myhelper: 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
⇒ $∠ A B C = \frac{230 °}{2} = 115 °$ $∠ A B C = \frac{230 °}{2} = 115 °$ ...(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 = 2∠ADC
⇒ 100° = 2∠ADC
∴ ∠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, $x = \frac{1}{2} (reflex ∠ B O D) = (\frac{1}{2} \times 210 °) = 105 °$ $x = \frac{1}{2} (reflex ∠ B O D) = (\frac{1}{2} \times 210 °) = 105 °$
∴ 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

⇒ $x = \frac{9.75}{2.5} = 3.9$ $x = \frac{9.75}{2.5} = 3.9$

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° +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°

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:
2∠B = 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 = $\frac{1}{2}$ ∠BAD.

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 = $\frac{1}{2}$ ∠BAD

Hence, ∠CBD + ∠CDB = $\frac{1}{2}$ ∠BAD.

RS AGGARWAL CLASS 9 CHAPTER 12 CIRCLES EXERCISE 12C