Selina Solutions for Class 6, maths Chapter 29

Class 6: Maths Chapter 29 solutions. Complete Class 6 Maths Chapter 29 Notes.

Contents

1 Selina Class 6 ICSE Solutions Mathematics : Chapter 29- The Circle
- 1.1 Exercise 29(A)
- 1.2 Exercise 29(B)
2 Download PDF
3 Chapterwise Selina Publishers ICSE Solutions for Class 6 Mathematics :
4 About Selina Publishers ICSE

Selina Class 6 ICSE Solutions Mathematics : Chapter 29- The Circle

Selina 6th Maths Chapter 29, Class 6 Maths Chapter 29 solutions

Exercise 29(A)

1. Use the figure given below to fill in the blanks:

(i) R is the …….. of the circle.

(ii) Diameter of a circle is …………

(iii) Tangent to a circle is …………

(iv) EF is a …….. of the circle

(v) ………. is a chord of the circle.

(vi) Diameter = 2 ×……….

(vii) …… is a radius of the circle.

(viii) If the length of RS is 5 cm, the length of PQ = ………

(ix) If PQ is 8 cm long, the length of RS = ……….

(x) AB is a …… of the circle

Solution:

(i) R is the centre of the circle

(ii) Diameter of a circle is PQ

(iii) Tangent to a circle is AB

(iv) EF is a secant of the circle

(v) CD is a chord of the circle

(vi) Diameter = 2 × radius

(vii) RS is a radius of the circle

(viii) If the length of RS is 5 cm, the length of PQ = 10 cm

(ix) If PQ is 8 cm long, the length of RS = 4 cm

(x) AB is a tangent of the circle

2. Draw a circle of radius 4.2 cm. Mark its centre as O. Take a point A on the circumference of the circle. Join AO and extend it till it meets point B on the circumference of the circle,

(i) Measure the length of AB.

(ii) Assign a special name to AB.

Solution:

(i) By measurement the length of AB = 8.4 cm

(ii) AB is diameter of the circle

3. Draw circle with diameter:

(i) 6 cm

(ii) 8.4 cm

In each case, measure the length of the radius of the circle drawn.

Solution:

(i) AB is the diameter of circle

AB = 6 cm and

OA is the radius of the circle

Selina Solutions Concise Mathematics Class 6 Chapter 29 - 3

The radius of the circle is,

1 / 2 × 6 = 3 cm

Therefore, OA = OB = 3 cm

(ii) AB is the diameter of circle

AB = 8.4 cm

Selina Solutions Concise Mathematics Class 6 Chapter 29 - 4

OA is the radius of circle

The radius of circle is,

1 / 2 × 8.4 = 4.2 cm

Therefore, OA = OB = 4.2 cm

4. Draw a circle of radius 6 cm. In the circle, draw a chord AB = 6 cm.

(i) If O is the centre of the circle, join OA and OB.

(ii) Assign a special name to △AOB

(iii) Write the measure of angle AOB.

Solution:

(ii) △AOB is an equilateral triangle

(iii) Since, △AOB is equilateral triangle

Hence, ∠AOB = 60⁰

5. Draw a circle of radius 4.8 cm and mark its centre as P.

(i) Draw radii PA and PB such that ∠APB = 45⁰

(ii) Shade the major sector of the circle

Solution:

(i) PA is the radius of circle

PA = 4.8 cm

(ii) ∠APB = 45⁰. P is the centre of the circle and PA and PB are radii of circle

(iii) Major sector of circle is shaded in the given figure

6. Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade the minor segment of the circle.

Solution:

(i) OP is the radius of the circle i.e OP = 3.6 cm

AB is the chord of the circle i.e AB = 5 cm

(ii) Minor segment of the circle is shaded in the given circle

7. Mark two points A and B, 4 cm a part. Draw a circle passing through B and with A as a centre

Solution:

In the given figure A is the centre of the circle and

AB is the radius of the circle i.e AB = 4 cm

8. Draw a line AB = 8.4 cm. Now draw a circle with AB as diameter. Mark a point C on the circumference of the circle. Measure angle ACB.

Solution:

By measurement we get,

∠ACB = 90⁰

Exercise 29(B)

1. Construct a triangle ABC with AB = 4.2 cm, BC = 6 cm and AC = 5 cm. Construct the circumcircle of the triangle drawn.

Solution:

Steps of Construction:

(i) Draw a line segment AB = 4.2 cm

(ii) Now, taking B as centre, draw an arc of length 6 cm from B using compass

(iii) Again taking A as centre, draw another arc bisecting the previous arc marked by B at point C such that AC = 5 cm

(iv) Join C to points A and B. The required triangle ABC is obtained

(v) Draw the perpendicular bisector of any two sides of the triangle. Let these intersects at point O.

(vi) Taking O as centre and OA or OB or OC as radius, draw a circle with the help of compass

This circle will pass through the vertices A, B and C

2. Construct a triangle PQR with QR = 5.5 cm, ∠Q = 60⁰ and angle R = 45⁰. Construct the circumcircle of the triangle PQR.

Solution:

Steps of Construction:

(i) Draw a △PQR in which QR = 5.5 cm.

(ii) With the help of compass, draw ∠Q = 60⁰ and ∠R = 45⁰

(iii) Both ∠Q and ∠R intersect at point P, thus forming a △PQR

(iv) Using compass, draw a perpendicular bisector of PR and QR which intersects at point O

(v) Now, taking O as centre and OP or OQ or OR as radius draw a circle with the help of compass

(vi) This circle will pass through the vertices P, Q and R

3. Construct a triangle ABC with AB = 5 cm, ∠B = 60⁰ and BC = 6.4 cm. Draw the incircle of the triangle ABC.

Solution:

Steps of Construction:

(i) Draw a line AB of length 5 cm

(ii) Taking B as centre, draw an angle B = 60⁰ using compass.

(iii) With the help of compass, draw an arc BC = 6.4 cm from point B as centre

(iv) Join the points A and C such that it forms a △ABC

(v) Now, from A and B cut the bisector of ∠A and ∠B which intersects each other at point D.

(vi) Taking D as centre, draw an incircle with the help of compass which touches all the three sides of △ABC

4. Construct a triangle XYZ in which XY = YZ = 4.5 cm and ZX = 5.4 cm. Draw the circumcircle of the triangle and measure its circumradius.

Solution:

Steps of Construction:

(i) Draw a line XY = 4.5 cm

(ii) With the help of compass, draw an arc of 4. 5 cm from Y as centre

(iii) Again with the help of compass, draw another arc of 5.4 cm from X as centre which intersects the previous arc made by Y

(iv) Now, mark the intersection as point Z and join it to both X and Y. The required triangle XYZ is obtained.

(v) Draw the bisectors of XZ and YZ which meet at point O

(vi) Taking O as centre and radius OX or OY or OZ draw a circle using compass

(vii) This circle will pass through the vertices X, Y and Z

5. Construct a triangle PQR in which, PQ = QR = RP = 5.7 cm. Draw the incircle of the triangle and measure its radius.

Solution:

Steps of Construction:

(i) Draw a line segment PQ = 5.7 cm

(ii) Taking Q as centre, draw an arc 5.7 cm from Q with the help of compass

(iii) Again with the help of compass, draw another arc of 5.7 cm from P as centre which intersects the previous arc made by Q

(iv) Mark the intersection point as R and joint it to both P and Q respectively

(v) Thus the required triangle RPQ is obtained

(vi) Now, from P and Q cut the bisector of ∠P and ∠Q, which intersect each other at point O

(vii) Taking P as centre, draw an incircle which touches all the three sides of △RPQ

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Selina Class 6 ICSE Solutions Mathematics : Chapter 29- The Circle

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