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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 19 - Constructions (Circles)

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Selina Textbook Solutions Chapter 19 - Constructions (Circles)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 19 - Constructions (Circles).

All Selina textbook questions of Chapter 19 - Constructions (Circles) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam. 

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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 19 - Constructions (Circles) Page/Excercise 19

Solution 1

Steps Of Construction:

i) Draw a circle with centre O and radius 3 cm.

ii) From O, take a point P such that OP = 5 cm

iii) Draw a bisector of OP which intersects OP at M.

iv) With centre M, and radius OM, draw a circle which intersects the given circle at A and B.

v) Join AP and BP.

AP and BP are the required tangents.

On measuring AP = BP = 4 cm

Solution 2

  1. Draw a circle of diameter 9 cm, taking O as the centre.
  2. Mark a point P outside the circle, such that PO = 7.5 cm.
  3. Taking OP as the diameter, draw a circle such that it cuts the earlier circle at A and B.
  4. Join PA and PB.
Thus, PA and PB are required tangents. PA = PB = 6 cm

Solution 3

Steps of Construction:

i) Draw a circle with centre O and radius BC = 5 cm

ii) Draw arcs making an angle of 180º - 45º = 135º at O such that AOB = 135º

iii) AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.

iv) AP and BP are the required tangents which make an angle of 45º with each other at P.

Solution 4

Steps of Construction:

i) Draw a circle with centre O and radius BC = 4.5 cm

ii) Draw arcs making an angle of 180º - 60º = 120º at O such that AOB = 120º

iii) AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.

iv) AP and BP are the required tangents which make an angle of 60º with each other at P.

Solution 5

Steps of construction:

i) Draw a line segment BC = 4.5 cm

ii) With centers B and C, draw two arcs of radius 4.5 cm which intersect each other at A.

iii) Join AC and AB.

iv) Draw perpendicular bisectors of AC and BC intersecting each other at O.

v) With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.

This is the required circumcircle of triangle ABC.

On measuring the radius OA = 2.6 cm

Solution 6

Steps of Construction:

i) Construction of triangle:

a) Draw a line segment BC = 7 cm

b) At B, draw a ray BX making an angle of 45o and cut off BE = AB - AC = 1 cm

c) Join EC and draw the perpendicular bisector of EC intersecting BX at A.

d) Join AC.

is the required triangle.

ii) Construction of incircle:

e) Draw angle bisectors of and intersecting each other at O.

f) From O, draw perpendiculars OL to BC.

g) O as centre and OL as radius draw circle which touches the sides of the . This is the required in-circle of .

On measuring, radius OL = 1.8 cm

Solution 7

Steps of Construction:

i) Draw a line segment BC = 5 cm

ii) With centers B and C, draw two arcs of 5 cm radius each which intersect each other at A.

iii) Join AB and AC.

iv) Draw angle bisectors of and intersecting each other at O.

v) From O, draw .

vi)Now with centre O and radius OL, draw a circle which will touch the sides of

On measuring, OL = 1.4 cm

Solution 8

Steps of construction:

i) Draw a line segment AB = 6 cm

ii) At A, draw a ray making an angle of 60o with BC.

iii) With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.

iv) Join BC.

is the required triangle.

v) Draw the perpendicular bisectors of AB and AC intersecting each other at O.

vi) With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.

vii) From O, draw .

Proof: In right and

OA = OB (radii of same circle)

Side OD = OD (common)

Solution 9

Steps of Construction:

i)Draw a line segment BC = 4 cm.

ii) At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.

iii) From E, draw another perpendicular line EY.

iv) From C, draw a ray making an angle of 45o with CB, which intersects EY at A.

v) Join AB.

is the required triangle.

vi) Draw perpendicular bisectors of sides AB and BC intersecting each other at O.

vii) With centre O, and radius OB, draw a circle which will pass through A, B and C.

Measuring the radius OB = OC = OA = 2 cm

Solution 10

i) O is called the circumcentre of circumcircle of .

ii) OA, OB and OC are the radii of the circumcircle.

iii) Yes, the perpendicular bisector of BC will pass through O.

Solution 11

i) O is called the incentre of the incircle of .

ii) OR and OQ are the radii of the incircle and OR = OQ.

iii) OC is the bisector of angle C

Solution 12

Steps of Construction:

i) Draw a line segment BC = 6 cm.

ii) With centre B and radius 8 cm draw an arc.

iii) With centre C and radius 5 cm draw another arc which intersects the first arc at A.

iv) Join AB and AC.

is the required triangle.

v) Draw the angle bisectors of intersecting each other at I. Then I is the incentre of the triangle ABC

vi) Through I, draw

vii) Now from D, cut off

viii) With centre I, and radius IP or IQ, draw a circle which will intersect each side of triangle ABC cutting chords of 2 cm each.

Solution 13

Steps of construction:

i) Draw a line segment BC = 6 cm

ii) With centers B and C, draw two arcs of radius 6 cm which intersect each other at A.

iii) Join AC and AB.

iv) Draw perpendicular bisectors of AC, AB and BC intersecting each other at O.

v) With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.

This is the required circumcircle of triangle ABC.

Solution 14

Steps of Construction:

i) Draw a line segment BC = 5.6 cm

ii) With centers B and C, draw two arcs of 5.6 cm radius each which intersect each other at A.

iii) Join AB and AC.

iv) Draw angle bisectors of and intersecting each other at O.

v) From O, draw .

vi)Now with centre O and radius OL, draw a circle which will touch the sides of

This is the required circle.

Solution 15

Steps of Construction:

i) Draw a regular hexagon ABCDEF with each side equal to 5 cm and each interior angle 120º.

ii) Join its diagonals AD, BE and CF intersecting each other at O.

iii) With centre as O and radius OA, draw a circle which will pass through the vertices A, B, C, D, E and F.

This is the required circumcircle.

Solution 16

Steps of Construction:

i) Draw a line segment AB = 5.8 cm

ii) At A and B, draw rays making an angle of 120o each and cut off AF = BC = 5.8 cm

iii) Again F and C, draw rays making an angle of 120o each and cut off FE = CD = 5.8 cm.

iv) Join DE. Then ABCDEF is the regular hexagon.

v) Draw the bisectors of intersecting each other at O.

vi) From O, draw

vii) With centre O and radius OL, draw a circle which touches the sides of the hexagon.

    This is the required in circle of the hexagon.

Solution 17

Steps of Construction:

(i) Draw a circle of radius 4 cm with centre O

(ii) Since the interior angle of regular hexagon is 60o, draw radii OA and OB such that

(iii) Cut off arcs BC, CD, EF and each equal to arc AB on given circle

(iv) Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.

The circle is the required circum circle, circumscribing the hexagon.

Solution 18

Steps of Construction:

i) Draw a line segment OP = 6 cm

ii) With centre O and radius 3.5 cm, draw a circle

iii) Draw the midpoint of OP

iv) With centre M and diameter OP, draw a circle which intersect the circle at T and S

v) Join PT and PS.

PT and PS are the required tangents. On measuring the length of PT = PS = 4.8 cm

Solution 19

i.

  1. Draw a line BC = 5.4 cm.
  2. Draw AB = 6 cm, such that mABC = 120°.
  3. Construct the perpendicular bisectors of AB and BC, such that they intersect at O.
  4. Draw a circle with O as the radius.

ii.

(e) Extend the perpendicular bisector of BC, such that

it intersects the circle at D.

(f) Join BD and CD.

(g) Here BD = DC.

Solution 20

 

  

Solution 21

 

  

Solution 22

 

  

Solution 23

Steps of construction:

 

 

  1. Draw AF measuring 5 cm using a ruler. 
  2. With A as the centre and radius equal to AF, draw an arc above AF.
  3. With F as the centre, and same radius cut the previous arc at Z
  4. With Z as the centre, and same radius draw a circle passing through A and F.
  5. With A as the centre and same radius, draw an arc to cut the circle above AF at B.
  6. With B as the centre and same radius, draw an arc to cut the circle at C.
  7. Repeat this process to get remaining vertices of the hexagon at D and E.
  8. Join consecutive arcs on the circle to form the hexagon.
  9. Draw the perpendicular bisectors of AF, FE and DE.
  10. Extend the bisectors of AF, FE and DE to meet CD, BC and AB at X, L and O respectively.
  11. Join AD, CF and EB.

 

These are the 6 lines of symmetry of the regular hexagon.

  

Solution 24

Steps for construction:

  1. Draw AB = 5 cm using a ruler.
  2. With A as the centre cut an arc of 3 cm on AB to obtain C.
  3. With A as the centre and radius 2.5 cm, draw an arc above AB.
  4. With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
  5. With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed.
  6. Join OB.
  7. Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
  8. With the M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
  9. Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.

 

 

  

 

 

QB = PB = 3 cm

That is, length of each tangent is 3 cm.  

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 19 - Constructions (Circles)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

Text Book Solutions

ICSE X - Mathematics

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