray (CY) making an angle of 90° with BC. This sets up two angles of the quadrilateral.
(iii) From B, cut off BA = 2 units on the ray BX. This determines point A.
(iv) Join AC. Now, draw the perpendicular bisector of AC. This bisector will meet ray CY at point D, as AD = DC means D must lie on the perpendicular bisector of AC.
(v) Join AD. This forms the side AD.
(vi) Join BD. On measuring, BD is 2.8 units. Thus, ABCD is the required quadrilateral.
The length of BD is 2.8 units.
In simple words: First, draw the line BC. From point B, draw a line at a 60-degree angle and mark point A, 2 units away. From point C, draw a line straight up at a 90-degree angle. Connect A and C. Then, find the middle of AC and draw a line that cuts it at 90 degrees. This line will cross the 90-degree line from C at a point, which will be D. Connect D to A and D to B. When you measure the line BD, it should be 2.8 units long.

🎯 Exam Tip: For quadrilaterals where AD=DC, point D must lie on the perpendicular bisector of the diagonal AC. This is a key locus concept to remember for such constructions.

Question 4. Without using set square or protractor, construct a quadrilateral ABCD in which BC = 2.2 units, ∠ABC = 90°, AB = 1.8 units, AD = DC and ∠ADC = 60°. Construct also the perpendicular bisector of BC to meet AD at X and measure XB.
Answer:
Steps of Construction:
(i) Draw a line segment AB = 1.8 units.
(ii) At B, draw a ray (BX) making an angle of 90° with AB. On this ray, cut off BC = 2.2 units. This forms the side BC.
(iii) Since AD = DC and ∠ADC = 60°, triangle ADC is an equilateral triangle. (This is an important deduction based on the given conditions).
(iv) With center A and C, and a radius equal to AC, draw arcs intersecting each other at D. This locates point D using the equilateral property.
(v) Join AD and CD to complete the triangle ADC.
(vi) Draw the perpendicular bisector of BC. This bisector will intersect the line segment AD at point X. The point X is found by the intersection of the locus.
(vii) Join BX.
On measuring, BX = 2.6 units.
In simple words: Start by drawing line AB. At B, draw a line straight up (90 degrees) and mark point C, 2.2 units away. Since AD=DC and angle ADC is 60 degrees, this means triangle ADC is an equilateral triangle. Use A and C as centers, with radius equal to AC, to find point D. Connect AD and CD. Next, find the middle of BC and draw a line that cuts it at a 90-degree angle. This line will cross AD at point X. Draw a line from B to X. When you measure BX, it should be 2.6 units.

🎯 Exam Tip: Recognizing special triangles like equilateral triangles from given side and angle properties simplifies construction. The perpendicular bisector of a segment is the locus of points equidistant from its endpoints.

Question 5. Draw a triangle ABC in which BC = 5.3 cm, the angle A = 38°, and the angle B = 56°. Using ruler and compasses only, construct the circle that passes through A, B and C.
Answer:
Steps of Construction:
First, calculate angle C: \( \angle C = 180° - (\angle B + \angle A) = 180° - (56° + 38°) = 180° - 94° = 86° \). Knowing all angles helps confirm construction.
(i) Draw a line segment BC = 5.3 cm.
(ii) At point B, draw a ray making an angle of 56° with BC. At point C, draw a ray making an angle of 86° with BC. These two rays will meet each other at point A, forming triangle ABC.
(iii) Now, draw the perpendicular bisectors of sides BC and CA. These bisectors will intersect each other at point O, which is the circumcenter of the triangle.
(iv) With O as the center, and a radius equal to OA (or OB or OC, as they will all be the same), draw a circle. This circle will pass through points A, B, and C. This is the required circumcircle.
In simple words: First, draw the line BC. Then, using your compass, draw angles at B (56 degrees) and C (86 degrees) to find point A, which forms the triangle ABC. Next, find the exact middle of two sides of the triangle (like BC and AC) and draw lines that cut these sides at a right angle (perpendicular bisectors). Where these two lines cross is the center of your circle, let's call it O. Use O as the center and the distance from O to any corner of the triangle (A, B, or C) as the radius. Draw your circle, and it will pass through all three corners.

🎯 Exam Tip: To construct a circumcircle, find the circumcenter by drawing the perpendicular bisectors of any two sides of the triangle. Their intersection is the center of the circle that passes through all three vertices.

Question 6. Draw a triangle with sides 3 units, \( 3\frac { 1 }{ 2 } \) units, and 4 units, without using set-square or protractor, construct the bisector of two of the angles of the triangle and let them meet at the point X. Construct also the perpendicular XL from X to the longest side.
Answer:
Steps of construction:
(i) Draw a line segment BC = \( 3\frac { 1 }{ 2 } \) units (which is 3.5 units). This will be one side of the triangle.
(ii) With B as the center and a radius of 3 units, draw an arc. With C as the center and a radius of 4 units, draw another arc. These arcs will intersect each other at point A, forming the triangle ABC.
(iii) Draw the angle bisector of angle B and the angle bisector of angle C. These two bisectors will intersect each other at point X. Point X is the incenter of the triangle.
(iv) From point X, draw a perpendicular line XL to the longest side AC. This segment XL represents the inradius if AC were the side to which inradius is drawn.
In simple words: First, draw the base of the triangle (3.5 units). Then, use your compass to draw arcs from each end of the base (3 units from one end, 4 units from the other) to find the third corner, A. This creates the triangle. Next, draw lines that cut two of the triangle's angles exactly in half (angle bisectors). Where these two lines cross is point X. Finally, draw a line from point X straight down to the longest side (AC) at a right angle. This line is XL.

🎯 Exam Tip: The intersection point of angle bisectors (incenter) is equidistant from the sides of the triangle. The perpendicular from this point to any side represents the inradius.

Question 7. Without using set-square or protractor, construct a triangle ABC in which AB = 1.6 units, BC = 2 units and ∠ABC = 120°. Find a point P outside the triangle such that ∠BAP = 90° and BP = CP. Measure the length of BP.
Answer:
Steps of construction:
(i) Draw a line segment BC = 2 units.
(ii) At point B, draw a ray (BX) making an angle of 120° with BC. On this ray, cut off BA = 1.6 units. This forms the side BA.
(iii) Join AC. This completes the triangle ABC.
(iv) At point A, draw a ray AY making an angle of 90° with AB. This ray represents the locus of points for which ∠BAP = 90°.
(v) Draw the perpendicular bisector of BC. This bisector represents the locus of points equidistant from B and C. The perpendicular bisector will intersect ray AY at point P. This P is the required point.
On measuring PB, it is 2.5 units.
In simple words: First, draw line BC. At B, draw a line at a 120-degree angle and mark point A, 1.6 units away. Connect A and C to make triangle ABC. Now, from A, draw a line upwards at a 90-degree angle to AB. This line shows all possible places where point P could be for angle BAP to be 90 degrees. Next, find the exact middle of BC and draw a line that cuts BC at a right angle. This line shows all possible places where point P could be to be equally far from B and C. The spot where these two lines cross is point P. If you measure the length of PB, it will be 2.5 units.

🎯 Exam Tip: A point P that forms a 90° angle with a given line segment (like BAP=90°) must lie on a line perpendicular to that segment. A point equidistant from two other points (BP=CP) must lie on the perpendicular bisector of the segment connecting them (BC).

Question 8. Draw a line ABC making AB and BC each 6 cm long. At B draw BD of length 5 cm, making angle ABD = 42°. Find by construction two points P, Q each of which is equidistant from B and D and 2.5 cm from the line ABC. Measure PA and QA. (Use of set-square is permitted. No explanation or proof is required bat all construction lines must be clearly shown.)
Answer:
Steps of construction:
(i) Draw a line segment ABC such that AB = BC = 6 cm. So, AC is a 12 cm line.
(ii) At B, draw a ray making an angle of 42° with AB. On this ray, cut off BD = 5 cm. This establishes point D.
(iii) Join BD and draw its perpendicular bisector (let's call it line 'n'). This bisector represents the locus of points equidistant from B and D.
(iv) Draw two parallel lines (let's call them 'l' and 'm') to the line ABC, each at a distance of 2.5 cm from ABC. These lines represent the locus of points 2.5 cm away from ABC. The lines 'l' and 'm' will intersect the perpendicular bisector 'n' at points P and Q, respectively. These P and Q are the required points.
(v) Now, join AP and AQ. On measuring, AP = 6.5 cm and AQ = 2.7 cm (approximately).
In simple words: Draw a straight line and mark points A, B, C so AB and BC are each 6 cm. From B, draw a line at a 42-degree angle and mark point D, 5 cm away. Then, draw a line that cuts BD exactly in half and at a right angle. This line is where all points equally far from B and D can be. Next, draw two lines that run parallel to ABC, one 2.5 cm above and one 2.5 cm below. Where these parallel lines cross the bisector of BD, those are your points P and Q. Measure the lines from A to P and A to Q. They should be about 6.5 cm and 2.7 cm respectively.

🎯 Exam Tip: Loci problems often combine multiple conditions. Identify the locus for each condition (e.g., perpendicular bisector for equidistant points, parallel lines for fixed distance from a line) and find their intersection(s).

Question 9. 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:
(i) Draw a line segment AB = 6 cm.
(ii) At point B, draw a ray (BX) making an angle of 60° with AB. On this ray, cut off BC = 5 cm. This completes the side BC.
(iii) Draw the angle bisector (BY) of angle B (∠ABC). This bisector is the locus of points equidistant from AB and BC.
(iv) With center A and a radius of 5 cm, draw an arc. This arc will intersect the angle bisector BY at point D. Point D is equidistant from AB and BC, and 5 cm from A.
(v) Join AD and CD. This completes the quadrilateral ABCD.
ABCD is the required quadrilateral. On measuring CD, it is 5.3 cm.
In simple words: First, draw line AB. At B, draw a line at a 60-degree angle and mark point C, 5 cm away. Then, draw a line that cuts angle B exactly in half; this line is where all points equally far from AB and BC will lie. Next, from A, draw an arc with a radius of 5 cm. Where this arc crosses the angle bisector, that's point D. Connect AD and CD. Finally, measure the length of CD, which should be 5.3 cm.

🎯 Exam Tip: The condition "D is equidistant from AB and BC" immediately tells you that D must lie on the angle bisector of ∠ABC. This is a crucial starting point for locating D.

Question 10. 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:
Steps of construction:
(i) Draw two intersecting lines (AB and CD) that meet at point O, forming an angle of 30° between them. (e.g., ∠BOD = 30°).
(ii) Draw the angle bisector (l) of ∠BOD and extend it to the opposite side. Also, draw the angle bisector (m) of ∠AOD. These bisectors represent points equidistant from the intersecting lines.
(iii) With center O and a radius of 2 cm, draw a circle. This circle represents the locus of points 2 cm away from the point of intersection O.
(iv) Mark the points where the circle intersects the angle bisectors 'l' and 'm'. There will be two points on bisector 'l' (P and Q) and two points on bisector 'm' (R and S).
Then P, Q, R and S are the four required points. There are 4 such points.
In simple words: Draw two lines that cross each other at point O, making a 30-degree angle. Then, draw lines that cut each of the four angles formed at O exactly in half (these are angle bisectors). Next, use O as the center and draw a circle with a radius of 2 cm. The spots where this circle crosses the angle bisector lines are your special points. You will find there are four such points.

🎯 Exam Tip: Remember that two intersecting lines create four angles, and thus four angle bisectors (two distinct lines, as opposite angles share a bisector). A circle centered at the intersection will cross these bisectors in four places, giving four solutions.

Question 11. 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:
Steps of construction:
(i) Draw a line segment AC = 5 cm.
(ii) With center A and a radius of 4 cm, draw an arc. With center C and a radius of 4 cm, draw another arc. These arcs will intersect each other at points B and D, as all sides of a rhombus are equal.
(iii) Join AB, BC, CD, and AD. ABCD is the required rhombus. On measuring ∠ABC, it is found to be 78°. This is one of the properties of the rhombus.
(iv) Draw the perpendicular bisector of BC. This bisector will intersect the line segment AD at point R. This point R is equidistant from B and C.
Now, measure the length of AR, which is 1.2 cm.
In simple words: First, draw the diagonal AC. Then, using A and C as centers and 4 cm as the radius, draw arcs that cross each other above and below AC. These crossing points are B and D. Connect A to B, B to C, C to D, and D to A to form the rhombus. Measure angle ABC, which should be 78 degrees. Next, find the middle of line BC and draw a line that cuts it at a right angle. This line will cross AD at point R. Measure the length of AR, which will be 1.2 cm.

🎯 Exam Tip: For constructing a rhombus with given side and diagonal, remember that all sides are equal. To find a point equidistant from two vertices (RB=RC), you need to construct the perpendicular bisector of the segment connecting them (BC).

Question 12. Without using set square or protractor, construct a 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:
Steps of construction:
(i) Draw a line segment AB = 6 cm.
(ii) At A, draw a ray (AX) making an angle of 45° with AB. On this ray, cut off AD = 6 cm. This establishes point D.
(iii) With center B and a radius of 3.6 cm, draw an arc. With center D and a radius of 5 cm, draw another arc. These arcs will intersect each other at point C. This completes the side BC and CD.
(iv) Join BC and CD. ABCD is the required quadrilateral.
(v) On measuring ∠BCD, it is 65°.
(vi) Join BD. This is a diagonal of the quadrilateral.
(vii) Draw the angle bisector of ∠BCD. This bisector will intersect the diagonal BD at point P. This point P is equidistant from BC and CD.
P is the required point.
In simple words: First, draw line AB. At A, draw a line at a 45-degree angle and mark point D, 6 cm away. From B, draw an arc with a radius of 3.6 cm. From D, draw another arc with a radius of 5 cm. Where these two arcs cross is point C. Connect BC and CD to finish the quadrilateral. Measure angle BCD, it should be 65 degrees. Next, draw the diagonal BD. Then, draw a line that cuts angle BCD exactly in half. This line will cross BD at point P. This point P is the same distance from lines BC and CD.

🎯 Exam Tip: When a point is required to be equidistant from two sides, it must lie on the angle bisector of the angle formed by those sides. This helps pinpoint its location accurately.

Question 13. By using ruler and compasses only, construct an isosceles triangle PQR in which QR = 4 cm, PQ = PR and ∠QPR = 90°. Locate the point X such that
(i) X is equidistant from the sides QR and PR.
(ii) X is equidistant from the points Q and R.
Answer:
Steps of construction:
(i) Draw a line segment QR = 4 cm.
(ii) Draw a semicircle on QR as its diameter. This semicircle will contain all points P such that ∠QPR = 90°.
(iii) Draw the perpendicular bisector of QR. This bisector will meet the semicircle at point P. Since P is on the perpendicular bisector, PQ = PR, making ∆PQR an isosceles triangle with ∠QPR = 90°.
(iv) Join PQ and PR. ∆PQR is the required isosceles triangle.
(v) Now, to locate X:
Draw the bisector of ∠Q (∠PQR).
The point X, which is equidistant from the sides QR and PR, must lie on the angle bisector of ∠R. The point X, which is equidistant from points Q and R, must lie on the perpendicular bisector of QR. The intersection of the angle bisector of ∠R and the perpendicular bisector of QR gives point X.
Alternatively, since PQ=PR, angle PQR = angle PRQ. The intersection of the angle bisector of ∠Q and the perpendicular bisector of QR would give point X.
In simple words: First, draw the line QR. Draw a semicircle above QR using QR as the diameter. Then, find the exact middle of QR and draw a line straight up until it touches the semicircle. This point is P. Connect P to Q and P to R to form the isosceles triangle PQR with a right angle at P. Now, to find point X, draw a line that cuts angle Q (PQR) exactly in half. Also, draw the line that cuts QR exactly in half and at a right angle (perpendicular bisector). The point where these two lines cross is X. This point X is equally far from the lines QR and PR, and also equally far from points Q and R.

🎯 Exam Tip: The locus of points equidistant from two sides is the angle bisector. The locus of points equidistant from two points is the perpendicular bisector of the segment joining them. The intersection of these loci gives the required point.

Question 14. Construct a triangle ABX such that AB = 4.5 cm and BX = 3 cm and ∠ABX = 35°. Complete the rhombus ABCD such that X is equidistant from AB and BC. Locate the point Y on the line BX such that Y is equidistant from A and B.
Answer:
Steps of construction:
(i) Draw a line segment AB = 4.5 cm.
(ii) At B, draw a ray (B to X) making an angle of 35° with AB. On this ray, cut off BX = 3 cm.
(iii) Since X is equidistant from AB and BC, X must lie on the angle bisector of ∠ABC.
(iv) Now, draw ray BC such that ∠XBC = ∠ABX = 35°. This means ∠ABC = 70°. This confirms X is on the angle bisector of ∠ABC.
(v) Cut off BC = AB = 4.5 cm. (All sides of a rhombus are equal).
(vi) With center C and a radius of AB (4.5 cm), draw an arc. With center A and a radius of AB (4.5 cm), draw another arc. These arcs will intersect each other at point D. This helps complete the rhombus.
(vii) Join CD and AD. Then ABCD is the required rhombus.
(viii) Draw the perpendicular bisector of AB. This bisector will meet the line BX at point Y. This point Y is equidistant from A and B.
In simple words: First, draw line AB. From B, draw a line at a 35-degree angle and mark point X, 3 cm away. Since X is equally far from lines AB and BC, it means that the line BX must cut angle ABC in half. So, draw another line from B, making an angle of 35 degrees on the other side of BX, and mark point C so that BC is the same length as AB (4.5 cm). Now, from C, draw an arc with a radius of 4.5 cm. From A, draw another arc with a radius of 4.5 cm. Where these arcs meet is point D. Connect the points to form rhombus ABCD. Finally, draw a line that cuts AB exactly in half and at a right angle. This line will cross the ray BX at point Y. This point Y is equally far from A and B.

🎯 Exam Tip: The property that a point is equidistant from two lines means it lies on the angle bisector. Also, for a point to be equidistant from two points, it must lie on the perpendicular bisector of the line segment connecting them. Apply these loci rules sequentially.

Self Evaluation And Revision (Latest Icse Questions)

Question 1. Use ruler and compasses only for the following questions : Construct triangle BCP, where CB = 5 cm, BP = 4 cm, ∠PBC = 45°. Complete the rectangle ABCD such that:
(i) P is equidistant from AB and BC; and
(ii) P is equidistant from C and D.
(iii) Measure and write down the length of AB.
Answer:
Steps of construction:
(i) Draw a line segment CB = 5 cm.
(ii) At B, draw a ray making an angle equal to 45° (∠PBC = 45°). On this ray, cut off BP = 4 cm. This forms the side BP.
(iii) Join PC. This completes triangle BCP.
(iv) To complete rectangle ABCD:
Since P is equidistant from AB and BC, it must lie on the angle bisector of ∠ABC. This means ∠PBA = ∠PBC = 45°. Therefore, ∠ABC = 90°.
(v) From C, draw a perpendicular CX. This forms a right angle at C.
(vi) From P, draw a perpendicular to CX, intersecting it at E.
(vii) Cut off ED = CE. This helps locate D.
(viii) With center D and radius BC (5 cm), and with center B and radius CD, draw arcs that intersect each other at A.
(ix) Join AB and AD. ABCD is the required rectangle. Since AB is opposite to CD, they are equal. Also, BC is opposite to AD. This ensures it's a rectangle.
On measuring AB, it is 5.5 cm.
In simple words: First, draw line CB. At B, draw a line at a 45-degree angle and mark point P, 4 cm away. Connect P and C. To make a rectangle ABCD, point P must be equally far from AB and BC, so angle ABC must be 90 degrees. Also, P must be equally far from C and D. Draw a line from C at a right angle (CX). From P, draw a line that is also at a right angle to CX, meeting it at E. Mark point D so ED is the same length as CE. Using D as a center, draw an arc with the length of BC. From B, draw an arc with the length of CD. Where these arcs meet is point A. Connect AB and AD to finish the rectangle. When you measure AB, it will be 5.5 cm.

🎯 Exam Tip: When a point (P) is equidistant from two non-parallel lines (AB and BC), it lies on the angle bisector. When it's equidistant from two points (C and D), it lies on the perpendicular bisector of the segment CD. Use these properties to precisely locate points.

Question 2. Ruler and compasses only may be used in this question. All construction lines and area must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct triangle ABC, in which BC = 8 cm, AB = 5 cm, angle 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 as P, the point which is equidistant from AB, BC and equidistant from B and C;
(v) Measure and record the length of PB.
Answer:
Steps of construction:
(i) Draw a line segment BC = 8 cm.
(ii) At B, draw a ray (BX) making an angle of 60° with BC. On this ray, cut off BA = 5 cm. This completes the side BA.
(iii) Join AC. This forms the required triangle ABC.
(iv) Construct the locus of points inside the triangle equidistant from BA and BC: This is the angle bisector of ∠ABC (let's call it BY).
(v) Construct the locus of points inside the triangle equidistant from B and C: This is the perpendicular bisector of BC (let's call it `l`).
(vi) Mark point P where the angle bisector BY and the perpendicular bisector `l` intersect. This point P is equidistant from AB and BC, and also equidistant from B and C.
(vii) Measure and record the length of PB. On measuring BP, it is 4.6 cm.
In simple words: First, draw the line BC. From B, draw a line at a 60-degree angle and mark point A, 5 cm away. Connect A and C to form triangle ABC. Now, draw a line that cuts angle ABC exactly in half (this is the angle bisector). Next, find the exact middle of BC and draw a line that cuts it at a right angle (this is the perpendicular bisector). Where these two lines cross is point P. This point P is equally far from the lines BA and BC, and also equally far from points B and C. Measure the length of PB, which should be 4.6 cm.

🎯 Exam Tip: Two key loci are often involved: the angle bisector (equidistant from two lines) and the perpendicular bisector (equidistant from two points). Finding their intersection is key to solving such problems.

Question 3. 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 assessment.
(i) Construct a triangle ABC, in in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangles ABC, which are equidistant from B and C.
(iii) Construct the locus of all the vertices of the triangle with BC as base, which are equal in area to triangle ABC.
(iv) Mark the point Q in your construction, which would make \( \triangle QBC \) equal in area to \( \triangle ABC \), and isosceles.
(v) Measure and record the length of CQ.
Answer:
(i) To construct triangle ABC:
- Draw a line segment BC of 6 cm.
- At point B, use a compass to construct an angle of 60° from BC, creating a ray.
- On this ray, mark point A such that BA is 9 cm long.
- Connect points A and C to complete triangle ABC. This forms the required triangle.
(ii) The locus of points inside the triangle that are equidistant from B and C is the perpendicular bisector of the line segment BC. Draw this bisector.
(iii) The locus of all vertices of a triangle with BC as base, having the same area as triangle ABC, is a line segment drawn through A and parallel to BC. Draw this line, labeled XY.
(iv) To mark point Q:
- Draw the perpendicular bisector of BC (from part ii).
- This perpendicular bisector will intersect the line XY (drawn in part iii) at a point. Label this point Q. Join QB and QC. Triangle QBC will be an isosceles triangle with the same area as triangle ABC.
(v) Measure CQ:
- When measured, the length of CQ is 8.4 cm.
In simple words: First, draw the triangle. Then, draw the line that's exactly in the middle of B and C, and another line through A that's parallel to BC. Where the middle line of BC crosses the parallel line, that's point Q. Then measure how long CQ is.

🎯 Exam Tip: When constructing triangles, ensure angles are precise and line segments are measured accurately to get correct final results for loci and measurements.

Question 4. State the locus of a point in a rhombus ABCD, which is equidistant :
(i) From AB and AD.
(ii) From the vertices A and C.
Answer:
(i) The locus of a point equidistant from sides AB and AD in a rhombus ABCD is the angle bisector of \( \angle DAB \). In a rhombus, the diagonal AC bisects \( \angle DAB \). Therefore, the diagonal AC is the required locus.
(ii) The locus of a point equidistant from vertices A and C is the perpendicular bisector of the line segment AC. In a rhombus, the diagonal BD is the perpendicular bisector of AC. Therefore, the diagonal BD is the required locus. The point where both diagonals intersect (point O) is equidistant from A, B, C, and D.
In simple words: For a rhombus, if a point is equally far from two sides that meet (like AB and AD), it must be on the line that cuts the angle between them in half (which is the diagonal AC). If a point is equally far from two corners (like A and C), it must be on the line that cuts the connection between them in half at a right angle (which is the diagonal BD).

🎯 Exam Tip: Remember that in a rhombus, diagonals serve as both angle bisectors for the vertices they connect and perpendicular bisectors for the other diagonal.

Question 5. Use graph paper for this question. Take 1 cm = 1 unit.
(i) Plot the points A (1,1), B (5,3) and C (2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure PA in cm.
Answer:
(i) Plot the points A (1,1), B (5,3) and C (2,7) on a graph paper. Connect these points to form triangle ABC. Make sure to label each point clearly.
(ii) The locus of points equidistant from A and B is the perpendicular bisector of the line segment AB. Draw this line using a compass and ruler.
(iii) The locus of points equidistant from AB and AC is the angle bisector of \( \angle BAC \). Construct this angle bisector using a compass and ruler.
(iv) Point P is found where the perpendicular bisector of AB (from part ii) intersects the angle bisector of \( \angle BAC \) (from part iii). This point P is equally distant from A and B, and also equally distant from the lines AB and AC.
(v) Measure PA:
- When measured, the length of PA is 2.5 cm.
In simple words: Draw the points and the triangle. Then, draw a line that cuts the line AB exactly in half and is straight up-and-down from it. Also, draw a line that splits the angle at A into two equal parts. Where these two new lines cross, that's point P. Finally, measure the distance from P to A.

🎯 Exam Tip: Always make your constructions on graph paper as neat and precise as possible, as small inaccuracies can lead to incorrect measurements.

Question 6. Construct \( \triangle ABC \), AB = 7 cm, BC = 8 cm and \( \angle ABC \) = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Answer:
First, construct triangle ABC:
- Draw a line segment BC of 8 cm.
- At point B, use a compass to construct a 60° angle ( \( \angle ABC \) ). Draw a ray for this angle.
- Along the ray of this 60° angle, mark point A such that AB is 7 cm.
- Connect points A and C to form triangle ABC.
(i) The locus of points equidistant from B and C is the perpendicular bisector of the line segment BC. Draw this bisector and label it 'l'.
(ii) The locus of points equidistant from AB and BC is the angle bisector of \( \angle ABC \). Draw this bisector and label it 'm'.
Point P is the intersection of the perpendicular bisector 'l' and the angle bisector 'm'. This point P satisfies both conditions: it is equally distant from B and C, and also equally distant from lines AB and BC.
(iii) Measure PB:
- When measured, the length of PB is 4.5 cm.
In simple words: First, draw the triangle. Then draw a line that cuts the line BC exactly in half and is straight up-and-down. Also, draw a line that splits the angle at B into two equal parts. Where these two new lines cross, that's point P. Finally, measure the distance from P to B.

🎯 Exam Tip: For construction problems, always list your steps clearly and show all construction arcs for full marks. The intersection of two loci precisely defines the point that meets both criteria.

Question 7. Construct an isosceles triangle ABC, such that AB = 6 cm, BC = AC = 4 cm. Bisect \( \angle C \) internally and make 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 line AB.
Answer:
First, construct isosceles triangle ABC:
- Draw a line segment AB of 6 cm.
- With A as the center and a radius of 4 cm, draw an arc.
- With B as the center and a radius of 4 cm, draw another arc. These two arcs will intersect at point C.
- Connect A to C and B to C to complete triangle ABC.
Next, find point P:
- Draw the angle bisector of \( \angle C \). This line will divide \( \angle C \) into two equal parts.
- On this angle bisector, measure and mark point P such that CP is 5 cm long.
Finally, find points Q and R:
- Draw a line 'l' that is parallel to AB and is 5 cm away from AB. This line represents all points 5 cm from AB.
- With P as the center and a radius of 5 cm, draw an arc. This arc represents all points 5 cm from P.
- The points where this arc intersects the parallel line 'l' are Q and R. These are the two required points.
In simple words: Draw the triangle first. Then, draw a line that cuts angle C in half. Mark point P on this line, 5 cm from C. Now, draw a line that is parallel to AB and 5 cm away from it. Draw a circle around P with a radius of 5 cm. The places where this circle crosses the parallel line are points Q and R.

🎯 Exam Tip: When constructing a point based on multiple distance conditions, it often involves finding the intersection of a circle (for distance from a point) and a parallel line (for distance from a line).

Question 8.
Using only a ruler and compasses, construct \( \angle ABC \) = 120°, where AB = BC = 5 cm.
(a) Mark two points D and E which satisfy the condition that they are equidistant from both BA and BC.
(b) In the above figure, join AE and EC. Describe the figures :
(i) ABCD,
(ii) ABD,
(iii) ABE
Answer:
First, construct triangle ABC:
- Draw a line segment BC of 5 cm.
- At point B, construct an angle of 120° ( \( \angle ABC \) ). Draw a ray for this angle.
- Along this ray, mark point A such that BA is 5 cm.
- Join A and C to form triangle ABC.
(a) Mark two points D and E equidistant from BA and BC:
- The locus of points equidistant from two intersecting lines BA and BC is the angle bisector of \( \angle ABC \). Draw this angle bisector, let's call it ray BY.
- Point D is where the angle bisector BY intersects the line segment AC.
- Point E is another point chosen anywhere further along the ray BY. Both D and E are equidistant from lines BA and BC.
(b) Join AE and EC. Describe the figures:
(i) Figure ABCD is a quadrilateral.
(ii) Figure ABD is a triangle.
(iii) Figure ABE is a triangle.
In simple words: Draw the triangle with a 120-degree angle. Then, draw a line that cuts the 120-degree angle at B exactly in half. Any point on this line is equally far from BA and BC. Mark points D (where this line crosses AC) and E (any other point on this line). After connecting A to E and E to C, name the shapes you see: ABCD, ABD, and ABE.

🎯 Exam Tip: Remember that the angle bisector is the locus for points equidistant from two lines. Carefully follow the construction steps to ensure accurate diagrams.

Question 9.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and \( \angle PBC \) = 45°.
Complete the rectangle ABCD such that
(i) P is equidistant from AB and AC.
(ii) P is equidistant from C and D.
Answer:
First, construct triangle BCP:
- Draw a line segment BC of 5 cm.
- At point B, draw a ray BX at an angle of 45° ( \( \angle PBC \) ).
- On this ray, mark point P such that BP is 4 cm.
- Join P and C to form triangle BCP.
Next, complete the rectangle ABCD:
- Draw a line CY perpendicular to BC at point C.
- From P, draw a line perpendicular to CY, and let it meet CY at E.
- Mark point D on the line CY such that ED = CE.
- Draw a line perpendicular to BC at B.
- Measure the length of CD, and mark point A on the perpendicular line at B such that BA = CD.
- Join A and D. ABCD is the required rectangle.
Based on this construction, point P satisfies the following properties:
- P is equidistant from BC and BA (meaning P lies on the angle bisector of \( \angle ABC \) ).
- P is equidistant from C and D (meaning P lies on the perpendicular bisector of CD).
In simple words: First, draw the triangle BCP using the given measurements. Then, make sure the shape is a rectangle by drawing perpendicular lines and making opposite sides equal. Once the rectangle ABCD is made, point P will be found to be equally far from lines BC and BA, and also equally far from points C and D.

🎯 Exam Tip: To construct a rectangle, ensure that all angles are 90 degrees and opposite sides are equal. Use perpendicular lines to achieve this accuracy.

Question 10.
A straight line 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:
First, draw the line segment:
- Draw a straight line segment AB that is 8 cm long.
(i) Locus of points equidistant from A and B:
- Construct the perpendicular bisector of the line segment AB. This line will pass through the midpoint of AB (let's call it O) and be perpendicular to AB. This perpendicular bisector is the required locus.
(ii) Locus of points always 4 cm from line AB:
- Draw two lines parallel to AB, one on each side of AB, with each line being exactly 4 cm away from AB. These two parallel lines form the required locus.
(iii) Mark two points X and Y, which are 4 cm from AB and equidistant from A and B. Name the figure AXBY:
- The points X and Y are located at the intersection of the perpendicular bisector of AB (from part i) and the two parallel lines (from part ii).
- Specifically, from the midpoint O of AB, measure 4 cm along the perpendicular bisector in both directions, marking points X and Y. These points X and Y are 4 cm from AB and are also equidistant from A and B.
- Join AX, XB, BY, and AY.
- The figure AXBY formed by connecting these points is a square.
In simple words: Draw a line 8 cm long. Then, draw the middle line that cuts it in half at 90 degrees. Also, draw two lines parallel to the first line, 4 cm above and below it. Where the middle line crosses the two parallel lines, mark points X and Y. Join A, X, B, Y. The shape AXBY is a square.

🎯 Exam Tip: The locus of points equidistant from two points is always their perpendicular bisector. The locus of points at a fixed distance from a line is always a pair of parallel lines.

Question 11.
Using ruler and compasses, construct
(i) A triangle ABC is 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) To construct triangle ABC:
- Draw a line segment BC of 3.4 cm.
- With B as the center and a radius of 5.5 cm, draw an arc.
- With C as the center and a radius of 4.9 cm, draw another arc. These two arcs will intersect at point A.
- Join A to B and A to C to form triangle ABC.
(ii) The locus of points equidistant from A and C is the perpendicular bisector of the line segment AC. Draw this bisector, and label it 'l'. This line 'l' is the required locus.
In simple words: First, draw the triangle using the three given side lengths. Then, draw a line that cuts the line AC exactly in half and is straight up-and-down from it. This line is the place where all points are equally far from A and C.

🎯 Exam Tip: For constructing a triangle with all three side lengths (SSS criterion), use compass arcs from two vertices to locate the third vertex accurately.

Question 12.
Use ruler and compasses only for this question :
(i) Construct \( \triangle ABC \), where AB = 3.5 cm, BC = 6 cm and \( \angle ABC \) = 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) To construct triangle ABC:
- Draw a line segment BC of 6 cm.
- At point B, construct an angle of 60° ( \( \angle ABC \) ). Draw a ray for this angle.
- Along this ray, mark point A such that AB is 3.5 cm.
- Join A and C to form triangle ABC.
(ii) The locus of points inside the triangle equidistant from BA and BC is the angle bisector of \( \angle ABC \). Construct this bisector.
(iii) The locus of points inside the triangle equidistant from B and C is the perpendicular bisector of the line segment BC. Construct this bisector.
(iv) Mark point P and measure PB:
- Point P is the intersection of the angle bisector of \( \angle ABC \) (from part ii) and the perpendicular bisector of BC (from part iii).
- This point P satisfies both conditions: it is equidistant from lines BA and BC, and it is also equidistant from points B and C.
- When measured, the length of PB is 3.5 cm.
In simple words: First, draw the triangle. Then, draw a line that cuts angle B in half, and another line that cuts line BC in half at a right angle. Where these two lines cross, that's point P. Then, measure how long PB is.

🎯 Exam Tip: When a point needs to satisfy multiple conditions (like being equidistant from both lines and points), you must construct the loci for each condition and find their intersection.

Question 13.
Using a ruler and compasses only with the following data :
AB = 3.5 cm, BC = 6 cm and \( \angle ABC \) = 120°
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC
(iii) Measure \( \angle BCP \).
Answer:
(i) To construct triangle ABC:
- Draw a line segment AB of 3.5 cm.
- At point B, construct an angle of 120° ( \( \angle ABC \) ). Draw a ray for this angle.
- With B as the center and a radius of 6 cm, draw an arc intersecting the 120° ray at point C.
- Join A and C to form triangle ABC.
(ii) To draw a circle with BC as diameter and find point P:
- First, find the midpoint of BC. Do this by drawing the perpendicular bisector of BC. Let the midpoint be O.
- With O as the center and radius OB (or OC), draw a circle. This circle will have BC as its diameter.
- Next, to find point P on the circumference that is equidistant from AB and BC, draw the angle bisector of \( \angle ABC \).
- The point where this angle bisector intersects the circumference of the circle is point P. This point P is equally distant from lines AB and BC.
- Join CP.
(iii) Measure \( \angle BCP \):
- When measured, \( \angle BCP \) is 90°. This is a known property because any angle inscribed in a semicircle is a right angle.
In simple words: First, draw the triangle with the given side lengths and angle. Then, draw a circle where line BC is the straight line across the middle (diameter). Next, draw a line that cuts angle B in half. Where this line crosses the circle, mark it as P. Finally, measure the angle formed by B, C, and P.

🎯 Exam Tip: Always remember that an angle subtended by a diameter at any point on the circumference of a circle is a right angle (90°).

Question 14.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and \( \angle BAC \) = 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:
First, construct triangle ABC:
- Draw a line segment AB of 5.5 cm.
- At point A, construct an angle of 105° ( \( \angle BAC \) ). Draw a ray for this angle.
- Along this ray, mark point C such that AC is 6 cm.
- Join B and C to complete triangle ABC.
(i) To construct the locus of points equidistant from BA and BC:
- This locus is the angle bisector of \( \angle ABC \). Construct this angle bisector.
(ii) To construct the locus of points equidistant from B and C:
- This locus is the perpendicular bisector of the line segment BC. Construct this perpendicular bisector.
(iii) Mark point P and measure PC:
- Point P is the intersection of the angle bisector of \( \angle ABC \) (from part i) and the perpendicular bisector of BC (from part ii).
- This point P satisfies both conditions: it is equidistant from lines BA and BC, and it is also equidistant from points B and C.
- When measured, the length of PC is 4.8 cm.
In simple words: First, draw the triangle with the given sides and angle. Then, draw a line that cuts angle B in half. Also, draw a line that cuts line BC in half at a right angle. Where these two lines cross, that's point P. Then, measure how long PC is.

🎯 Exam Tip: For constructing a 105° angle, you can combine a 60° and a 45° angle, or bisect the angle between 90° and 120° angles.

Question 15.
Use ruler and compasses only for the following questions. All construction lines and arcs must be clearly shown.
(i) Construct a \( \triangle ABC \) in which BC = 6.5 cm, \( \angle ABC \) = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are 4 cm from AB and equidistant from A and B. Measure XY.
Answer:
(i) To construct triangle ABC:
- Draw a line segment BC of 6.5 cm.
- At point B, construct an angle of 60° ( \( \angle ABC \) ). Draw a ray for this angle.
- With B as the center and a radius of 5 cm, draw an arc intersecting the 60° ray at point A.
- Join A and C to form triangle ABC.
(ii) To construct the locus of points at a distance of 3.5 cm from A:
- With A as the center and a radius of 3.5 cm, draw a circle. This circle is the required locus.
(iii) To construct the locus of points equidistant from AC and BC:
- Draw the angle bisector of \( \angle ACB \). To do this, draw an arc from C to cut AC and BC at two points (say D and E). Then, with the same radius, draw arcs from D and E to intersect at a point (say H). Join C and H. The line CH is the angle bisector of \( \angle ACB \) and is the required locus.
(iv) To mark 2 points X and Y and measure XY:
- The points X and Y are located at the intersection of the circle (from part ii, which is the locus of points 3.5 cm from A) and the angle bisector CH (from part iii, which is the locus of points equidistant from AC and BC).
- When measured, the length of XY is 5 cm.
In simple words: First, draw the triangle. Then, draw a circle around point A with a radius of 3.5 cm. Next, draw a line that splits angle C in half. The places where this circle and the angle-splitting line cross are points X and Y. Finally, measure how long the line XY is.

🎯 Exam Tip: When a question has multiple sub-parts for construction, ensure each step is distinct and clearly executed, and that all necessary loci are properly identified and drawn before finding intersection points.