Circles
Class 9th Mathematics Part Ii MHB Solution
- Distance of chord AB from the center of a circle is 8 cm. Length of the chord AB is 12…
- Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the…
- Radius of a circle is 34 cm and the distance of the chord from the center is 30 cm,…
- Radius of a circle with center O is 41 units. Length of a chord PQ is 80 units, find…
- In figure 6.9, center of two circles is O. Chord AB of bigger circle intersects the…
- Prove that, if a diameter of a circle bisects two chords of the circle then those two…
- Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the…
- In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the…
- Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC…
- Construct ΔABC such that ∠B = 100^0 , BC = 6.4, ∠C = 50^0 and construct its incircle.…
- Construct ΔPQR such that ∠70^0 , ∠R = 50^0 , QR = 7.3cm, and construct its…
- Construct ΔXYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.…
- In ΔLMN, LM = 7.2cm, ∠M = 105^0 , MN = 6.4cm, then draw ΔLMN and construct its…
- Construct ΔDEF such that DE = EF = 6 cm, ∠F = 45^0 and construct its circumcircle.…
- Radius of a circle is 10 cm and distance of a chord from the center is 6 cm. Hence the…
- The point of concurrence of all angle bisectors of a triangle is called the ......…
- The circle which passes through all the vertices of a triangle is called ..... Choose…
- Length of a chord of a circle is 24 cm. If distance of the chord from the center is 5…
- The length of the longest chord of the circle with radius 2.9 cm is ..... Choose…
- Radius of a circle with center O is 4 cm. If l(OP) = 4.2 cm, say where point P will…
- The lengths of parallel chords which are on opposite sides of the center of a circle…
- Construct incircle and circumcircle of an equilateral ΔDSP with side 7.5 cm. Measure…
- Construct ΔNTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 110^0 and draw incircle and…
- In the figure 6.19, C is the center of the circle. seg QT is a diameter CT = 13, CP =…
- In the figure 6.20, P is the center of the circle. chord AB and chord CD intersect on…
- In the figure 6.21, CD is a diameter of the circle with center O. Diameter CD is…
Practice Set 6.1
Question 1.Distance of chord AB from the center of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of the circle.
Answer:
Given that OP = 8 cm
And AB = 12 cm
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
∴ AP = PB = 6 cm
In the right angled ΔOAP using Pythagoras theorem,
⇒ OA2 = OP2 + AP2
⇒ OA2 = 82 + 62
⇒ OA2 = 64 + 36
⇒ OA2 = 100
⇒ OA = 10cm
So, the diameter of the circle is (2×10) = 20cm (Diameter = 2×Radius).
Question 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 center.
Answer:
Given that diameter = 26cm
Radius = Diameter / 2 = 26 /2 = 13cm
So, OA = 13cm
And AB = 24 cm
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
∴ AP = PB = 12 cm
In the right angled ΔOAP using Pythagoras theorem,
⇒ OA2 = OP2 + AP2
⇒ 132 = OP2 + 122
⇒ 169 = OP2 + 144
⇒ OP2 = 25
⇒ OP = 5cm
So, the distance of chord from the center is 5cm.
Question 3.
Radius of a circle is 34 cm and the distance of the chord from the center is 30 cm, find the length of the chord.
Answer:
Given that
Radius = 34cm
So, OA = 34cm
And OP = 30 cm
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
∴ AP = PB,
AB = 2PB
In the right angled ΔOAP using Pythagoras theorem,
⇒ OA2 = OP2 + AP2
⇒ 342 = 302 + AP2
⇒ 1156 = 900 + AP2
⇒ AP2 = 256
⇒ AP = 16cm
So, the length of chord is 16× 2 = 32cm (AB = 2AP)
Question 4.
Radius of a circle with center O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the center of the circle.
Answer:
Given that
Radius = 41 units
So, OP = 41 units
And PQ = 80 units
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
∴ PM = MQ = 40 cm
In the right angled ΔOAP using Pythagoras theorem,
⇒ OP2 = OM2 + PM2
⇒ 412 = OM2 + 402
⇒ 1681 = OM2 + 1600
⇒ OM2 = 81
⇒ OM = 9 units
So, the distance of chord from the center is 9 units.
Question 5.
In figure 6.9, center of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ
Answer:
We draw a perpendicular on chord AB from O.
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
Therefore,
AM = MB …….(1)
OM is also perpendicular to chord PQ of smaller circle.
Therefore,
PM = MQ ………….(2)
Subtracting (2) from (1)
AM-PM = MB-MQ
⇒ AP = BQ
Hence Proved.
Question 6.
Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
Answer:
We draw a circle with center O and AB, CD are the chords of this circle. Diameter PQ bisects AB and CD at M and N respectively.
We know that the line from the center bisecting the chord is perpendicular to the chord.
Therefore,
∠ OMA = ∠ OMB = 90°
Also, ∠ ONC = ∠ OND = 90°
∠ OMA + ∠ ONC = 90° + 90° = 180°
Hence the two chords, AB and CD are parallel to each other.
Practice Set 6.2
Question 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 center of the circle?
Answer:
Given radius of circle is 10cm
OA = OD = 10cm
AB = CD = 16cm
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
CQ = QD = 8cm
In right angled ΔOQD using the Pythagoras theorem
OD2 = OQ2 + QD2
102 = OQ2 + 82
100 = OQ2 + 64
OQ2 = 36
OQ = 6cm
Therefore the chord CD is at 6cm from the center.
We know that Congruent chords of a circle are equidistant from the center of the circle.
As AB and CD are equal in length, they are equidistant.
∴ OP = OQ = 6cm
Question 2.
In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the center. Find the lengths of the chords.
Answer:
Given radius of circle is 13cm
OA = OD = 13cm
OQ = OP = 16cm
We know that a perpendicular drawn from the centre of a circle on its chord bisects
the chord.
CQ = QD
CD = 2×QD
In right angled ΔOQD using the Pythagoras theorem
OD2 = OQ2 + QD2
132 = 52 + QD2
169 = 25 + QD2
QD2 = 144
QD = 12cm
Therefore the length of chord CD = 2×12 = 24cm
We know that The chords of a circle equidistant from the center of a circle are congruent
As AB and CD are equidistant, they are equal in length.
∴ AB = CD = 24cm
Question 3.
Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC is the bisector of ∠NPM.
Answer:
Given that PM = PN
We know that Congruent chords of a circle are equidistant from the center of the circle.
Therefore, AC = CB ………………(1)
Also,
A perpendicular drawn from the centre of a circle on its chord bisects
the chord.
CB bisects PN as PB = BN,
Similarly, CA bisects PM as PA = AM.
In ΔAPC and ΔBPC,
∠CAP = ∠ CBP = 90°
PC = PC (common side)
AC = CB (From eq (1))
∴ ΔAPC≅ ΔBPC (RHS congruence)
∴ ∠ APC = ∠ BPC (by CPCT)
Hence proved that PC is the bisector of ∠ NPM.
Practice Set 6.3
Question 1.Construct ΔABC such that ∠B = 1000, BC = 6.4, ∠C = 500 and construct its incircle.
Answer:
Steps of Construction:
1.Construct ΔABC of given dimensions.
2.Draw bisectors of two angles, ∠A and ∠B.
3.Denote the point of intersection as O.
4.Draw perpendicular OP on AB.
5.Draw a circle with O as center and OP as radius.
Question 2.
Construct ΔPQR such that ∠700, ∠R = 500, QR = 7.3cm, and construct its circumcircle.
Answer:
Steps of Construction:
1.Construct ΔPQR of given dimensions.
2.Draw perpendicular bisectors of two sides, QR and PR.
3.Denote the point of intersection as O.
4.Draw a circle with O as center and OP as radius.
Question 3.
Construct ΔXYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.
Answer:
Steps of Construction:
1.Construct ΔXYZ of given dimensions.
2.Draw bisectors of two angles, ∠X and ∠Y.
3.Denote the point of intersection as O.
4.Draw perpendicular OA on XY.
5.Draw a circle with O as center and OA as radius.
Question 4.
In ΔLMN, LM = 7.2cm, ∠M = 1050, MN = 6.4cm, then draw ΔLMN and construct its circumcircle.
Answer:
Steps of Construction:
1.Construct ΔLMN of given dimensions.
2.Draw perpendicular bisectors of two sides, LM and MN.
3.Denote the point of intersection as O.
4.Draw a circle with O as center and OM as radius.
Question 5.
Construct ΔDEF such that DE = EF = 6 cm, ∠F = 450 and construct its circumcircle.
Answer:
Steps of Construction:
1.Construct ΔDEF of given dimensions.
2.Draw perpendicular bisectors of two sides, DE and EF.
3.Denote the point of intersection as O.
4.Draw a circle with O as center and OD as radius.
Problem Set 6
Question 1.Choose correct alternative answer and fill in the blanks.
Radius of a circle is 10 cm and distance of a chord from the center is 6 cm. Hence the length of the chord is .........
A. 16 cm
B. 8 cm
C. 12 cm
D. 32 cm
Answer:
Given that
Radius = 10cm
So, OA = 10cm
And OP = 6 cm
We know that a perpendicular drawn from the centre of a circle on its chord bisects
the chord.
∴ AP = PB,
AB = 2PB
In the right angled ΔOAP using Pythagoras theorem,
⇒ OA2 = OP2 + AP2
⇒ 102 = 62 + AP2
⇒ 100 = 36 + AP2
⇒ AP2 = 64
⇒ AP = 8cm
So, the length of chord is 8× 2 = 16cm (AB = 2AP)
The correct answer is A.
Question 2.
Choose correct alternative answer and fill in the blanks.
The point of concurrence of all angle bisectors of a triangle is called the ......
A. centroid
B. circumcenter
C. incentre
D. orthocenter
Answer:
Incenter is defined as the point of occurrence of all angle bisectors of a triangle. Incircle is the corresponding circle formed with incenter as the center.
Question 3.
Choose correct alternative answer and fill in the blanks.
The circle which passes through all the vertices of a triangle is called .....
A. circumcircle
B. incircle
C. congruent circle
D. concentric circle
Answer:
Circle passing through all the vertices of a triangle is called circumcircle of the triangle and the center of the circle is called the circumcenter of the triangle.
Question 4.
Choose correct alternative answer and fill in the blanks.
Length of a chord of a circle is 24 cm. If distance of the chord from the center is 5 cm, then the radius of that circle is ….....
A. 12 cm
B. 13 cm
C. 14 cm
D. 15 cm
Answer:
Given that OP = 5 cm
And AB = 24cm
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
∴ AP = PB = 12 cm
In the right angled ΔOAP using Pythagoras theorem,
⇒ OA2 = OP2 + AP2
⇒ OA2 = 52 + 122
⇒ OA2 = 25 + 144
⇒ OA2 = 169
⇒ OA = 13cm
Hence, The correct option is B.
Question 5.
Choose correct alternative answer and fill in the blanks.
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
Answer:
The longest chord of a circle is its diameter.
The length of diameter is 2× radius = 2×2.9 = 5.8cm
Hence, The correct answer is D.
Question 6.
Choose correct alternative answer and fill in the blanks.
Radius of a circle with center O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie.
A. on the center
B. Inside the circle
C. outside the circle
D. on the circle
Answer:
The longest distance of a point, from the center, within the circle is its radius = 4cm.
For distance = 4cm , it is on the circle.
For distance>4.2cm, it is outside the circle.
Hence, The correct answer is C.
Question 7.
Choose correct alternative answer and fill in the blanks.
The lengths of parallel chords which are on opposite sides of the center 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
Answer:
Let, length of AB = 6cm and length of CD = 8cm
Radius of circle = 5cm
OB = OC = 5cm
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
AM = MB = 3cm
Also, CN = ND = 4cm
In ΔOMB,
⇒ OB2 = OM2 + MB2
⇒ 52 = OM2 + 32
⇒ OM2 = 25-9
⇒ OM2 = 16
⇒ OM = 4cm
In ΔONC,
⇒ OC2 = ON2 + CN2
⇒ 52 = ON2 + 42
⇒ ON2 = 25-16
⇒ ON2 = 9
⇒ ON = 3cm
Distance between AB and CD = OM + ON = 4 + 3 = 7cm
Hence the correct option is D.
Question 8.
Construct incircle and circumcircle of an equilateral ΔDSP 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.
Answer:
Steps of Construction:
1.Construct ΔDSP of given dimensions.
2.Draw perpendicular bisectors of two sides, DS and SP.
3.Denote the point of intersection as O.
4.Draw a circle with O as center and OD as radius.
As we know that for an equilateral triangle, the incenter is same as the circumcenter.
5.Taking O as center, and the OM as the radius draw a circle.
Circumradius = OP = 4.3cm
Inradius = OM = 2.1cm
For an equilateral triangle, Circumradius = 2×Inradius.
Question 9.
Construct ΔNTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 1100 and draw incircle and circumcircle of it.
Answer:
Steps of Construction:
1. Construct ΔNTS of given dimensions.
2.Draw bisectors on ∠T and ∠S.
3.The point of intersection as B.
4.Draw perpendicular on NT from B.
5. With B as center and length of perpendicular as radius, draw a circle.
6.Draw perpendicular bisectors of ST and TN.
7.The point of intersection is A.
8. With A as center, and AS as radius draw a circle.
Question 10.
In the figure 6.19, C is the center of the circle. seg QT is a diameter CT = 13, CP = 5, find the length of chord RS.
Answer:
We join C to S to form ΔCPS.
CP = 5 (given)
Radius of circle = CT = 13 (given)
Therefore CS = 13 (radius of the same circle)
In ΔCPS,
⇒ CS2 = CP2 + PS2
⇒ 132 = 52 + PS2
⇒ PS2 = 169-25
⇒ PS2 = 144
⇒ PS = 12 units
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
PS = RP = 12 units
RS = 2×PS = 2×12 = 24 units
Question 11.
In the figure 6.20, P is the center of the circle. chord AB and chord CD intersect on the diameter at the point E If ∠AEP ≅ ∠DEP then prove that AB = CD.
Answer:
In the figure, we join PA and PD. Draw perpendiculars on AB and CD from P as PM and PN respectively.
∠AEP = ∠ DEP (given)
So, ∠PEN = ∠PEM (M and N are points on line AE and ED respectively)
In ΔPEN and ΔPEM,
∠PNE = ∠PME = 90°
∠PEN = ∠PEM
PE = PE (common)
Therefore, Δ PEN ≅ ΔPEM (by AAS congruence)
∴ PN = PM (by CPCT)
We know that The chords of a circle equidistant from the center of a circle are congruent.
So, AB = CD.
Hence proved.
Question 12.
In the figure 6.21, CD is a diameter of the circle with center O. Diameter CD is perpendicular to chord AB at point E. Show that ΔABC is an isosceles triangle.
Answer:
We know that a perpendicular drawn from the center of a circle on its chord bisects
the chord.
So, AE = EB
In ΔACE and ΔBCE,
AE = EB
∠AEC = ∠BEC = 90°
CE = CE (common)
Δ ACE ≅ Δ BCE (By SAS congruence)
Therefore, AC = BC (by CPCT)
Hence proved that ABC is an isosceles triangle.