RS Aggarwal 2017 Solutions for Class 9 Math Chapter 11 - Circles

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Page No 400:

Question 5:

Answer:

Given: Two distinct circles
To prove: Two distinct circles cannot intersect each other in more than two points.
Proof: Suppose that two distinct circles intersect each other in more than two points.
∴ These points are non-collinear points.
Three non-collinear points determine one and only one circle.
∴ There should be only one circle.
This contradicts the given, which shows that our assumption is wrong.
Hence, two distinct circles cannot intersect each other in more than two points.

Page No 401:

Question 12:

Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres.

Answer:

Given: OA = 10 cm, O'A = 8 cm and AB = 12 cm
$A D = (\frac{AB}{2}) = (\frac{12}{2}) = 6 cm$
Now, in right angled ΔADO, we have:
OA² = AD² + OD²
⇒ OD² = OA² - AD²
             = 10² - 6²
             = 100 - 36 = 64
∴ OD = 8 cm

Similarly, in right angled ΔADO', we have:
O'A² = AD² + O'D²
⇒ O'D²= O'A² - AD²
              = 8² - 6²
              = 64 - 36
              = 28
⇒ $O' D = \sqrt{28} = 2 \sqrt{7}$ cm
Thus, OO' = (OD + O'D)
      = $(8 + 2 \sqrt{7}) cm$
Hence, the distance between their centres is $(8 + 2 \sqrt{7}) cm$ .

Page No 401:

Question 13:

Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB.

Answer:

Given: Two equal circles intersect at point P and Q.
A straight line passes through P and meets the circle at points A and B.
To prove: QA = QB
Construction: Join PQ.

Proof:
Two circles will be congruent if and only if they have equal radii.
Here, PQ is the common chord to both the circles.
Thus, their corresponding arcs are equal (if two chords of a circle are equal, then their corresponding arcs are congruent).
So, arc PCQ = arc PDQ
∴ ∠QAP = ∠QBP (Congruent arcs have the same degree in measure)
Hence, QA = QB (In isosceles triangle, base angles are equal)

Page No 402:

Question 14:

If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.

Answer:

Given: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at points L and M.
To prove: AB || CD
Proof: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at L and M.

Then OL ⊥ AB
Also, OM ⊥ CD
∴ ∠ ALM = ∠ LMD = 90^o
Since alternate angles are equal, we have:
AB|| CD

Page No 402:

Question 15:

In the adjoining figure, two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Answer:

Two circles with centres A and B of respective radii 5 cm and 3 cm touch each other internally.
The perpendicular bisector of AB meets the bigger circle at P and Q.
Join AP.

Let PQ intersect AB at point L.
Here, AP = 5 cm
Then AB = (5 - 3) cm = 2 cm
Since PQ is the perpendicular bisector of AB, we have:
$A L = (\frac{A B}{2}) = (\frac{2}{2}) = 1 cm$
Now, in right angled ΔPLA, we have:
AP² = AL² + PL²
⇒ PL² = AP² - AL²
            = 5² - 1²
            = 25 - 1 = 24
⇒ $PL = \sqrt{24} = 2 \sqrt{6} cm$
Thus PQ = 2 × PL
              = $(2 \times 2 \sqrt{6}) = 4 \sqrt{6} cm$
Hence, the required length of PQ is $4 \sqrt{6} cm$ .

Page No 402:

Question 16:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠ACD = y° and ∠AOD = x°, prove that x = 3y.

Answer:

We have:
OB = OC, ∠BOC = ∠BCO = y
External ∠OBA = ∠BOC + ∠BCO = (2y)
Again, OA = OB, ∠OAB = ∠OBA = (2y)
External ∠AOD = ∠OAC + ∠ACO
Or x = ∠OAB + ∠BCO
Or x = (2y) + y = 3y
Hence, x = 3y

Page No 402:

Question 17:

In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥ AB and OQ ⊥ AC, prove that PB = QC.

Answer:

Given: AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC

To prove: PB = QC
Proof:
AB = AC      (Given)
⇒ $\frac{1}{2} AB = \frac{1}{2} AC$
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ MB = NC            ...(i)
Also, OM = ON    (Equal chords of a circle are equidistant from the centre)
and OP = OQ (Radii)
⇒ OP - OM = OQ - ON
∴ PM = QN    ...(ii)
Now, in ΔMPB and ΔNQC, we have:
MB = NC            [From (i)]
∠PMB = ∠QNC     [90° each]
PM = QN                [From (ii)]
i.e., ΔMPB ≅ ΔNQC    (SAS criterion)
∴ PB = QC        (CPCT)

Page No 402:

Question 18:

In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.

Answer:

Given: BC is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD.
TO prove: AB = CD
Construction: Draw OL ⊥ AB and OM ⊥ CD.

Proof:
In ΔOLB and ΔOMC, we have:
∠OLB = ∠OMC        [90° each]
∠OBL = ∠OCD         [Alternate angles as AB || CD]
OB = OC                      [Radii of a circle]
∴ ΔOLB ≅ ΔOMC   (AAS criterion)
Thus, OL = OM   (CPCT)
We know that chords equidistant from the centre are equal.
Hence, AB = CD

Page No 402:

Question 19:

Answer:

(i) Join BO.

In ΔBOC, we have:
OC = OB (Radii of a circle)
⇒ ∠OBC = ∠OCB
∠OBC = 30°                 ...(i)
In ΔBOA, we have:
OB = OA (Radii of a circle)
⇒∠OBA = ∠OAB    [∵ ∠OAB = 40°]
⇒∠OBA = 40°       ...(ii)
Now, we have:
∠ABC = ∠OBC + ∠OBA
          = 30° + 40°    [From (i) and (ii)]
∴ ∠ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., ∠AOC = 2∠ABC
           = (2 × 70°) = 140°
(ii)

Here, ∠BOC = {360° - (90° + 110°)}
            = (360° - 200°) = 160°
We know that ∠BOC = 2∠BAC
$\Rightarrow ∠ B A C = \frac{∠ B O C}{2} = (\frac{160 °}{2}) = 80 °$
Hence, ∠BAC = 80°

Page No 421:

Question 2:

In the given figure, O is the canter of the circle and ∠AOB = 70°.
Calculate the values of (i) ∠OCA, (ii) ∠OAC.

Answer:

(i)
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
Thus, ∠AOB = 2∠OCA
$\Rightarrow ∠ O C A = (\frac{∠ A O B}{2}) = (\frac{70 °}{2}) = 35 °$

(ii)
OA = OC (Radii of a circle)
∠OAC = ∠OCA [Base angles of an isosceles triangle are equal]
= 35°

AB and CD are two chords of a circle which intersect each other at P outside the circle.
AB = 6 cm, BP = 2 cm and PD = 2.5 cm
∴ AP × BP = CP × DP
⇒ 8 × 2 = (CD + 2.5) × 2.5 [∵ CP = CD + DP]
Let CD = x cm
Thus, 8 × 2 = (CD + 2.5) × 2.5
⇒ 16 = 2.5x + 6.25
⇒ 2.5x = (16 - 6.25) = 9.75

⇒ $x = \frac{9.75}{2.5} = 3.9$

Hence, CD = 3.9 cm

Page No 441:

Question 15:

In the given figure, O is the centre of a circle. If ∠AOD = 140° and ∠CAB = 50°, calculate
(i) ∠EDB,
(ii) ∠EBD.

Answer:

O is the centre of the circle where ∠AOD = 140° and ∠CAB = 50°.
(i) ∠BOD = 180° – ∠AOD
              = (180° – 140°) = 40°
We have the following:
OB = OD (Radii of a circle)
∠OBD = ∠ODB

In ΔOBD, we have:
∠BOD + ∠OBD + ∠ODB = 180°
⇒ ∠BOD + ∠OBD + ∠OBD = 180°      [∵ ∠OBD = ∠ODB]
⇒ 40° +2∠OBD = 180°
⇒ 2∠OBD = (180° – 40°) = 140°
⇒ ∠OBD = 70°
Since ABCD is a cyclic quadrilateral, we have:
∠CAB + ∠BDC = 180°
⇒ ∠CAB + ∠ODB + ∠ODC = 180°
⇒ 50° + 70° + ∠ODC = 180°
⇒ ∠ODC = (180° – 120°) = 60°
∴ ∠ODC = 60°
∠EDB = (180° – (∠ODC + ∠ODB)
          = 180° – (60° + 70°)
          = 180° – 130° = 50°
∴ ∠EDB = 50°

(ii) ∠EBD = 180° - ∠OBD
                  = 180° - 70°
                 = 110°

Page No 442:

Question 24:

In the given figure, ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD. Show that the points A, B, C, D lie on a circle.

Answer:

ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD.
Draw DE ⊥ AB and CF ⊥ AB.
In ΔADE and ΔBCF, we have:
∠ADE = ∠ADC - 90° = ∠BCD - 90° = ∠BCF (Given: ∠ADC = ∠BCD)
AD = BC (Given)
and ∠AED = ∠BCF = 90°
∴ ΔADE ≅ ΔBCF (By AAS congruency)
⇒ ∠A = ∠B
Now, ∠A + ∠B + ∠C + ∠D = 360°
⇒ 2∠B + 2∠D = 360°
⇒ ∠B + ∠D = 180°
Hence, ABCD is a cyclic quadrilateral.

Page No 442:

Question 25:

In the given figure, ∠BAD = 75°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.

Answer:

We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
i.e., ∠BAD = ∠DCF = 75°
⇒ ∠DCF = x = 75°

Again, the sum of opposite angles in a cyclic quadrilateral is 180°.
Thus, ∠DCF + ∠DEF = 180°
⇒ 75° + y = 180°
⇒ y = (180° - 75°) = 105°

Hence, x = 75° and y = 105°

Page No 442:

Question 26:

The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

Answer:

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles.
Let OL ⊥ AB such that LO produced meets CD at M.

Then we have to prove that CM = MD
Clearly, ∠1 = ∠2   [Angles in the same segment]
∠2 + ∠3 = 90°   [∵ ∠OLB = 90°]
∠3 + ∠4= 90°    [∵ LOM is a straight line and ∠BOC = 90°]
∴ ∠2 + ∠3 = ∠3 + ∠4 ⇒∠2 = ∠4
Thus, ∠1 = ∠2 and ∠2 = ∠4 ⇒ ∠1 = ∠4
∴ OM = CM and, similarly, OM = MD
Hence, CM = MD

Question 4:

In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB = ?
(a) 30°
(b) 15°
(c) 60°
(d) 90°
Figure

Answer:

(c) 60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference.
Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°

Page No 448:

Question 5:

In the given figure, O is the centre of a circle. If ∠OAB = 40° and C is a point on the circle, then ∠ACB = ?
(a) 40°
(b) 50°
(c) 80°
(d) 100°

Answer:

(b) 50°
OA = OB
⇒ ∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
∴ $∠ ACB = \frac{1}{2} ∠ AOB = (\frac{1}{2} \times 100) ° = 50 °$

Page No 448:

Question 6:

In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then the distance of CD from AB is
(a) 8 cm
(b) 15 cm
(c) 18 cm
(d) 6 cm

Answer:

(a) 8 cm
Join OC. Then OC = radius = 17 cm

$CL = \frac{1}{2} CD = (\frac{1}{2} \times 30) cm = 15 cm$
In right ΔOLC, we have:
OL² = OC² - CL² = (17)² - (15)² = (289 - 225) = 64
$\Rightarrow OL = \sqrt{64} = 8 cm$
∴ Distance of CD from AB = 8 cm

Page No 449:

Question 13:

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is
(a) 3 cm
(b) $3 \sqrt{2} cm$
(c) $3 \sqrt{3} cm$
(d) 6 cm
Figure

Answer:

(c) $3 \sqrt{3} cm$

Let ΔABC be an equilateral triangle of side 9 cm.
Let AD be one of its medians.

Then AD ⊥ BC and BD = 4.5 cm
∴ $AD = \sqrt{{AB}^{2} - {BD}^{2}} = \sqrt{{(9)}^{2} - {(\frac{9}{2})}^{2}} = \sqrt{81 - \frac{81}{4}} = \sqrt{\frac{324 - 81}{4}} = \sqrt{\frac{243}{4}} = \frac{9 \sqrt{3}}{2} cm$
Let G be the centroid of ΔABC.
Then AG : GD = 2 : 1
∴ Radius = AG = $\frac{2}{3} AD = (\frac{2}{3} \times \frac{9 \sqrt{3}}{2}) cm = 3 \sqrt{3} cm$

Page No 449:

Question 14:

The angle in a semicircle measures
(a) 45°
(b) 60°
(c) 90°
(d) 36°
Figure

Answer:

Page No 449:

Question 15:

Angles in the same segment of a circle area are
(a) equal
(b) complementary
(c) supplementary
(d) none of these
Figure

Answer:

(a) equal
The angles in the same segment of a circle are equal.

Page No 450:

Question 24:

Answer:

(c) 115°
Take a point D on the remaining part of the circumference.
Join AD and CD.

Then $∠ ADC = \frac{1}{2} ∠ AOC = (\frac{1}{2} \times 130 °) = 65 °$
In cyclic quadrilateral ABCD, we have:
∠ABC + ∠ADC = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC + 65° = 180°
⇒ ∠ABC = (180° - 65°) = 115°

Page No 451:

Question 30:

In the given figure, AOB is a diameter of a circle and CD || AB. If ∠BAD = 30°, then ∠CAD = ?
(a) 30°
(b) 60°
(c) 45°
(d) 50°

Answer:

(a) 30°
∠ADC = ∠BAD = 30° (Alternate angles)
∠ADB = 90° (Angle in semicircle)
∴ ∠CDB = (90° + 30°) = 120°
But ABCD being a cyclic quadrilateral, we have:
∠BAC + ∠CDB = 180°
⇒ ∠BAD + ∠CAD + ∠CDB = 180°
⇒ 30° + ∠CAD + 120° = 180°
⇒ ∠CAD = (180° - 150°) = 30°

Page No 451:

Question 31:

In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?
(a) 50°
(b) 40°
(c) 25°
(d) 80°

Answer:

(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE.

Then $∠ AEC = \frac{1}{2} ∠ AOC = (\frac{1}{2} \times 100 °) = 50 °$
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior ∠CBD = ∠AEC = 50°

Page No 452:

Question 37:

In the given figure, O is the centre of a circle and ∠AOB = 130°. Then, ∠ACB = ?
(a) 50°
(b) 65°
(c) 115°
(d) 155°

Answer:

(c) 115°
Join AB.
Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle.

∴∠AOB = 2∠ADB
$\Rightarrow ∠ ADB = \frac{1}{2} ∠ AOB = (\frac{1}{2} \times 130 °) = 65 °$
In cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°
⇒ 65° + ∠ACB = 180°
∴∠ACB = (180° - 65°) = 115°

Page No 454:

Question 44:

Is ABCD a cyclic quadrilateral?
I. Points A, B, C and D lie on a circle.
II. ∠B + ∠D = 180°.

(a) if the given question can be answered by any one of the statements but not the other;
(b) if the given question can be answered by using either statement alone;
(c) if the given question can be answered by using both the statements together but cannot be answered by using either statement;
(d) if the given question cannot be answered by using both the statements together.

Answer:

(b) if the given question can be answered by using either statement alone
We have,
I. If A, B, C and D lie on a circle, then ABCD is a cyclic quadrilateral.
II. ∠A + ∠B + ∠C + ∠D = 360° and ∠B + ∠D = 180°
⇒ ∠A + ∠C = 180°
∴ A, B, C and D lie on a circle.
Thus, each one of I and II gives answer to the given question.
Hence, the correct option is (b).

Page No 454:

Question 45:

Is ∆ABC right-angled at B?
I. ABCD is a cyclic quadrilateral.
II. ∠D = 90°.

(a) if the given question can be answered by any one of the statements but not the other;
(b) if the given question can be answered by using either statement alone;
(c) if the given question can be answered by using both the statements together but cannot be answered by using either statement;
(d) if the given question cannot be answered by using both the statements together.

Answer:

(c) The given question can be answered by using both the statements together but cannot be answered by using either of the statements.
I gives ∠B + ∠D = 180° and ∠A + ∠C = 180°.
II gives ∠D = 90°.
∴ I and II together give ∠B = 90°.
Thus, the given question can be answered by using both statements.

Page No 454:

Question 46:

Assertion: The circle drawn taking any one of the equal sides of an isosceles right triangle as diameter bisects the base.
Reason: The angle in a semicircle is 1 right angle.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both the assertion and the reason are true and the reason is a correct explanation of the assertion.

Assertion (A): Let ABC be a triangle in which AB = AC and let O be the mid point of AB. With O as centre and OA as radius, draw a circle meeting BC at D.

In Δ ADB, ∠ADB =90° (Angle in a semicircle)
Also, ∠ADB + ∠ADC = 180°
⇒ ∠ADC = 90°
In ΔADB and ΔADC, we have:
AB = AC (Given)
AD = AD (Common)
and ∠ADB = ∠ADC = 90°
∴ ΔADB ≅ ΔADC (RHS criterion)
∴ BD = DC
Hence, the given circle bisects the base.
Thus, assertion (A) is true.
Reason (R): Let ∠BAC be an angle in a semicircle with centre O and diameter BC.

Now, the angle subtended by arc BOC at the centre is ∠BOC = 180°
∴ ∠BOC = 2∠BAC
⇒ 2∠BAC = 180°
⇒ ∠BAC = 90°
Thus, reason (R) is true.
Clearly, reason (R) gives assertion (A).
Hence, the correct option is (a).

Page No 454:

Question 47:

Assertion: The radius of a circle is 10 cm and the length of one of its chords is 16 cm. Then, the distance of the chord from the centre is 6 cm.
Reason: The perpendicular from the centre of a circle to a chord (other than the diameter) bisects the chord.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both the assertion and the reason are true and the reason is a correct explanation of the assertion.
Assertion(A): Let O be the centre of a circle and AB be the chord. Draw OL ⊥ AB.
Then L is the mid point of AB.
Now, OA = 10 cm and AL = $\frac{1}{2} AB = 8 cm$

Thus, $OL = \sqrt{{OA}^{2} - {AL}^{2}} = \sqrt{{(10)}^{2} - {(8)}^{2}} = \sqrt{36} = 6 cm$
i.e., assertion(A) is true.
Clearly, reason (R) gives assertion (A).

Page No 455:

Question 48:

Assertion: In a circles of radius 13 cm, there is a chord of length 10 cm at a distance of 12 cm from the centre of the circle.
Reason: A unique circle can be drawn to pass through three give non-collinear points.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both the assertion and the reason are true, but the reason is not a correct explanation of the assertion.
Clearly, reason (R) is true.
Assertion (A):
OA = 13 cm and OL = 12 cm

AL² = OA² - OL²
⇒ AL² = (13)² - (12)² = (169 - 144) = 25
⇒ AL = 5 cm
⇒ AB = 2 × AL = 10 cm
Thus, assertion (A) is true.
∴ Reason (R) and assertion (A) are both true, but reason (R) does not give assertion (A).

Page No 455:

Question 49:

Assertion: In the given figure, ∠ABC = 70° and ∠ACB = 30°. Then, ∠BDC = 70°.

Reason: In the given figure, ∠AOC = 130°, then ∠ABC = 115°.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) The assertion is false and the reason is true.

Reason (R):
We know that $∠ ADC = \frac{1}{2} ∠ AOC = (\frac{1}{2} \times 130 °) = 65 °$
Also, ∠ABC + ∠ADC = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC = (180° - 65°) = 115°, which is true.
∴ Reason (R) is true.

Assertion(A):
∠ABC + ∠BAC + ∠BCA = 180°
⇒ 70° + ∠BAC + 30° = 180°
⇒ ∠BAC = (180° - 100°) = 80°
∴ ∠BDC = ∠BAC = 80° (Angles in the same segment)
This is false.
Thus, assertion (A) is false and reason (R) is true.

Page No 455:

Question 50:

Assertion: A cyclic parallelogram is a square.
Reason: Diameter is the largest chord in a circle.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) The assertion is false and the reason is true.
Clearly, assertion (A) is false and reason (R) is true.
Hence, the correct answer is option (d).

Page No 455:

Question 51:

Assertion: If two circles intersect at two points, then the line joining their centres is perpendicular to the common chord.
Reason: The perpendicular bisectors of two chords of a circle intersect at its centre.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Clearly, Assertion (A) and Reason (R) are both correct and Reason (R) is the correct explanation of Assertion (A).

Page No 455:

Question 52:

Write T for true and F for false
(i) The degree measures of a semicircle is 180°.
(ii) The perimeter of a circle is called its circumference.
(iii) A circle divides the plane into three parts.
(iv) Let O be the centre of a circle with radius r. Then a point P such that OP < r is called an interior point of the circle.
(v) A circle can have only a finite number of equal chords.

Answer:

(i) True.

(ii) True.

(iii) True, a circle divides a plane into three parts:
a) Points outside the circle.
b) Points inside the circle.
c) Points on the circle.

(iv) True.

(v) False. On a circle, there are infinite points. So, we can draw an infinite number of chords of a given length. Hence, a circle has infinite number of equal chords.

Page No 456:

Question 53:

Match the following columns:

Column I	Column II
(a) Angle in a semicircle measures	(p) 40°
(b) In the given figure, O is the centre of a circle. If ∠AOB = 120°, then ∠ACB = ?	(q) 80°
(c) In the given figure, O is the centre of a circle. If ∠POR = 90° and ∠POQ = 110°, then ∠QPR = ?	(r) 90°
(d) In cyclic quadrilateral ABCD, it is given that ∠ADC = 130° and AOB is a diameter of the circle through A, B, C and D. Then, ∠BAC = ?	(s) 60°

(a) .....,
(b) .....,
(c) .....,
(d) .....,

Answer:

(a) Angle in a semicircle measures 90°.
(b) Now, chord AB subtends ∠AOB at the centre and ∠ACB at a point C of the remaining part of a circle.
       ∴ $∠ AOB = 2 ∠ ACB$
       $\Rightarrow ∠ ACB = \frac{1}{2} ∠ AOB = (\frac{1}{2} \times 120 °) = 60 °$
(c) ∠QOR = {360° - (110° + 90° )} = (360° - 200°) = 160°
     ∴ ∠QPR = $\frac{1}{2} ∠ QOR = (\frac{1}{2} \times 160 °) = 80 °$
(d) Since ∠ADC + ∠ABC = 180° (Opposite angles in a cyclic quadrilateral)
      ⇒ 130° + ∠ABC = 180°
      ∴ ∠ABC = (180° - 130°) = 50°
      and ∠ACB = 90°    (Angle in a semicircle)
      In ΔABC, we have:
      ∠ABC + ∠ACB + ∠BAC = 180°   (Angle sum property of a triangle)
      ⇒ 50° + 90° + ∠BAC = 180°
      ⇒ ∠BAC = (180° - 140°) = 40°

Hence, (a) - (r), (b) - (s), (c) - (q) and (d) - (p)

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Question 54:

Page No 466:

Question 19:

In the given figure, O is the centre of a circle and ∠BCO = 30°. Find x and y.

Answer:

In Δ EOC, we have:
∠EOC + ∠OEC + ∠ECO = 180°     (Angle sum property of a triangle)
⇒ ∠EOC + 90° + 30° = 180°       [∵ ∠OEC = 90° and ∠ECO = ∠BCO = 30°]
⇒ ∠EOC = (180° - 120°) = 60°
Thus, ∠COD = (90° - 60°) = 30°
Since arc CD subtends ∠COD at the centre and ∠CBD on the remaining part of the circle.
i.e., $∠ CBD = \frac{1}{2} ∠ COD = (\frac{1}{2} \times 30 °) = 15 °$
Hence, y = ∠CBD = 15°

Similarly, arc AD subtends ∠AOD at the centre and ∠ABD on the remaining part of the circle.
i.e., $∠ ABD = \frac{1}{2} ∠ AOD = (\frac{1}{2} \times 90 °) = 45 °$        [∵ ∠AOD = 90°]
⇒ ∠ABE = (45° + 15°) = 60°
⇒ ∠AEB = 90°
Now, in Δ ABE, we have:
∠ABE + ∠AEB + ∠BAE = 180°
⇒ 60° + 90° + x = 180°
⇒ x = (180° - 150°) = 30°
Hence, x = 30° and y = 15°

Page No 466:

Question 20:

PQ and RQ are the chords of a circle equidistant from the centre. Prove that the diameter passing through Q bisects ∠PQR and ∠PSR.

Answer:

We know that the chords of a circle that are equidistant from the centre are equal.
∴ PQ = RQ
Now, in ΔPQS and ΔRQS, we have:
PQ = RQ      (Equal chords)
∠QPS = ∠QRS = 90°    (Angles in semicircle)
QS = QS                        (Common hypotenuse)
Thus, ΔPQS ≅ ΔRQS    (RHS criterion)
i.e., ∠PQS = ∠ RQS and ∠PSQ = ∠ RSQ   (CPCT)
Hence, the diameter QOS bisects ∠PQR and ∠PSR.

Page No 466:

Question 21:

Prove that there is one and only one circle passing through three non-collinear points.

Answer:

Given: Three non-collinear points P, Q and R
To prove: There is one and only one circle passing through the points P, Q and R.
Construction: Join PQ and QR.
Draw the perpendicular bisectors AB of PQ and CD of QR.
Let the perpendicular bisectors intersect at the point O.
Now, join OP, OQ and OR. A circle is obtained that passes through the points P, Q and R.

Proof:
We know that each and every point on the perpendicular bisector of a line segment is equidistant from its end points.
Thus, OP = OQ           (∵ O lies on the perpendicular bisector of PQ)
Also, OQ = OR (∵ O lies on the perpendicular bisector of QR)
∴ OP = OQ = OR
Let OP = OQ = OR = r    (Radius of a circle)
Now, draw a circle C(O, r) with O as the centre and r as the radius.
Then, circle C(O, r) passes through the points P, Q and R.
Now, we have to show that this is the only circle that passes through P, Q and R.

Let us suppose that there is an another circle C(O′, t) that passes through P, Q and R.
Then O′ will lie on the perpendicular bisectors AB and CD.
But O is the point of intersection of the perpendicular bisectors AB and CD.
∴ O′ must coincide with the point O.      (Since two lines can not intersect at more than one point)
As O′P = t and OP = r and O′ coincides with O, we get t = r.
∴ C(O, r) and C(O, t) are congruent.
Hence, there is one and only one circle passing through three given non-collinear points.

Page No 466:

Question 22:

In the give figure, OPQR is square. A circle drawn with centre O cuts the square in X and Y. Prove that OX = QY.

Answer:

Given: OPQR is a square. A circle with centre O cuts the square at X and Y.
To prove: QX = QY
Construction: Join OX and OY.

Proof:
In ΔOXP and ΔOYR, we have:
∠OPX = ∠ORY      (90° each)
OX = OY                (Radii of a circle)
OP = OR                (Sides of a square)
∴ ΔOXP ≅ ΔOYR    (By RHS congruency rule)
⇒ PX = RY                 (by CPCT)
⇒ PQ − PX = QR − RY   (∵ PQ = QR, sides of the given square)
Hence, QX = QY

Page No 466:

Question 23:

In the given figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisectors of ∠BAC.

Answer:

Given: AB and AC are two equal chords of a circle with centre O.
To prove: ∠OAB = ∠OAC

Proof:
In ΔOAB and ΔOAC, we have:
AB = AC         (Given)
OA = OA        (Common)
OB = OC         (Radii of a circle)
∴ ΔOAB ≅ ΔOAC (By SSS congruency rule)
i.e.,∠OAB = ∠OAC    (CPCT)
Hence, point O lies on the bisector of ∠BAC.

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