Circles — Exercise 15.1

Using the given information, find the value of x in the following Figure :

Answer

∠ADB = ∠ACB = 50° (∵ angles in same segment are equal.)

We know that sum of angles in a triangle is 180°.

Considering △ADB,

⇒ ∠ADB + ∠DAB + ∠ABD = 180°
⇒ 50° + 42° + x° = 180°
⇒ x° + 92° = 180°
⇒ x° = 180° - 92°
⇒ x° = 88°

Hence, the value of x = 88°.

Using the given information, find the value of x in the following figure :

Answer

From figure,

∠ACB = ∠ADB = 45° (∵ angles in same segment are equal.)

∴ ∠DCB = 32° + 45° = 77°.

Since, sum of opposite angles of a parallelogram = 180°.

∴ ∠DCB + x° = 180°
⇒ 77° + x° = 180°
⇒ x° = 180° - 77°
⇒ x° = 103°.

Hence, the value of x = 103°.

Using the given information, find the value of x in the following figure :

Answer

Considering △ABC and △ADC,

∠ABC = ∠ADC = 20° (∵ angles in same segment are equal.)

∠DOC = ∠DOB = 90° (As DA is perpendicular to BC)

We know that sum of angles in a triangle is 180°.

Considering △DOC,

⇒ ∠ODC + ∠DOC + ∠OCD = 180°
⇒ 20° + 90° + x° = 180°
⇒ x° + 110° = 180°
⇒ x° = 180° - 110°
⇒ x° = 70°

Hence, the value of x = 70°.

Using the given information, find the value of x in the following figure :

Answer

Considering △ABC and △DBC,

∠BAC = ∠BDC = x° (∵ angles in same segment are equal.)

We know that sum of angles in a triangle is 180°.

Considering △ABC,

⇒ ∠BAC + ∠ABC + ∠BCA = 180°
⇒ x° + 69° + 31° = 180°
⇒ x° + 100° = 180°
⇒ x° = 180° - 100°
⇒ x° = 80°

Hence, the value of x = 80°.

Using the given information, find the value of x in the following figure :

Answer

Considering △ACB and △CDB,

∠CAB = ∠CDB = x° (∵ angles in same segment are equal.)

Considering △ACP,

∠CPB = ∠APD = 120° (∵ vertically opposite angles are equal.)

Since exterior angle in a triangle is equal to the sum of the opposite interior angles,

⇒ ∠CAP + ∠ACP = ∠APD
⇒ x° + 70° = 120°
⇒ x° = 120° - 70°
⇒ x° = 50°

Hence, the value of x = 50°.

Using the given information, find the value of x in the following figure :

Answer

From figure,

∠DAB = ∠BCD = 25° (∵ angles in same segment are equal)

In △DAP,

Exterior angle ∠CDA = ∠DAP + ∠DPA ....(i)

From figure,

∠DAP = ∠DAB = 25°.

Putting values in equation (i),

⇒ x° = 25° + 35°
⇒ x° = 60°.

Hence, the value of x = 60°.

If O is the center of the circle, find the value of x in the following figure (using the given information) :

Answer

Considering △ACB and △ADB,

∠ACB = ∠ADB = x° (∵ angles in same segment are equal.)

Considering △ACB,

∠ABC = 90° (∵ angle in semicircle is 90°.)

Since sum of angles in a triangle is equal to 180°.

⇒ ∠BAC + ∠ACB + ∠ABC = 180°.
⇒ 40° + x° + 90° = 180°
⇒ x° + 130° = 180°
⇒ x° = 50°.

Hence, the value of x = 50°.

If O is the center of the circle, find the value of x in the following figure (using the given information) :

Answer

∠ADB = ∠ACB = x° (∵ angles in same segment are equal.)

Considering △AOD,

∠ODA = ∠OAD = 62° (∵ OD = OA, and angles of equal sides are equal.)

∴ x = 62°.

Hence, the value of x = 62°.

If O is the center of the circle, find the value of x in the following figure (using the given information) :

Answer

Angle of a circle is 360°.

⇒ ∠AOB + ∠AOC + ∠BOC = 360°
⇒ ∠AOB + 80° + 130° = 360°
⇒ ∠AOB + 210° = 360°
⇒ ∠AOB = 360° - 210°
⇒ ∠AOB = 150°.

From figure,

⇒ ∠AOB = 2∠ACB (∵ angle subtended by an arc at the center = double the angle subtended by it any point on the remaining part of the circle.)

⇒ 150° = 2∠ACB
⇒ ∠ACB = $\dfrac{150°}{2}$
⇒ ∠ACB = 75°
⇒ x° = 75°.

Hence, the value of x = 75°.

If O is the center of the circle, find the value of x in the following figure (using the given information) :

Answer

AC subtends reflex angle ∠AOC at center and ∠ABC at point B. Since angle subtended at center is double the angle subtended at any other part of the circle.

⇒ Reflex ∠AOC = 2∠ABC .....(i)

From figure,

∠ABC + ∠CBD = 180° (As they are linear pair.)

∠ABC + 75° = 180°
∠ABC = 180° - 75°
∠ABC = 105°.

Putting value of ∠ABC = 105° in (i)

x° = 2(105°)
x° = 210°

Hence, the value of x = 210.

If O is the center of the circle, find the value of x in the following figure (using the given information) :

Answer

From figure,

⇒ ∠AOC + ∠COB = 180°
⇒ ∠COB = 180° - ∠AOC
⇒ ∠COB = 180° - 135°
⇒ ∠COB = 45°.

BC subtends ∠COB = 45° at center and ∠CDB = x° at point D. Since angle subtended at center is double the angle subtended at any other part of the circle.

⇒ ∠COB = 2∠CDB

⇒ 45° = 2x°

⇒ x° = $\dfrac{45\degree}{2} = 22\dfrac{1\degree}{2}.$

Hence, the value of x = $22\dfrac{1\degree}{2}$.

If O is the center of the circle, find the value of x in the following figure (using the given information) :

Answer

Label the points as shown in the figure below:

Arc AD subtends ∠AOD = 70° at center.

⇒ ∠AOD = 2∠ABD (∵ angle subtended by an arc at the center = double the angle subtended by it any point on the remaining part of the circle.)

⇒ 70° = 2∠ABD
⇒ ∠ABD = $\dfrac{70°}{2}$
⇒ ∠ABD = 35°

∠ABM = ∠ABD.

Considering △ABM,

Since sum of angles in a triangle is equal to 180°.

⇒ ∠AMB + ∠ABM + ∠BAM = 180°.
⇒ 90° + 35° + x° = 180°
⇒ x° + 125° = 180°
⇒ x° = 55°.

Hence, the value of x = 55.

In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC.

Answer

From figure,

∠DAC = ∠ACB = 35° (∵ alternate angles are equal.)

∠DAC = ∠DBC = 35° (∵ angles in same segment are equal.)

Hence, the value of ∠DBC = 35°.

In the figure (ii) given below, it is given that O is the center of the circle and ∠AOC = 130°. Find ∠ABC.

Answer

From figure,

⇒ ∠AOC + Reflex ∠AOC = 360°
⇒ 130° + Reflex ∠AOC = 360°
⇒ Reflex ∠AOC = 360° - 130°
⇒ Reflex ∠AOC = 230°.

Arc AC subtends Reflex ∠AOC at center and ∠ABC at another point of circle.

⇒ Reflex ∠AOC = 2 ∠ABC

⇒ 2∠ABC = 230°

⇒ ∠ABC = $\dfrac{230°}{2}$

⇒ ∠ABC = 115°.

Hence, the value of ∠ABC = 115°.

In the figure (i) given below, calculate the values of x and y.

Answer

In the figure,

ABCD is a cyclic quadrilateral.

⇒ ∠B + ∠D = 180°
⇒ 40° + 45° + y = 180°
⇒ 85° + y = 180°
⇒ y = 180° - 85°
⇒ y = 95°.

Considering △ABD and △ACD,

∠ABD = ∠ACD = 40° (∵ angles in same segment are equal.)

x = 40°.

Hence, the value of x = 40° and y = 95°.

In the figure (ii) given below, O is the center of the circle. Calculate the values of x and y.

Answer

From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ 120° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - 120°
⇒ Reflex ∠AOB = 240°.

Arc AB subtends Reflex ∠AOB at center and ∠ADB at point D of circle.

⇒ Reflex ∠AOB = 2 ∠ADB
⇒ 2∠ADB = 240°
⇒ ∠ADB = $\dfrac{240°}{2}$
⇒ ∠ADB = 120°
⇒ y° = 120°

Arc AB subtends ∠AOB at center and ∠ACB at point C of circle.

⇒ ∠AOB = 2 ∠ACB
⇒ 2∠ACB = 120°
⇒ ∠ACB = $\dfrac{120°}{2}$
⇒ ∠ACB = 60°
⇒ x° = 60°

Hence, the value of x = 60 and y = 120.

In the figure (i) given below, M, A, B, N are points on a circle having center O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON.

Answer

Considering △YBN,

Since sum of angles in a triangle is equal to 180°.

⇒ ∠NYB + ∠YNB + ∠YBN = 180°.
⇒ 50° + 20° + ∠YBN = 180°
⇒ 70° + ∠YBN = 180°
⇒ ∠YBN = 180° - 70°
⇒ ∠YBN = 110°.

From figure,

∠MBN = ∠YBN = 110°.

Considering △MAN and △MBN,

∠MAN = ∠MBN = 110° (∵ angles in same segment are equal.)

MN subtends Reflex ∠MON at center and ∠MAN at point A of circle.

⇒ Reflex ∠MON = 2 ∠MAN = 2 × 110° = 220°.

Hence, ∠MAN = 110° and Reflex ∠MON = 220°.

In the figure (ii) given below, O is the center of the circle. If ∠AOB = 140° and ∠OAC = 50°, find

(i) ∠ACB

(ii) ∠OBC

(iii) ∠OAB

(iv) ∠CBA.

Answer

(i) From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ 140° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - 140°
⇒ Reflex ∠AOB = 220°.

Arc AB subtends Reflex ∠AOB at center and ∠ACB at point C of circle.

⇒ Reflex ∠AOB = 2 ∠ACB
⇒ 2∠ACB = 220°
⇒ ∠ACB = $\dfrac{220°}{2}$
⇒ ∠ACB = 110°.

Hence, the value of ∠ACB = 110°.

(ii) In Quadrilateral OABC,

⇒ ∠OAC + ∠ACB + ∠BOA + ∠OBC = 360°
⇒ 50° + 110° + 140° + ∠OBC = 360°
⇒ 300° + ∠OBC = 360°
⇒ ∠OBC = 360° - 300°
⇒ ∠OBC = 60°

Hence, the value of ∠OBC = 60°.

(iii) In △OAB,

OA = OB (Radius of the circle)

∠OAB = ∠OBA = x (∵ angles of equal sides in isosceles triangle are equal.)

Sum of angles in a triangle are equal,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°.
⇒ 140° + x + x = 180°
⇒ 140° + 2x = 180°
⇒ 2x = 40°
⇒ x = 20°.

Hence, the value of ∠OAB = 20°.

(iv) ∠CBA = ∠OBC - ∠OBA

⇒ ∠CBA = 60° - 20° ⇒ ∠CBA = 40°

Hence, the value of ∠CBA = 40°.

In the figure (i) given below, A, B, C and D are points on the circle with center O. Given that ∠ABC = 62°, find

(i) ∠ADC

(ii) ∠CAB

Answer

(i) From figure,

⇒ ∠ADC = ∠ABC (Angles in same segment are equal)

⇒ ∠ADC = 62°.

Hence, ∠ADC = 62°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∠ACB = 90°

Using angle sum property,

⇒ ∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 90° + 62° = 180°

⇒ ∠CAB + 152° = 180°

⇒ ∠CAB = 180° - 152°

⇒ ∠CAB = 28°.

Hence, ∠CAB = 28°.

In the figure (ii) given below, AB is a diameter of the circle whose center is O. Given that ∠ECD = ∠EDC = 32°, calculate

(i) ∠CEF

(ii) ∠COF

Answer

(i) In △EDC,

∠ECD = ∠EDC = 32° (Given)

Since sum of angles of triangle = 180°.

⇒ ∠DEC + ∠ECD + ∠EDC = 180°
⇒ ∠DEC + 32° + 32° = 180°
⇒ ∠DEC = 180° - 64°
⇒ ∠DEC = 116°.

Since, ∠CEF and ∠DEC are linear pair,

∴ ∠CEF + ∠DEC = 180°
⇒ ∠CEF + 116° = 180°
⇒ ∠CEF = 180° - 116°
⇒ ∠CEF = 64°

Hence, ∠CEF = 64°.

(ii) ∠FDC = ∠EDC = 32°. (From figure)

Arc FC subtends ∠COF at center and ∠FDC at point D of circle so,

⇒ ∠COF = 2 ∠FDC
⇒ ∠COF = 2 × 32°
⇒ ∠COF = 64°

Hence, the value of ∠COF = 64°.

In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find :

(i) ∠PRB

(ii) ∠PBR

(iii) ∠BPR

Answer

(i) ∠PRB = ∠PAB = 35° (∵ angles in same segment are equal.)

Hence, the value of ∠PRB = 35°

(ii) From figure,

∠APB = 90° (∵ angle in semicircle is 90°.)

⇒ ∠APB + ∠BPQ = 180° (∵ angles form a linear pair).
⇒ 90° + ∠BPQ = 180°
⇒ ∠BPQ = 90°.

Exterior angle in a triangle is equal to the sum of opposite two interior angles.

In △PBQ,

Ext. ∠PBR = ∠PQB + ∠BPQ = 25° + 90° = 115°.

Hence, the value of ∠PBR = 115°

(iii) In △PRQ,

Ext. ∠APR = ∠PRQ + ∠PQR = ∠PRB + ∠PQR = 35° + 25° = 60°.

∠APB = 90° (∵ angle in semicircle is 90°.)

From figure,

∠BPR = ∠APB - ∠APR = 90° - 60° = 30°.

Hence, the value of ∠BPR = 30°

In the figure (ii) given below, it is given that ∠ABC = 40° and AD is a diameter of the circle. Calculate ∠DAC.

Answer

Consider △ABC and △ADC,

∠ABC = ∠ADC = 40° (∵ angles in same segment are equal.)

In △ADC,

∠DCA = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠DAC + ∠ADC + ∠DCA = 180°.
⇒ ∠DAC + 40° + 90° = 180°
⇒ ∠DAC + 130° = 180°
⇒ ∠DAC = 180° - 130°
⇒ ∠DAC = 50°.

Hence, the value of ∠DAC = 50°.

In the figure (i) given below, P and Q are centers of two circles intersecting at B and C. ACD is a straight line. Calculate the value of x.

Answer

Arc AB subtends ∠APB at center and ∠ACB at the point C on the circle.

∴ ∠APB = 2∠ACB

⇒ ∠APB = 2∠ACB
⇒ 130° = 2∠ACB
⇒ ∠ACB = 65°.

From figure,

∠ACB + ∠BCD = 180°. (∵ both angles form a linear pair)

⇒ 65° + ∠BCD = 180°
⇒ ∠BCD = 180° - 65°
⇒ ∠BCD = 115°.

In circle with center Q,

⇒ ∠BQD + Reflex ∠BQD = 360°
⇒ x° + Reflex ∠BQD = 360°
⇒ Reflex ∠BQD = 360° - x°.

Arc BD subtends reflex ∠BQD at center and ∠BCD at the point C on the circle.

∴ Reflex ∠BQD = 2∠BCD

⇒ 360° - x° = 2 × 115°
⇒ 360° - x° = 230°
⇒ x° = 360° - 230°
⇒ x° = 130°.

Hence, the value of x = 130.

In the figure (ii) given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate

(i) ∠CAB

(ii) ∠OAC.

Answer

(i) From figure,

AC = BC so,

∠CBA = ∠CAB (As angles of equal sides are equal)

In △ABC,

∠CAB + ∠CBA + ∠ACB = 180°
2∠CAB + 56° = 180°
2∠CAB = 180° - 56°
2∠CAB = 124°
∠CAB = 62°.

Hence, ∠CAB = 62°.

(ii) OC is the radius of the circle. OC bisects ∠ACB.

∠OCA = $\dfrac{1}{2}$∠ACB = $\dfrac{1}{2} \times$ 56° = 28°.

Now in △OCA,

OA = OC (Radius of the same circle)

∠OAC = ∠OCA = 28°.

Hence, ∠OAC = 28°.

In the figure (i) given below, chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Answer

Consider △AEC and △EBC,

∠EAC = ∠EBC = 65° (∵ angles in same segment are equal.)

In △AEC,

∠AEC = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠AEC + ∠EAC + ∠ACE = 180°.
⇒ 90° + 65° + ∠ACE = 180°
⇒ ∠ACE + 155° = 180°
⇒ ∠ACE = 180° - 155°
⇒ ∠ACE = 25°.

We know that,

Alternate angles are equal.

∴ ∠DEC = ∠ACE = 25°.

Hence, the value of ∠DEC = 25°.

In the figure (ii) given below, C is a point on the minor arc AB of the circle with centre O. Given ∠ACB = p°, ∠AOB = q°, express q in terms of p. Calculate p if OACB is a parallelogram.

Answer

From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ q° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - q°.

Arc AB subtends reflex ∠AOB at center and ∠ACB at the point C on the circle.

∴ Reflex ∠AOB = 2∠ACB

⇒ 360° - q° = 2 × p°
⇒ 360° - q° = 2p°
⇒ 2p° + q° = 360°
⇒ q° = 360° - 2p°
⇒ q° = 2(180° - p°)
⇒ q = 2(180 - p).

Given, OABC is a parallelogram, then

Opposite angles are equal.

∴ ∠AOB = ∠ACB

⇒ p° = q°
⇒ p° = 360° - 2p°
⇒ 3p° = 360°
⇒ p° = 120°.

Hence, q = 2(180 - p) and the value of p = 120.

In the figure (i) given below, straight lines AB and CD pass through the center O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in

(i) ∠CDE

(ii) ∠OBE.

Answer

(i) In △CED,

∠CED = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠CED + ∠DCE + ∠CDE = 180°.
⇒ 90° + 40° + ∠CDE = 180°
⇒ ∠CDE + 130° = 180°
⇒ ∠CDE = 180° - 130°
⇒ ∠CDE = 50°.

Hence, the number of degrees in ∠CDE = 50.

(ii) From figure,

∠AOD + ∠DOB = 180° (∵ they form linear pair)
⇒ 75° + ∠DOB = 180°
⇒ ∠DOB = 180° - 75°
⇒ ∠DOB = 105°.

In △DOB,

∠ODB = ∠CDE = 50°

We know that sum of angles of a triangle is 180°.

⇒ ∠DOB + ∠ODB + ∠DBO = 180°.
⇒ 105° + 50° + ∠DBO = 180°
⇒ ∠DBO + 155° = 180°
⇒ ∠DBO = 180° - 155°
⇒ ∠DBO = 25°.

From figure,

∠OBE = ∠DBO

∴ ∠OBE = 25°.

Hence, the number of degrees in ∠OBE = 25.

In the figure (ii) given below, I is the incentre of △ABC. AI produced meets the circumcircle of △ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate

(i) ∠BCD

(ii) ∠CBD

(iii) ∠DCI

(iv) ∠BIC.

Answer

(i) Join BI and CI as shown in the figure below:

In △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180° (∵ sum of angles = 180°.)
⇒ ∠BAC + 55° + 65° = 180°
⇒ ∠BAC + 120° = 180°
⇒ ∠BAC = 180° - 120°
⇒ ∠BAC = 60°.

I is the incentre,

∴ I lies on the bisectors of angle of the △ABC,

∴ ∠BAD = ∠CAD = $\dfrac{60°}{2}$ = 30°.

∠BCD = ∠BAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠BCD = 30°

(ii) Similarly,

∠CBD = ∠CAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠CBD = 30°

(iii) The line CI bisects ∠C (∵ I lies on the bisectors of angle of the △ABC).

∴ ∠BCI = $\dfrac{65°}{2} = 32\dfrac{1}{2}°$.

From figure,

∠DCI = ∠BCD + ∠BCI = $30° + 32\dfrac{1}{2}° = 62\dfrac{1}{2}°$.

Hence, the value of ∠DCI = $62\dfrac{1}{2}°$.

(iv) ∠IBC = $\dfrac{55°}{2} = 27\dfrac{1}{2}°$

∠ICB = $\dfrac{65°}{2} = 32\dfrac{1}{2}°$

∠BIC = 180° - (∠IBC + ∠ICB)

$= 180° - \Big(\dfrac{55°}{2} + \dfrac{65°}{2}\Big) \\[1em] = 180° - \Big(\dfrac{120°}{2}\Big) \\[1em] = 180° - 60° \\[1em] = 120°.$

Hence, the value of ∠BIC = 120°.

O is the circumcentre of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A.

Answer

From the below figure:

Arc BC subtends ∠BOC at center and ∠BAC at the point A on the circle.

∴ ∠BOC = 2∠A

In △OBD and △ODC,

OD = OD (Common side)

BD = CD (As D is the mid-point of BC)

OB = OC (Radius of the same circle)

∴ △OBD ≅ △ODC (SSS rule of congruency).

∴ ∠BOD = ∠COD (As corresponding part of congruent triangles are congruent.)

Since, ∠BOD = ∠COD so,

∠BOD = $\dfrac{1}{2}$∠BOC ....(i)

∠BOC = 2∠A
∠A = $\dfrac{1}{2}$ ∠BOC .....(ii)

From (i) and (ii) we get,

∠BOD = ∠A

Hence, proved that ∠BOD = ∠A.

In the adjoining figure, AB and CD are equal chords. AD and BC intersects at E. Prove that AE = CE and BE = DE.

Answer

In △AEB and △CED,

∠A = ∠C (∵ angles in same segment of a circle are equal.)

∠B = ∠D (∵ angles in same segment of a circle are equal.)

AB = CD (Given)

∴ △AEB ≅ △CED (By ASA axiom)

As corresponding part of congruent triangles are congruent hence,

AE = CE and BE = DE.

Hence, proved that AE = CE and BE = DE.

In the figure (i) given below, AB is a diameter of a circle with center O. AC and BD are perpendiculars on a line PQ. BD meets the circle at E. Prove that AC = ED.

Answer

Join AE.

∠AEB = 90° (∵ angle in semicircle is 90°.)

∠AED = 90° (∵ ∠AEB and ∠AED form a linear pair.)

Hence, we can say that,

DE is also perpendicular to AE, since DE is also perpendicular to PQ hence,

AE || PQ.

Since, CA and DE both are perpendicular to PQ hence,

CA || DE.

Hence, proved that ACDE is a rectangle.

In rectangle opposite sides are equal so,

AC = DE.

Hence, proved that AC = DE.

In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.

Answer

Join OC and OD as shown in the figure below:

AC subtends angle AOC at centre and ∠ABC at point B.

∴ ∠AOC = 2∠ABC = 2 × 25° = 50°.

From figure,

∠OCD = ∠AOC (Alternate angles)

Hence, ∠OCD = 50°.

In △OCD,

OC = OD (Both are radius of the circle)

so, ∠ODC = ∠OCD.

Since, sum of angles of a triangle is 180°.

⇒ ∠COD + ∠OCD + ∠ODC = 180°
⇒ ∠COD + 50° + 50° = 180°
⇒ ∠COD + 100° = 180°
⇒ ∠COD = 80°.

CD subtends ∠COD at center and ∠CED at point E of the circle.

∴ ∠COD = 2∠CED
⇒ 80° = 2∠CED
⇒ ∠CED = 40°.

Hence, ∠CED = 40°.

In the adjoining figure, O is the center of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D. Prove that ∠ABC = 2∠OAD.

Answer

Join AD.

Arc AC subtends ∠AOC at the center and ∠ADC at the point D of the circle.

∴ ∠AOC = 2∠ADC (As angle at center = double the angle at the remaining part of the circle)

∠OAD = ∠ADC (∵ alternate angles are equal.)

∴ ∠AOC = 2∠OAD .....(i)

Since, opposite angles are equal in parallelogram,

∴ ∠ABC = ∠AOC

Putting values of ∠AOC in eqn (i) we get,

∠ABC = 2∠OAD.

Hence, proved that ∠ABC = 2∠OAD.

In figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.

Answer

Given, two chords AQ and BC intersect each other at P inside the circle. AB and CQ are joined and AB = AP.

To prove : CP = CQ

Construction : Join AC.

Proof :

In △ABP and △CQP

∠B = ∠Q (∵ angles in same segment are equal)

∠BAP = ∠PCQ (∵ angles in same segment are equal)

∠BPA = ∠CPQ (∵ vertically opposite angles are equal.)

∴ △ABP ~ △CQP (By AAA axiom of similarity.)

Since, triangles are similar hence, the ratio of the corresponding sides are equal.

$\dfrac{AB}{CQ} = \dfrac{AP}{CP}$

We know, AB = AP,

$\therefore \dfrac{AP}{CQ} = \dfrac{AP}{CP} \\[1em] CQ = CP.$

Hence, proved that CQ = CP.

In the figure (ii) given below, AB = AC = CD, ∠ADC = 38°. Calculate

(i) ∠ABC

(ii) ∠BEC.

Answer

(i) In △ACD,

AC = CD

∴ ∠CAD = ∠ADC = 38° (∵ angles of equal sides in triangle are equal)

∠ACB = ∠CAD + ∠ADC = 38° + 38° = 76° (∵ exterior angle = sum of two opposite interior angles.)

In △ABC,

AB = AC

∴ ∠ABC = ∠ACB = 76° (As angles of equal sides in triangle are equal)

Hence, the value of ∠ABC = 76°.

(ii) We know that sum of angles in a triangle = 180°.

⇒ ∠BAC + ∠ABC + ∠ACB = 180°
⇒ ∠BAC + 76° + 76° = 180°
⇒ ∠BAC + 152° = 180°
⇒ ∠BAC = 180° - 152°
⇒ ∠BAC = 28°.

∠BEC = ∠BAC (∵ angles in same segment are equal.)

∠BEC = 28°.

Hence, the value of ∠BEC = 28°.

In the figure (i) given below, CP bisects ∠ACB. Prove that DP bisects ∠ADB.

Answer

From figure,

∠ACB = ∠ADB (∵ angles in same segment are equal.) ....(i)

∠ACP = ∠ADP (∵ angles in same segment are equal.) ....(ii)

We know,

∠ACP = $\dfrac{1}{2}$∠ACB (∵ CP bisects ∠ACB.) .....(iii)

Using values from eq (i) and eq (ii) and putting in eq (iii) we get,

∠ADP = $\dfrac{1}{2}$∠ADB

Hence, proved that DP bisects ∠ADB.

In the figure (ii) given below, BD bisects ∠ABC. Prove that $\dfrac{AB}{BD} = \dfrac{BE}{BC}$.

Answer

Join CD as shown in the figure below:

In △ABE and △BCD,

∠A = ∠D (∵ angles in same segment are equal.)

∠ABE = ∠DBC (As BD is bisector of ∠ABC)

△ABE ~ △BCD (AA rule of similarity).

Since, ratio of corresponding sides of similar triangles are equal,

∴ $\dfrac{AB}{BD} = \dfrac{BE}{BC}.$

Hence, proved.

In the figure (i) given below, chords AB and CD of a circle intersect at E.

(i) Prove that triangles ADE and CBE are similar.

(ii) Given DC = 12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.

Answer

(i) In △CBE and △ADE,

∠B = ∠D (∵ angles in same segment are equal.)

∠BEC = ∠DEA (∵ vertically opposite angles are equal.)

△CBE ~ △ADE. (By AA axiom)

Hence, proved that △CBE ~ △ADE.

(ii) Given, DC = 12 cm.

From figure,

⇒ DC = DE + EC
⇒ 12 = 4 + EC
⇒ EC = 12 - 4
⇒ EC = 8 cm.

Chords AB and CD intersect each other at E.

Considering △BEC and △AED,

∠BEC = ∠DEA (∵ vertically opposite angles are equal.)

∠CBE = ∠EDA (∵ both angles are subtended on circle by arc AC and angles in same segment are equal.)

Hence, △BEC ~ △AED.

Since triangles are similar hence the ratio of the corresponding sides are equal.

$\therefore \dfrac{BE}{DE} = \dfrac{EC}{EA} \\[1em] \Rightarrow BE \times EA = EC \times DE \\[1em] BE \times 16 = 8 \times 4 \\[1em] BE = \dfrac{32}{16} \\[1em] BE = 2.$

Hence, the length of BE = 2 cm.

In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6 cm, PB = 4 cm, and CD = 14 cm (PC > PD).

Answer

In △APD and △CPB,

∠DAB = ∠DCB (∵ angles in same segment are equal.)

∠APD = ∠CPB (∵ vertically opposite angles are equal.)

△APD ~ △CPB. (By AA axiom)

Hence, proved that △APD ~ △CPB.

Chords AB and CD intersect each other at P.

Since △APD ~ △CPB, Hence, the ratio of corresponding sides are equal.

$\therefore \dfrac{AP}{CP} = \dfrac{PD}{PB}$

∴ AP × PB = CP × PD .....(i)

From figure,

CD = CP + PD

Let CP = x cm.

⇒ 14 = x + PD
⇒ PD = (14 - x) cm.

Putting values in eq (i)

⇒ 6 × 4 = x × (14 - x)

⇒ 24 = 14x - x²

⇒ x² - 14x + 24 = 0

⇒ x² - 12x - 2x + 24 = 0

⇒ x(x - 12) - 2(x - 12) = 0

⇒ (x - 2)(x - 12) = 0

⇒ x - 2 = 0 or x - 12 = 0

⇒ x = 2 or x = 12.

Since, given PC > PD so, CP = 12 cm.

Hence, the length of CP = 12 cm.

In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE.

Answer

Join A and B as shown in the figure below:

From figure,

Since, ∠CDE = 90° so, ∠ADB = 90° (∵ vertically opposite angles are equal.)

In right angle triangle △ADB, by pythagoras theorem,

$\Rightarrow AB^2 = AD^2 + BD^2 \\[1em] \Rightarrow 5^2 = AD^2 + 4^2 \\[1em] \Rightarrow 25 = AD^2 + 16 \\[1em] \Rightarrow AD^2 = 25 - 16 \\[1em] \Rightarrow AD^2 = 9 \\[1em] \Rightarrow AD = 3 \text{ cm}.$

Chords AE and CB intersect each other at D.

In △ADB and △CDE,

∠BAD = ∠DCE (∵ angles in same segment are equal.)

∠ADB = ∠CDE (∵ vertically opposite angles are equal.)

△ADB ~ △CDE. (By AA axiom)

Since △ADB ~ △CDE, Hence, the ratio of corresponding sides are equal.

$\therefore \dfrac{AD}{CD} = \dfrac{BD}{DE}$

∴ AD × DE = CD × BD

⇒ 3 × DE = 9 × 4

⇒ DE = $\dfrac{36}{3}$

⇒ DE = 12 cm.

Hence, the length of DE = 12 cm.

In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.

Answer

We know that ∠PQR = 90° as angle in semicircle is equal to 90°.

So, △PQR is a right angled triangle.

By pythagoras theorem,

$PR^2 = PQ^2 + QR^2 \\[1em] PR^2 = 7^2 + 6^2 \\[1em] PR^2 = 49 + 36 \\[1em] PR^2 = 85.$

In △PRS,

∠PSR = 90° as angle in semicircle is equal to 90°.

So, △PRS is a right angled triangle.

By pythagoras theorem,

$PR^2 = PS^2 + RS^2 \\[1em] 85 = PS^2 + 2^2 \\[1em] PS^2 = 85 - 4 \\[1em] PS^2 = 81 \\[1em] PS = 9.$

Perimeter of PQRS = PQ + QR + RS + SP = 7 + 6 + 2 + 9 = 24 cm.

Hence, the perimeter of cyclic quadrilateral is 24 cm.

In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of △DPC.

Answer

In △ABP and △DPC,

∠APB = ∠DPC (∵ vertically opposite angles are equal.)

∠ABP = ∠DCP (∵ angles in same segment are equal.)

△APB ~ △DPC. (By AA axiom)

We know that ratio of the area of similar triangles is the ratio of their corresponding sides.

$\therefore \dfrac{\text{Area of △APB}}{\text{Area of △DPC}} = \dfrac{AB^2}{CD^2} \\[1em] \Rightarrow \dfrac{24}{\text{Area of △DPC}} = \dfrac{64}{25} \\[1em] \Rightarrow \text{Area of △DPC} = \dfrac{24 \times 25}{64} \\[1em] \Rightarrow \text{Area of △DPC} = \dfrac{600}{64} \\[1em] \Rightarrow \text{Area of △DPC} = \dfrac{75}{8} \\[1em] \Rightarrow \text{Area of △DPC} = 9\dfrac{3}{8}. \\[1em]$

Hence, the area of △DPC = $9\dfrac{3}{8}$ cm².

In adjoining figure, AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P.

(i) Prove that Δ PAB ∼ Δ PCD.

(ii) Find the length of CD.

(iii) Find the area of Δ PAB : area of Δ PCD.

Answer

(i) As we know that,

If two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

∴ PA.PD = PB.PC

$\Rightarrow \dfrac{PA}{PC} = \dfrac{PB}{PD}$

⇒ ∠APB = ∠CPD (Vertically opposite angles)

If the corresponding sides of two triangles are proportional and one angle are equal, then the two triangles are similar.

Hence, proved that Δ PAB ∼ Δ PCD (By SAS rule of similarity).

(ii) Since, Δ PAB ∼ Δ PCD

$\therefore \dfrac{PA}{PC} = \dfrac{AB}{CD}$

Substituting the values, we get :

$\Rightarrow \dfrac{7.5}{5} = \dfrac{9}{CD} \\[1em] \Rightarrow 1.5 = \dfrac{9}{CD} \\[1em] \Rightarrow CD = \dfrac{9}{1.5} \\[1em] \Rightarrow CD = 6.$

Hence, the length of CD = 6 cm.

(iii) As we know that,

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$\therefore \dfrac{\text{Area of ΔPAB}}{\text{Area of ΔPCD}} = \dfrac{PA^2}{PC^2} \\[1em] = \dfrac{(7.5)^2}{5^2} \\[1em] = \dfrac{56.25}{25} \\[1em] = \dfrac{9}{4}$

Hence, the area of ΔPAB : area of ΔPCD = 9 : 4.

In the figure (i) given below, QPX is the bisector of ∠YXZ of the triangle XYZ. Prove that XY : XQ = XP : XZ.

Answer

In △XYQ and △XPZ,

∠Q = ∠Z (∵ angles in same segment are equal as arc XY subtends both the angles at the circle.)

∠YXQ = ∠PXZ (∵ QPX is the bisector of ∠YXZ hence it divides the angle in two equal halves.)

△XYQ ~ △XPZ. (By AA axiom)

Since triangles are similar hence, the ratio of corresponding sides are similar,

$\therefore \dfrac{XY}{XP} = \dfrac{XQ}{XZ} \\[1em] \Rightarrow \dfrac{XY}{XQ} = \dfrac{XP}{XZ} \\[1em] \Rightarrow XY : XQ = XP : XZ \\[1em]$

Hence, proved that XY : XQ = XP : XZ.

In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that

(i) ∠PAD = ∠PCB

(ii) PA × PB = PC × PD.

Answer

(i) From figure,

∠PAD + ∠DAB = ∠PCB + ∠BCD (∵ both are equal to 180°)....(i)

∠DAB = ∠BCD (∵ angles in same segment are equal.)

Putting this value of ∠DAB in (i) we get,

⇒ ∠PAD + ∠BCD = ∠PCB + ∠BCD
⇒ ∠PAD = ∠PCB + ∠BCD - ∠BCD
⇒ ∠PAD = ∠PCB

Hence, proved that ∠PAD = ∠PCB.

(ii) In △PBC and △PAD,

∠PAD = ∠PCB (Proved above.)

∠P = ∠P (Common angle.)

△PBC ~ △PAD. (By AA axiom)

Since triangles are similar hence, the ratio of corresponding sides are similar,

$\therefore \dfrac{PC}{PA} = \dfrac{PB}{PD} \\[1em] \Rightarrow PA \times PB = PC \times PD.$

Hence proved that PA × PB = PC × PD.