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.
|Maharashtra Class 9|
Math Part 2 Solution
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.
(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
OD2 + DB2 = OB2
=> OB = √82 + 62
∴ 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.
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)
OD = √OB2 – DB2
= √132 – 122
= √169 – 144
= 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
OB2 = OD2 + DB2
= DB = √OB2 – OD2
= √342 – 302
= 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.
PQ = 80 units = RQ = 40 units
Radius = OQ = 41 units
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
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.
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?
∵ 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
= 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.
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.
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
Construct △ABC such that ∠B =100°, BC = 6.4 cm, ∠C = 50° and construct it’s in circle.
(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.
(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.
(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.
In △LMN, LM = 7.2 cm, ∠M = 105°, MN = 6.4 cm, then draw △LMN and construct its circum circle.
(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.
Construct △DEF such that DE = EF = 6 cm, ∠F = 45° and construct its circum circle.
(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
(A) 16 cm.
(ii) The point of concurrence of all angle bisectors of a triangle is called the……
(iii) The circle which passes through all the vertices of a triangle is called…..
(C) Congruent circle
(D) Concentric circle
(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
(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
(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
(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
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.
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.
(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.
Construct △NTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it.
(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.
Given CT = 13
∴ CS = 13
CP = 5
CS2 = CP2 + PS2
= PS = √132 – 52 = √169 – 25
∴ 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
PQ ⊥ CD & PR ⊥ AB is drawn.
∴ In △PQE & △PRE,
∠QEP = ∠REP (given)
∠PQE = ∠PRE = 90°
∴ △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.
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.