Students often turn to S Chand Class 10 Maths Solutions ICSE Chapter 13 Loci Ex 13(c) to clarify doubts and improve problem-solving skills.
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.
Solution:
(i) Steps of construction :
(a) Draw a line AB and take a point C out side it.
(b) From C, draw an arc with suitable radius which intersects. AB at E and F.
(c) With centres E and F, draw arcs intersecting each other at G.
(d) Join CG meeting AB at N.
Then CN ⊥ AB.
(e) Join CA.
(ii) Draw the angle bisector l ∠CAB which intersects the perpendicular CN at P.
P is the required point equidstant from AB and AC.
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.
Solution:
Steps of construction :
(i) Draw a line segment GH = 2.5 units (cm).
(ii) Draw its perpendicular bisector l and cut off XF = 3 units (cm).
(iii) Join FG and FH.
On measuring FG, it is 3.4 units (cm).
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.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 2.5 units.
(ii) At B, draw a ray making an angle of 60° and at C, a ray making an angle of 90°.
(iii) Cut off BA = 2 units.
(iv) Join AC and draw its perpendicular bisector which meets CY at D.
(v) Join AD.
(vi) Join BD, on measuring it, it is 2.8 units. Then ABCD is the required quadrilateral.
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.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 1.8 units.
(ii) At B, draw a ray by making an angle of 90° and cut off BC = 2.2 units.
∵ ∠D = 60° and DA = DC
△ADC is an equilateral triangle.
(iii) With centre A and C and radius AC, draw arcs intersecting each other at D.
(iv) Join AD and CD.
(v) Draw the perpendicular bisector of BC which intersects AD at X.
(vi) Join BX.
On measuring, BX = 2.6 units.
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.
Solution:
Steps of Construction :
In △ABC, ∠B = 56°, ∠A = 38°
∴ ∠C = 180° – (∠B + ∠A)
= 180° – (56° + 38°)
= 180° – 94° = 86°
(i) Draw a line segment BC = 5.3 cm.
(ii) At B, draw a ray making an angle of 56° and at C, a ray making an angle of 86°. Which meet eachother at A.
(iii) Now draw the perpendicular bisectors of sides BC and CA which intersect eachother at O.
(iv) With centre O and radius OA or OB or OC, draw a circle which passes through A, B and C.
This is the required circle.
Question 6.
Draw a triangle with sides 3 units, 3\(\frac { 1 }{ 2 }\) units, and 4 units, without using setsquare 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.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3\(\frac { 1 }{ 2 }\) units.
(ii) With centre B and radius 3 units and with centre C and radius 4 units draw arcs intersecting each other at A.
(iii) Join AB and AC.
(iv) Draw the angle bisector of ∠B and ∠C intersecting each other at X.
(v) From X, draw a perpendicular XL on the longest side AC.
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.
Solution:
Steps of construetion :
(i) Draw a line segment BC = 2 units.
(ii) Draw a ray BX at B making an angle of 120° and cut off BA =1.6 units.
(iii) Join AC.
(iv) At A, draw a ray AY making an angle of 90°.
(v) Draw the perpendicular bisector of BC which intersects AY at P.
P is the required point and join PB and PC on measuring PB, it is 2.5 units.
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.)
Solution:
Steps of construction :
(i) Draw a line segment AB such that AB =BC = 6 cm.
(ii) Draw a ray AX at A making an angle of 42° and cut off AD = 5 cm.
(iii) Join BD and draw its perpendicular bisector n.
(iv) Draw parallel lines l and m parallel to ABC at a distance of 2.5 cm from ABC. Which intersect the perpendicular bisector n at P and Q.
Then P and Q are the required points.
Now Join AP and AQ.
On measuring AP = 6.5 cm and AQ = 2.7 cm (approx)
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.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BC = 5 cm.
(iii) Draw the angle bisector BY of ∠B.
(iv) With centre A and radius 5 cm. draw an arc which intersects BY at D.
(v) Join AD and CD.
ABCD is the required quadrilateral.
On measuring CD, it is 5.3 cm.
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?
Solution:
Steps of constuction :
(i) AB and CD are two intersecting lines making an angle of 30° at O.
(ii) Draw the angle bisector P of ∠BOD and produce it to other side also and also the bisector m ∠AOD.
(iii) With centre O and radius 2 cm, mark points P and Q on l. Similarly mark the points R and S on bisector m.
Then P, Q, R and S are four required points.
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.
Solution:
Steps of construction :
(i) Draw a line segment AC = 5 cm.
(ii) With centres A, and C and radius 4 cm, draw arcs intersecting each other at B and D.
(iii) Join AB, BC, CD and AD. ABCD is the required rhombus on measuring ∠ABC, it is 78°.
(iv) Draw the perpendicular bisector of BC which intersects AD at R.
Now AR = 1.2 cm.
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.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) At A, draw a ray making 45° and cut off AD = 6 cm.
(iii) With centre B and radius 3.6 cm and with centre D and radius 5 cm, draw arcs intersecting each other at C.
(iv) Join BC and CD.
ABCD is the required quadrilateral.
(v) On measuring ∠BCD, it is 65°.
(vi) Join BD.
(vii) Draw the bisector of ∠BCD which intersects BD at P.
P is the required point.
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.
Solution:
Steps of construction :
(i) Draw a line segment QR = 4 cm.
(ii) Draw a semicircle on QR as diameter.
(iii) Draw the perpendicular bisector of QR meeting the semicircle at P.
(iv) Join PQ and PR.
△PQR is the required isosceles triangle.
(v) Now draw the bisector of ∠Q which meets the perpendicular bisector of QR at X.
X is the required point which is equidistant from QR and PR and also from Q and R.
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.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.5 cm.
(ii) At B, draw a ray making an angle of 35° and cut off BX = 3 cm.
(iii) As X is equidistant from AB and BC.
∴ X will be on the angle bisector of ∠ABC.
(iv) Now draw BC such that ∠XBC = ∠ABX = 35°.
(v) Cut off BC = AB.
(vi) Now with centre C and A, and radius AB draw arcs intersecting each other at D.
(vii) Join CD and AD.
Then ABCD is the required rhombus.
(viii) Draw the perpendicular bisector of AB which meets BX at Y.
Y is the required point which is equidistant from A and B.
Question 15.
Construct ∠AOB = 75°, Mark a point P equidistant from OA and OB such that its distance from another given line CD is 3 cm.
Solution:
Steps of construction :
(i) Draw an angle AOB = 75° and a line CD.
(ii) Draw the angle bisector OE of ∠AOB.
(iii) Take a point L and draw a perpendicular and cut off LM = 3 cm:
(iv) Through M, draw a line P parallel to CD which intersects OE at P.
P is the required point.
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.
Solution:
Steps of construction :
(i) Draw a line segment CB = 5 cm.
(ii) At B, draw an angle equal to 45° and cut off BP = 4 cm.
(iii) Join PC.
Then BCP is the required triangle.
(iv) At B draw an angle PBA such that ∠PBA = ∠PBC = 45°.
(v) From C draw a perpendicular CX.
(vi) From P, draw a perpendicular on CX intersecting it at E.
(vii) Cut off ED = CE.
(viii) With centre D and radius BC and with centre B and riadius CD, draw arcs which intersect each other at A.
(ix) Join AB and AD.
ABCD is the required rectangle.
On measuring AB, it is 5.5 cm.
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 insiur ate 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.
Solution:
Steps of construction :
(i) Draw a line segment BC = 8 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BA = 5 cm.
(iii) Join AC.
(iv) Draw the angle bisector BY of ∠ABC.
(v) Draw the perpendicular bisector of BC which intersects BY at P.
P is the required point which is equidistant from BA and BC and also from B and C.
On measuring BP, it is 4.6 cm.
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 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 △QBC equal in area to △ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw an angle of 60° and cut off BA = 9 cm.
(iii) Join AC.
Then △ABC is the required triangle.
(iv) From A, draw a perpendicular AL on BC.
(v) Through A, draw a line XY parallel to BC. Then XY is the required locus of vertices of △ABC.
(vi) Draw the perpendicular bisector of BC which meets XY at Q.
Q is the required locus.
Join QB and QC.
Then Q is the point such that △QBC is an isosceles triangle and area (△QBC = area (△ABC)
On measuring CQ, it is 8.4 cm.
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.
Solution:
Steps of consturction :
(i) ABCD is a rhombus in which AB = BC = CD = DA and AC is its one diagonal
∵ The locus of point is equidistant from AB and AD
∴ It will be on the bisector of ∠DAB which is AC
∵ The locus is equidistant from A and C
∴ It lies on the perpendicular bisector which is diagonal BD
∵ Diagonal AC and BC bisect each other at O
∴ O is the required locus of point
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.
Solution:
Steps of construction:
(i) Plot the points A (1,1),B (5,3) and C (2,7) on the graph and join them to form a triangle ABC.
(ii) Draw the perpendicular bisector of AB.
(iii) Draw the angle bisector of ∠A intersecting the perpendicular bisector at P.
P is the required point which is equidistant from A and B and also AB and AC On measuring PA, it is 2.5 cm.
Question 6.
Construct △ABC, 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.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Solution:
Steps of construction :
(i) Draw a line segment BC = 8 cm.
(ii) Draw a ray BX at B making an angle of 60° and cut off BA = 7 cm.
(iii) Join CA.
△ABC is the required triangle.
(iv) Draw the perpendicular bisector l of BC.
(v) Draw the angle bisector m of ∠ABC which intersects l at P.
Then P is the required point which is equidistant from B and C as well as from BA and BC.
On measuring the length of PB, it is 4.5 cm.
Question 7.
Construct an isosceles triangle ABC, such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠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.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) With centre A and B and radius 4 cm, draw arcs intersecting each other at C.
(iii) Join AC and BC.
(iv) Draw the angle bisector m of ∠C and take a point P on it such that CP = 5 cm.
(v) Draw a line l parallel to AB at a distance of 5 cm.
(vi) With centre P and radius 5 cm, draw arcs intersecting the line l at Q and R.
Then points Q and R are the required points.
Question 8.
Using only a ruler and compasses, construct ∠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
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray BX making an angle of 120° and cut off BA = 5 cm.
(iii) Join AC.
(iv) Draw the angle bisector BY of ∠B and perpendicular bisector of BC intersecting each other at D.
(v) Join DA and DC.
(vi) Take one more point E on BY.
(vii) (vii) Join EA and EC.
Now (i) Figure ABCD is a quadrilateral,
(ii) ABD is a triangle,
(iii) ABE is a triangle.
Question 9.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
Complete the rectangle ABCD such that
(i) P is equidistant from AB and AC.
(ii) P is equidistant from C and D.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BP = 4 cm.
(iii) Join PC.
(iv) Now draw perpendicular CY at C.
(v) From P, draw perpendicular l meeting CY at E.
(vi) Cut off ED = CE.
(vii) Draw perpendicular at B and cut off BA = CD.
(viii) Join AD.
Then ABCD is the required rectangle in which P is equidistant from BC and BA and also from C and D.
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.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.
(ii) Draw its perpendicular bisector l which bisects AB at O.
(iii) From O, cut off OX = OY = 4 cm.
(iv) Join AX, XB, BY and AY.
Points X and Y are the required points which are equidistant from A and B and at a distance of 4 cm from the line AB.
Since OA = OB = OX = OY = 4 cm or AB = XY
∴ AXBY is a square in shape.
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.
Solution:
Steps of construction :
(i) Draw BC = 3.4 cm and mark the arcs of AB = 5.5 cm and CA =4.9 cm from B and C respectivoly intersecting each other at A.
(ii) Join A, B and C. ABC is the required triangle.
(iii) Draw ⊥ bisector l of AC.
(iv) l is the required locus.
Question 12.
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.
Solution:
Steps of construction :
1. Draw a line BC = 6 cm and an angle CBX = 60°. Cut off AB = 3.5 cm. Join AC, △ABC is the required triangle.
2. Draw ⊥ bisector of BC and bisector of ∠B.
3. Bisector of ∠B meets bisector of BC at P
∴ BP is the required length, where PB = 3.5 cm
4. P is the point which is equidistant from BA and BC, also equidistant from B and C.
Question 13.
Using a ruler and compasses only :
(i) Construct a triangle ABC with the following data :
AB = 3.5 cm, BC = 6 cm and ∠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 ∠BCP.
Solution:
Steps of construction :
(i) Draw AB = 3.5.
(ii) At B, draw ∠ABX = 120°.
(iii) With B as center draw an arc of radii 6 cm at C.
(iv) Join A and C.
(v) Draw the perpendicular bisector of line BC and draw a circle with BC as diameter.
(vi) Draw angle bisector of ∠B.
Meets the circle at P
∴ P is the required point ∠BCP = 90°
(vii) Join CP.
Question 14.
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.
Solution:
Steps of construction :
Construct the triangle ABC with AB = 5.5 cm
∠BAC = 105° and AC = 6 cm
(i) Points which are equidistant from BA and BC lies on the bisector of ∠ABC.
(ii) Points equidistant from B and C lies on the perpendicular bisector of BC. Draw perpendicular bisector of BC. The required point P is the point of intersection of the bisector of ∠ABC and the perpendicular bisector of BC.
(iii) Required length of PC = 4.8 cm.
Question 15.
Use ruler and compasses only for the following questions. All construction lines and arcs must be clearly shown.
(i) Construet a △ABC in which BC = 6.5 cm, ∠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 at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
(i) Steps of construction :
(1) Draw BC = 6.5 cm using a ruler.
(2) With B as the centre and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
(3) With Q as the centre, and same radius, cut the previous arc at P.
(4) Join PB and extend it.
(5) With B as the centre and radius 5 cm, draw an arc that cuts the arc PB to obtain point A.
(6) Join AC to obtain △ABC.
(ii) Steps of construction :
(1) With A as the centre and radius 3.5 cm, draw a circle.
(2) The circle is the required locus of points at a distance of 3.5 cm from A.
(iii) Steps of construction :
(1) With C as the centre and with radius of a length less than CA or BC draw an arc to cut the line segments AC and BC at D and E respectively.
(2) With the same radius or a suitable and with D as the centre, draw an arc of a circle.
(3) With the same radius and with E as the centre draw an arc such that the two arcs intersect at H
(4) Join C and H.
(5) CH is the bisector of ∠ACB and is the locus of the points equidistant from AC and BC
(iv) Steps for construction :
(1) We know that the points at a distance of 3.5 cm from A will surely lie on the circle with centre A.
(2) Also, the points on the angle bisector CH are the points equidistant from AC and BC.
(3) Mark X and Y which are at the intersection of the circle and the angle bisector CH.
(4) Measure XY.
XY = 5 cm.