# Maharashtra Board Class 9 Math Part 2 Chapter 6 Circle Solution

## Maharashtra Board Class 9 Math Part 2 Solution Chapter 6 – Circle

Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 6: Circle. Marathi or English Medium Students of Class 9 get here Circle full Exercise Solution.

 Std Maharashtra Class 9 Subject Math Part 2 Solution Chapter Circle

### Practice Set 6.1

(1) Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of the circle.

Solution: (1) We know, perpendicular from centre bisects the chord.

∴ If OD is distance drawn from centre, then

AD = DB = ½ AB = ½ × 12 = 6 cm

In △ODB,

OD2 + DB2 = OB2

=> OB = √82 + 62

= √100

= 10cm

∴ Diameter = 2×10 = 20cm.

(2) Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.

Solution: Diameter = 26 cm

Radius = 26/2 = 13 cm

∴ OB = 13 cm

And AB = 34 cm

∴ AD = DB = 24/2 = 12 cm (Perpendicular from centre bisects the chord)

In △ODB,

OD = √OB2 – DB2

= √132 – 122

= √169 – 144

= √25

= 5 cm

(3) Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord

Solution: In △ODB

OB2 = OD2 + DB2

= DB = √OB2 – OD2

= √342 – 302

= √256

= 16 cm

∴ AB = 2DB = 2×16 = 32 cm

(4) Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.

Solution: PQ = 80 units = RQ = 40 units

Radius = OQ = 41 units

In △ORQ,

OQ2 = RO2 + RQ2

=> RO = √412 – 402

= 9 cm

(5) In figure 6.9, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ

Solution: Let, OR be perpendicular distance from the centre to the chord.

∴ In smaller circle, PR = RQ — (I)

And in large circle

AR = RB

= AR + PR = RQ + BQ

= AP + RQ = RQ + BQ (By (i))

= AP = BQ

(6) Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.

Solution: Let, LM be the diameter that bisects two chords AB and PQ

∴ LM is perpendicular to both AB and PQ

∴ ∠EDB = ∠DEP = 90°

But, these are alternate angles

∴ AB || PQ

### Practice Set – 6.2

(1) Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle?

Solution:

∵ The two chord are congruent.

∴ They are equidistant from the centre

∵ Radius = 10 cm and chord length = 16 cm

∴ By Pythagoras triplet,

Distance from centre = √102 – (16/2)2

=√100 – 64

= √36

= 6 cm.

(2) In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of the chords.

Solution:

Radius = 13 cm and distance = 5 cm

∵ The chords are equal, they are equidistant from the entre.

∴ Length of chords = 2 √132 – 52

= 2 √169 – 25

= 2 √144

= 2 √12×12

= 2 x 12

= 24 cm.

(3) Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of NPM.

Solution: We draw perpendicular CA and CB on PM and PN.

∵ PM = PN, ∴ AC = BC

And AM = AP & BN = PB

In △ACP & △BCP,

AC = BC, ∠CAP = ∠CBP = 90° and AP = BP (∵ PM = PN)

∴ △ACP ≅ △BCP (S-A-S test)

∴ ∠APC = ∠BPC

∴ PC bisects ∠NPM

### Practice Set – 6.3

(1)

Construct ABC such that B =100°, BC = 6.4 cm, C = 50° and construct it’s in circle.

Solution: Steps:

(1) Draw a triangle with given measures.

(2) Construct bisect of ∠B and ∠C. Let, they meet at I.

(3) Draw perpendicular IM on BC

(4) Taking I as centre & IM as radius, we draw a circle.

(2) Construct PQR such that P = 70°, R = 50°, QR = 7.3 cm. and construct its circum circle.

Solution: Steps:

(1) Draw △PQR of given measures.

(2) We draw perpendicular bisects of birds PR & QR

(3) The perpendicular at O. Join OP.

(4) With O as centre and OP as radius, draw a circle.

(3) Construct XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct it’s in circle.

Solution: Steps:

(1) Construct a triangle XYZ of given measures.

(2) Draw bisects of ∠Y and ∠Z which meet at I

(3) Draw IM ⊥ YZ

(4) With I as centre & IM as radius, draw a circle.

(4)

In LMN, LM = 7.2 cm, M = 105°, MN = 6.4 cm, then draw LMN and construct its circum circle.

Solution: Steps:

(1) Draw triangle LMN of given measures

(2) Draw perpendicular bisects of sides LM & MN that meet at C

(3) Join CN and taking C as centre & CN as radius, construct a circle.

(5)

Construct DEF such that DE = EF = 6 cm, F = 45° and construct its circum circle.

Solution: Steps:

(1) Draw a triangle of given measures

(2) Construct perpendicular bisects of sides DE and EF that meet at C.

(3) Join CE with C as centre & CE as radius construct a circle.

### Problem Set – 6

(1) Choose correct alternative answer and fill in the blanks.

(i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence the length of the chord is………

(A) 16 cm

(B) 8 cm

(C) 12 cm

(D) 32 cm

Solution:

(A) 16 cm.

(ii) The point of concurrence of all angle bisectors of a triangle is called the……

(A) centroid

(B) circumcentre

(C) incentre

(D) Orthocenter

Solution:

(C) Incentre.

(iii) The circle which passes through all the vertices of a triangle is called…..

(A) circumcircle

(B) incircle

(C) Congruent circle

(D) Concentric circle

Solution:

(A) Circumcircle.

(iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is….

(A) 12 cm

(B) 13 cm

(C) 14 cm

(D) 15 cm

Solution:

(B) 13 cm.

(v) The length of the longest chord of the circle with radius 2.9 cm is…..

(A) 3.5 cm

(B) 7 cm

(C) 10 cm

(D) 5.8 cm

Solution:

(D) 5.8 cm.

(vi) Radius of a circle with centre O is 4 cm. If l (OP) = 4.2 cm, say where point P will lie.

(A) On the centre

(B) Inside the circle

(C) Outside the circle

(D) On the circle

Solution:

(C) Outside the circle.

(vii) The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is…..

(A) 2 cm

(B) 1 cm

(C) 8 cm

(D) 7 cm

Solution: In △OBC

BC = 3 cm & OC = 3 cm

∴ OB = √52 – 32

= √25 – 9

= 4 cm

OA = √52 – 42

= √25 – 16

= 3 cm

∴ AB = OA + OB = 4 + 3 = 7 cm

(D) 7 cm.

(2)

Construct incircle and circumcircle of an equilateral D DSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.

Solution: Steps:

(1) Construct △DSP of given measures

(2) Draw angle bisectors of ∠P and ∠S that meet at I

(3) Draw perpendicular IM on SP

(4) Taking I as centre & IM as radius, draw a circle.

(5) Construct perpendicular bisectors of DP & SP which meet at I.

(6) With I as centre, draw a circle.

(3)

Construct NTS where NT = 5.7 cm, TS = 7.5 cm and NTS = 110° and draw incircle and circumcircle of it.

Solution: Steps:

(1) Construct a △NTS of given measures.

(2) Draw angle bisectors of ∠T and ∠S which meet at I

(3) IM ⊥ TS is drawn. Taking I as centre & IM a radius draw a circle.

(4) Draw perpendicular bisectors of sides TS and NT. Which meet at C

(5) Taking C as centre & radius CN, draw a circle.

(4) In the figure 6.19, C is the centre of the circle.

seg QT is a diameter

CT = 13, CP = 5, find the length of chord RS.

Solution:

Given CT = 13

∴ CS = 13

CP = 5

In △CPS

CS2 = CP2 + PS2

= PS = √132 – 52 = √169 – 25

= 12

∴ RS = 2 PS = 24

(5) In the figure 6.20, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E.

If AEP ≅ ∠DEP then prove the AB = CD

Solution:

PQ ⊥ CD & PR ⊥ AB is drawn.

∴ In △PQE & △PRE,

∠QEP = ∠REP (given)

∠PQE = ∠PRE = 90°

PE common.

∴ △PQE ≅ △PRE (A-S-A test)

∴ PR = QP

∴ AB = CD because equal chords are equidistant from the centre.

(6) In the 6.21, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to Chord AB at point E. Show that ABC is an isosceles triangle.

Solution:

In △AEC and △BEC,

∠CEB = ∠CEA = 90°

CE is common

AE = EB (∵ OE bisects AB)

∴ △AEC ≅ △BEC (S-A-S) test

∴ AC = BC

∴ ABC is an isosceles triangle.

Updated: September 11, 2021 — 6:25 pm