SELINA Solution Class 9 Chapter 17 Circle Exercise 17D

Question 1

The radius of a circle is 13 cm and the length of one of its chords is 24 cm.
Find the distance of the chord from the center.

Sol:

To find: OM

Given that AB = 24 cm
Since OM ⊥ AB
⇒ OM bisects AB
So, AM = 12 cm
In right ⇒ OMA,
OA² = OM²  + AM²
OM² = OA² - AM²
OM² = 13² - 12²
OM² = 25
OM² = 5 cm
Hence, the distance of the chord from the centre is 5 cm.

Question 2

Prove that equal chords of congruent circles subtend equal angles at their center.

Sol:

Given: AB and CD are two equal chords of a congruent circle with centres O and O respectively.

To prove: ∠AOB = ∠CO'D

Proof:
In ΔOAB and ΔO'CD
OA = O'C                            ...[ ∵ Radii of congruent circles ]
OB = O'D                            ...[ ∵ Radii of congruent circles ]
AB = CD                              ...[ Given ] 
ΔOAB ≅ ΔO'CD                   ...[ By SSS congruence criterion ]
∠AOB = ∠CO'D                   ...[ c. p. c. t ]

Question 3

Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points?

Sol:

So, the circle can have 0, 1 or 2 points in common.
The maximum number of common points is 2. 

Question 4

Suppose you are given a circle. Describe a method by which you can find the center of this circle.

Sol:

To draw the center of a given circle : 
1. Draw the circle.
2. Take any two different chords AB and CD of this circle and draw perpendicular bisector of these chords.
3. let these perpendicular bisectors meet at point O.

So, O will be the center of the given circle.

Question 5

Given two equal chords AB and CD of a circle with center O, intersecting each other at point P.
Prove that:
(i) AP = CP
(ii) BP = DP 

Sol:

In ΔOMP and ΔONP,
OP = OP                            ...( common sides )
∠OMP = ∠ONP                 ...( both are right angles )
OM = OM                          ...( side both the chords are equal, so the distance of the chords from the centre are also equal )
ΔOMP ≅ ΔONP                 ...( RHS congruence criterion )
⇒  MP = PN                       ...(c.p.c.t )  ....( a )

(i) Since AB = CD              ...( given )
⇒  AM = CN                     ...( drawn from the centre to the chord bisects the chord )
⇒  AM + MP = CN + NP  .....( from a )
⇒  AP = CP                       ....( b )

(ii) Since AB = CD
⇒ AP + BP = CP + DP
⇒  BP = DP                     ....( from  b )
Hence proved.

Question 6

In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively.
Calculate the distance between the chords, if they are on:
(i) the same side of the center.
(ii) the opposite sides of the center.

Sol:

Given that AB = 16 cm and CD = 12 cm
So, AL = 8 cm and CM = 6 cm   ....( ⊥ from the center to the chord bisects the chord )

In right triangle OLA and OMC,
By Pythagoras theorem,
OA² = OL² + AL² and OC² = OM² + Cm² 
10² = OL² + 8² and 10² = OM² + 6²
OL² = 100 - 64 and OM² = 64
OL² = 6 cm and OM² = 8 cm

(i) In the first case, distance between AB and CD is
LM= OM - OL = 8 - 6 = 2 cm

(ii) In the second case , distance between AB and CD is
LM = OM + OL = 8 + 6 = 14 cm

Question 7

In the given figure, O is the center of the circle with radius 20 cm and OD is perpendicular to AB. If AB = 32 cm,
find the length of CD.

Sol:

To find: CD

Given : AB = 32 cm
⇒ AC = 16 cm( Since Perpendicular is drawn from the centre to the chord, bisects the chord )

In Right ΔOCA,
OA² = OC² + AC²        ....( By Pythagoras theorem )
⇒ OC² = OA² - AC²
⇒ OC² = 20² - 16²
⇒ OC²= 144
⇒ OC = 12cm
Since OD = 20 cm and OC = 12 cm
⇒ CD = OD - OC
          = 20 - 12 = 8 cm.

Question 8

In the given figure, AB and CD are two equal chords of a circle, with centre O. If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.

Sol:

It is given in the question that point.

P is the mid-point of the chord AB and Point Q is the mid-point of the CD.
⇒ ∠APO = 90°      ...( as the straight line drawn from the center of a  circle to bisect a chord, which is not a diameter, is at the right angle to the chord. )

As chords, AB and CD are equal therefore they are equidistant from the center i.e; PO = OQ      ...( ∵ Equal chords of a circle are equidistant from the center)

Now, the ΔPOQ is an isosceles triangle with OP = OQ as its two equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
⇒ ∠POQ + ∠OPQ + ∠PQO = 180°
⇒ ∠OPQ + ∠POQ + 150° = 180° ...( Given: ∠POQ = 150° )
⇒ 2∠OPQ = 180° - 150°  ...( As, ∠OPQ = ∠PQO ) 
⇒ 2∠OPQ = 30°
⇒ ∠OPQ = 15°

As ∠APO = 90°
⇒ ∠APQ + ∠OPQ = 90°
⇒ ∠APQ = 90° - 15°     ....( As, ∠OPQ = 15° )
⇒ ∠APQ = 75°.

Question 9

In the given figure, AOC is the diameter of the circle, with centre O. If arc AXB is half of arc BYC, find ∠BOC.

Sol:

Given :
1. AOC is the diameter.
2. Arc AXB = ${\displaystyle \frac{1}{2}}$ Arc BYC

From Arc AXB = ${\displaystyle \frac{1}{2}}$ Arc BYC We can see that
Arc AXB : Arc BYC = 1 : 2
⇒ ∠BOA : ∠BOC = 1 : 2

Since AOC is the diameter of the circle hence,
∠AOC = 180°
Now,
Assume that ∠BOA = x° and ∠BOC = 2x°
∠AOC = ∠BOA + ∠BOC = 180°
⇒ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Hence, ∠BOA = 60° and ∠BOC = 120°.

Question 10

The circumference of a circle, with center O, is divided into three arcs APB, BQC, and CRA such that:
${\displaystyle \frac{\text{arc APB}}{2}=\frac{\text{arc BQC}}{3}=\frac{\text{arc CRA}}{4}}$

Find ∠BOC.

Sol:

From the given conditions given in the question
We can draw the circle with arc APB, arc BQC, and arc CRA

The given equation is
${\displaystyle \frac{\text{arc APB}}{2}=\frac{\text{arc BQC}}{3}=\frac{\text{arc CRA}}{4}}$
Let 
${\displaystyle \frac{\text{arc APB}}{2}=\frac{\text{arc BQC}}{3}=\frac{\text{arc CRA}}{4}}$ = k ( say )
then 
Arc APB = 2k, Arc BQC = 3k, Arc CRA = 4k
or
Arc APB : Arc BQC : Arc CRA = 2 : 3 : 4
⇒ ∠AOB : ∠BOC : ∠AOC = 2 : 3 : 4
and therefore,
and ∠AOB = 2k°, ∠BOC = 3k°, and ∠AOC = 4k°
Now,
Angle in a circle is 360°
So, 2k + 3k + 4k = 360°
⇒ 9k = 360°
⇒ k = 40°
Hence,
∠BOC = 3 x 40° = 120°.