RS AGGARWAL CLASS 9 CHAPTER 12 CIRCLES EXERCISE 12C

 EXERCISE 12C

PAGE NO-482

Question 1:

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that DBC = 60° and ∠BAC = 40°. Find (i) ∠BCD, (ii) ∠CAD.

Answer 1:

(i) ∠BDC = ∠BAC = 40°  (Angles in the same segment)
     In
ΔBCD, we have:
    ∠BCD + ∠DBC + ∠BDC = 180°  (Angle sum property of a triangle)        
     ⇒ ∠BCD + 60° + 40° = 180°
     ⇒ ∠BCD = (180° - 100°) = 80°

(ii) ∠CAD = ∠CBD  (Angles in the same segment)
               = 60°

Question 2:

In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If PSR = 150°, find ∠RPQ.

Answer 2:


In cyclic quadrilateral PQRS, we have:
∠PSR + ∠PQR = 180°
⇒ 150° + ∠PQR = 180°
⇒ ∠PQR = (180° – 150°) = 30°
∴ ∠PQR = 30°                ...(i)
Also, ∠PRQ = 90° (Angle in a semicircle)                 ...(ii)
Now, in ΔPRQ, we have:
∠PQR + ∠PRQ + ∠RPQ = 180°
⇒ 30° + 90° + ∠RPQ = 180°   [From(i) and (ii)]
⇒ ∠RPQ = 180° – 120° = 60°
∴ ∠RPQ = 60°

Question 3:

In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is extended to P, find ∠PBC.

Answer 3:

Reflex ∠AOC + ∠AOC = 360
Reflex ∠AOC + 130 + x = 360
Reflex ∠AOC = 360 − 130
Reflex ∠AOC = 230

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc AC subtends reflex ∠AOC at the centre and ∠ABC at B on the circle.

∴ ∠AOC = 2∠ABC
ABC=230°2=115°          
...(1)

Since ABP is a straight line, ∠ABC + PBC = 180
⇒ ∠PBC = 180 − 115
⇒ ∠PBC = 65            ...(2)

Hence, ∠PBC = 65.

Question 4:

In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If ABC = 92° and ∠FAE = 20°, find ∠BCD.

Answer 4:


Given: ABCD is a cyclic quadrilateral.

Then ABC + ADC = 180°
⇒ 92° + ADC = 180°
ADC = (180° – 92°) = 88°
Again, AE parallel to CD.
Thus, EAD = ADC = 88°  (Alternate angles)
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
BCD = DAF
BCD = EAD + EAF
            =  88° + 20° = 108°
Hence, BCD = 108°

Question 5:

In the given figure, BD = DC and CBD = 30°, find m(∠BAC).

Answer 5:



BD = DC
BCD = CBD = 30°
In ΔBCD, we have:
BCD + CBD + CDB = 180°  (Angle sum property of a triangle)
⇒ 30° + 30° + CDB = 180°
CDB = (180° – 60°) = 120°
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, CDB + BAC = 180°
⇒ 120° + BAC = 180°
BAC = (180° – 120°) = 60°
BAC = 60°

Question 6:

In the given figure, O is the centre of the given circle and measure of arc ABC is 100°. Determine ∠ADC and ∠ABC.

Answer 6:


We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.
Thus, AOC = 2ADC
⇒ 100° = 2ADC
ADC = 50°
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
Thus, ADC +ABC = 180°
⇒ 50° + ABC = 180°
ABC = (180° – 50°) = 130°
ADC = 50° and ABC = 130°

PAGE NO-483

Question 7:

In the given figure, ABC is equilateral. Find (i) ∠BDC, (ii) ∠BEC.

Answer 7:


(i)
Given: ΔABC is an equilateral triangle.
i.e., each of its angle = 60°
BAC = ABC = ACB = 60°
Angles in the same segment of a circle are equal.
i.e., BDC = BAC = 60°
BDC = 60°
(ii)
The opposite angles of a cyclic quadrilateral are supplementary.
Then in cyclic quadrilateral ABEC, we have:
BAC + BEC = 180°
⇒ 60° + BEC = 180°
BEC = (180° – 60°) = 120°
BDC = 60° and BEC = 120°

Question 8:

In the adjoining figure, ABCD is a cyclic quadrilateral in which BCD = 100° and ∠ABD = 50°. Find ∠ADB.

Answer 8:


Given: ABCD is a cyclic quadrilateral.
DAB + DCB = 180°   ( Opposite angles of  a cyclic quadrilateral are supplementary)
DAB + 100° = 180°
DAB = (180° – 100°) = 80°
Now, in ΔABD, we have:
DAB + ABD + ADB = 180°
⇒ 80° + 50° + ADB = 180°
ADB = (180° – 130°) = 50°
Hence, ADB = 50°

Question 9:

In the given figure, O is the centre of a circle and BOD = 150°. Find the values of x and y.

Answer 9:


O is the centre of the circle and BOD = 150°.
Thus, reflex angle BOD = (360° – 150°) = 210°
Now, x=12reflexBOD=12×210°=105°
x = 105°
Again, x + y = 180° (Opposite angles of a cyclic quadrilateral)
⇒ 105° + y = 180°
y = (108° - 105°)= 75°
y = 75°
Hence,
x = 105° and y = 75°

Question 10:

In the given figure, O is the centre of the  circle and DAB = 50°. Calculate the values of x and y.

Answer 10:


O is the centre of the circle and DAB = 50°.
OA = OB (Radii of a circle)
OBA = OAB = 50°
In ΔOAB, we have:
OAB + OBA + AOB = 180°
⇒ 50° + 50° +AOB = 180°
AOB = (180° – 100°) = 80°
Since AOD is a straight line, we have:
x = 180°AOB
       = (180° – 80°) = 100°
i.e., x = 100°
The opposite angles of a cyclic quadrilateral are supplementary.
ABCD is a cyclic quadrilateral.
Thus, DAB + BCD = 180°
BCD = (180° – 50°) = 130°
y = 130°
Hence, x = 100° and y = 130°

Question 11:

In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If CBF = 130° and ∠CDE = x°, find the value of x.

Answer 11:



ABCD is a cyclic quadrilateral.
We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
CBF = CDA
CBF = (180°x)
⇒ 130° = 180°x   [∵ CBF = 130°]
x = (180° – 130°) = 50°
Hence, x = 50°

PAGE NO-484

Question 12:

In the given figure, AB is a diameter of a circle with centre O and DO || CB.
If BCD = 120°, calculate
(i) ∠BAD
(ii) ∠ABD
(iii) ∠CBD
(iv) ∠ADC.
Also, show that ∆OAD is an equilateral triangle.

Answer 12:

We have,
AB is a diameter of the circle where O is the centre, DO || BC and BCD = 120°.
(i)
Since ABCD is a cyclic quadrilateral, we have:
BCD + BAD = 180°
⇒ 120° + BAD = 180°
BAD = (180° – 120°) = 60°
BAD = 60°
(ii)
BDA = 90° (Angle in a semicircle)
In Δ ABD, we have:
BDA + BAD + ABD = 180°
⇒ 90° + 60° + ABD = 180°
ABD = (180° – 150°) = 30°
ABD = 30°
(iii)
OD = OA (Radii of a circle)
ODA = OAD
          = BAD
= 60°
ODB = 90° - ODA = (90° - 60°) = 30°
Here, DO || BC (Given; alternate angles)
CBD = ODB = 30°
∠CBD = 30°
(iv)
ADC = ADB + CDB
            = 90° + 30° = 120°
In ΔAOD, we have:
ODA + OAD +AOD = 180°
⇒ 60° + 60° + AOD = 180°
AOD = 180° – 120° = 60°

Since all the angles of ΔAOD are of 60° each, ΔAOD is an equilateral triangle.

Question 13:

Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 6 cm, BP = 2 cm and PD = 25 cm, find CD.

Answer 13:


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=9.752.5=3.9

Hence, CD = 3.9 cm

Question 14:

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

Answer 14:


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° +2OBD = 180°
⇒ 2OBD = (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°

Question 15:

In the given figure ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.

Answer 15:


ABC is an isosceles triangle.
Here, AB = AC
∴ ∠ACB = ∠ABC   ...(i)
So, exterior ∠ADE = ∠ACB 
                             = ∠ABC   [from(i)]
∴ ∠ADE = ∠ABC  (Corresponding angles)
Hence, DE || BC

Question 16:

In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that AEB is isosceles.

Answer 16:


AB and CD are two parallel chords of a circle. BDE and ACE are two straight lines that intersect at E.
If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
∴ Exterior ∠EDC = ∠A   ...(i)
  Exterior ∠DCE = ∠B    ...(ii)
Also, AB parallel to CD.
Then, ∠EDC = ∠B  (Corresponding angles)
and ∠DCE = ∠A  (Corresponding angles)
∴ ∠A = ∠B     [From(i) amd (ii)]
Hence, ΔAEB is isosceles.

Question 17:

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

Answer 17:


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-485

Question 18:

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 18:


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.

Question 19:

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Answer 19:

Let ABCD be the cyclic quadrilateral and PO, QO, RO and SO be the perpendicular bisectors of sides AB, BC, CD and AD.


We know that the perpendicular bisector of a chord passes through the centre of the circle.
Since, AB, BC, CD and AD are the chords of a circle, PO, QO, RO and SO pass through the centre.
i.e., PO, QO, RO and SO are concurrent.
Hence, the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Question 20:

Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.

Answer 20:

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O.

The diagonals of a rhombus bisect each other at right angles.
i.e., ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
Now, circles with AB, BC, CD and DA as diameter passes through O (angle in a semi-circle is a right angle).
Hence, the circle with four sides of a rhombus as diameter, pass through O, i.e., the point of intersection of its diagonals.

Question 21:

ABCD is a rectangle. Prove that the centre of the circle thought A, B, C, D is the point of intersection of its diagonals.

Answer 21:

Given: ABCD is a cyclic rectangle whose diagonals intersect at O.
To prove: O is the centre of the circle.
Proof:

Here, ∠BCD = 90°     [Since it is a rectangle]
So, BD is the diameter of the circle (if the angle made by the chord at the circle is right angle, then the chord is the diameter).
Also, diagonals of a rectangle bisect each other and are equal.
∴ OA =  OB = OC = OD
BD is the diameter.
∴ BO and OD are the radius.
Thus, O is the centre of the circle.
Also, the centre of the circle is circumscribing the cyclic rectangle.
Hence, O is the point of intersection of the diagonals of ABCD.

Question 22:

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.

Answer 22:

Let A, B and C be the given points.
With B as the centre and a radius equal to AC, draw an arc.
With C as the centre and AB as radius, draw another arc intersecting the previous arc at D.
Then D is the desired point.
Proof: Join BD and CD.

In ΔABC and ΔDCB, we have:
AB =  DC
AC = DB
BC =  CB
i.e., ΔABC ≅ ΔDCB
⇒ ∠BAC = ∠CDB
Thus, BC subtends equal angles ∠BAC and ∠CDB on the same side of it.
∴ Points A, B, C and D are cyclic.

Question 23:

In a cyclic quadrilateral ABCD, if (B − ∠D) = 60°, show that the smaller of the two is 60°.

Answer 23:

In cyclic quadrilateral ABCD, we have:
B + D = 180°            ...(i)     (Opposite angles of a cyclic quadrilateral )
B - D = 60°               ...(ii)     (Given)
From (i) and (ii), we get:
2B  = 240°
B = 120°
∠D = 60°
Hence, the smaller of the two angles is 60°.

Question 24:

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 24:

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 25:

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.

Answer 25:

Draw two right triangles ACB and ADB in a circle with centre O, where AB is the diameter of the circle.

Join CO.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc CB subtends ∠COB at the centre and ∠CAB at A on the circle.

∴ ∠COB = 2∠CAB           ...(1)

Also, arc CB subtends ∠COB at the centre and ∠CDB at D on the circle.

∴ ∠COB = 2∠CDB           ...(2)

Equating (1) and (2),
2∠CAB = 2∠CDB
CAB = ∠CDB

Hence, ∠BAC = ∠BDC.

Question 26:

ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = 12BAD.

Answer 26:

In the given figure, ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D.

Join AC and BD.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc CD subtends ∠CAD at the centre and ∠CBD at B on the circle.

∴ ∠CAD = 2∠CBD           ...(1)

Also, arc CB subtends ∠CAB at the centre and ∠CDB at D on the circle.

∴ ∠CAB = 2∠CDB           ...(2)

Adding (1) and (2), we get
CAD + CAB = 2(∠CBD + CDB)
BAD = 2(∠CBD + CDB)
CBD + ∠CDB = 12BAD

Hence, ∠CBD + ∠CDB = 12BAD.

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