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Circles Class 9th Mathematics Part Ii MHB Solution

Circles Class 9th Mathematics Part Ii MHB Solution
Practice Set 6.1
  1. Distance of chord AB from the center of a circle is 8 cm. Length of the chord AB is 12…
  2. Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the…
  3. Radius of a circle is 34 cm and the distance of the chord from the center is 30 cm,…
  4. Radius of a circle with center O is 41 units. Length of a chord PQ is 80 units, find…
  5. In figure 6.9, center of two circles is O. Chord AB of bigger circle intersects the…
  6. Prove that, if a diameter of a circle bisects two chords of the circle then those two…
Practice Set 6.2
  1. Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the…
  2. In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the…
  3. Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC…
Practice Set 6.3
  1. Construct ΔABC such that ∠B = 100^0 , BC = 6.4, ∠C = 50^0 and construct its incircle.…
  2. Construct ΔPQR such that ∠70^0 , ∠R = 50^0 , QR = 7.3cm, and construct its…
  3. Construct ΔXYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.…
  4. In ΔLMN, LM = 7.2cm, ∠M = 105^0 , MN = 6.4cm, then draw ΔLMN and construct its…
  5. Construct ΔDEF such that DE = EF = 6 cm, ∠F = 45^0 and construct its circumcircle.…
Problem Set 6
  1. Radius of a circle is 10 cm and distance of a chord from the center is 6 cm. Hence the…
  2. The point of concurrence of all angle bisectors of a triangle is called the ......…
  3. The circle which passes through all the vertices of a triangle is called ..... Choose…
  4. Length of a chord of a circle is 24 cm. If distance of the chord from the center is 5…
  5. The length of the longest chord of the circle with radius 2.9 cm is ..... Choose…
  6. Radius of a circle with center O is 4 cm. If l(OP) = 4.2 cm, say where point P will…
  7. The lengths of parallel chords which are on opposite sides of the center of a circle…
  8. Construct incircle and circumcircle of an equilateral ΔDSP with side 7.5 cm. Measure…
  9. Construct ΔNTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 110^0 and draw incircle and…
  10. In the figure 6.19, C is the center of the circle. seg QT is a diameter CT = 13, CP =…
  11. In the figure 6.20, P is the center of the circle. chord AB and chord CD intersect on…
  12. 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.