RD Sharma Solutions Class 9 Maths Chapter 16 – Circles (Updated for 2021-22)

RD Sharma Solutions Class 9 Maths Chapter 16 – Circles: RD Sharma Solutions Class 9 Maths Chapter 16 enable them also to study all topics more effectively & clear their doubts quickly. The students who want to obtain higher marks in the exams must regularly practice mathematics subject exercises with the help of RD Sharma Solutions Class 9 Maths chapter 16 Circles. This also builds confidence of the students and helps them to revise each important topic during the exam.

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RD Sharma Class 9 Solutions Chapter 16

Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 16 Circles

RD Sharma class 9 chapter 16 exercise 16a

RD Sharma class 9 chapter 16 exercise 16b

RD Sharma class 9 chapter 16 exercise 16c

RD Sharma class 9 chapter 16 exercise 16d

Access answers of RD Sharma Solutions Class 9 Maths Chapter 16

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length 12 (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = 12 (AB).

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length 34 (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A₁, A₂, A₃ and A₄.
(iii) JoinA₄B.
(iv) From A3, draw a line parallel to A₄B which meets AB at C.
C is the required point and AC = 34 (AB).

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = 12 x 108°
∴ We shall bisect it.

(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =12 x 90° = 45°.

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ 12 ∠DCA + 12 ∠DCB = 180° x 12 = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x 12 = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + 12 x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.

(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 12°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.

(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.

(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.

(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.

(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.

(vi) 22 12°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 12°

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm

(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise

• RD Sharma class 9 chapter 16 exercise 16a: This exercise is based on the topics related to the position of a point with respect to a circle, Concentric Circles, Length of an arc, Circular Disc, Minor & Major Arc, & many more. 
• RD Sharma class 9 chapter 16 exercise 16b: This exercise includes topics related to the congruence of circles & arcs. Practicing RD Sharma class 9 chapter 16 exercise 16b questions enables the students to build an in-depth understanding of the basics of a circle. 
• RD Sharma class 9 chapter 16 exercise 16c: This exercise includes topics based on a few important results on equal chords. The solutions of RD Sharma class 9 chapter 16 exercise 16c are helpful to solve several circle problems. The students will study more circles concepts in detail in order to obtain better marks in the exam.
• RD Sharma class 9 chapter 16 exercise 16d: This exercise includes topics based on arcs & angles subtended by them. Each answer is created by subject experts that assist the students to understand this concept in a better and fun way. The students can practice RD Sharma class 9 chapter 16 exercise 16d & clear their doubts to secure excellent marks in the final mathematics exam.

Important Topics in the Exercise

RD Sharma Solutions Class 9 Maths Chapter 16 Circles include some important topics that are listed below:

• Introduction of Circle
• Position of a point with respect to a circle
• Circular Disc
• Concentric Circles
• The degree measure of an arc
• Chord and segment of a circle
• Congruence of circles and arcs

This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 16. If you have any doubts regarding the CBSE Class 9 Maths exam, ask in the comments.

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