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,
OD^{2} + DB^{2} = OB^{2}
=> OB = √8^{2} + 6^{2}
= √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 = √OB^{2} – DB^{2}
= √13^{2} – 12^{2}
= √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
OB^{2} = OD^{2} + DB^{2}
= DB = √OB^{2} – OD^{2}
= √34^{2} – 30^{2}
= √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,
OQ^{2} = RO^{2} + RQ^{2}
=> RO = √41^{2} – 40^{2}
= 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 = √10^{2} – (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 √13^{2} – 5^{2}
= 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 = √5^{2} – 3^{2}
= √25 – 9
= 4 cm
In △OAD
OA = √5^{2} – 4^{2}
= √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
CS^{2} = CP^{2} + PS^{2}
= PS = √13^{2} – 5^{2} = √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.