ICSE Solutions for Selina Concise Chapter 17 Circles Class 10 Maths
Exercise 17(A)
1. In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30o and 40o respectively. Find ∠AOC Show your steps of working.
Solution
Firstly, let’s join AC.
And, let ∠OAC = ∠OCA = x [Angles opposite to equal sides are equal]
So, ∠AOC = 180° – 2x
Also,
∠BAC = 30° + x
∠BCA = 40° + x
Now, in ∆ABC
∠ABC = 180° – ∠BAC – ∠BCA [Angles sum property of a triangle]
= 180° – (30° + x) – (40o + x)
= 110° – 2x
And, ∠AOC = 2∠ABC
[Angle at the center is double the angle at the circumference subtend by the same chord]
180° – 2x = 2(110° – 2x)
⇒ 2x = 40°
⇒ x = 20°
Thus, ∠AOC = 180° – 2×20° = 140°
2. In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Solution
(i) In ∆ABD,
∠DAB + ∠ABD + ∠ADB = 180°
⇒ 65° + 70° + ∠ADB = 180°
⇒ 135° + ∠ADB = 180°
⇒ ∠ADB = 180° – 135° = 45°
Now,
∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
As ∠ADC is the angle of semi-circle for AC as the diameter of the circle.
(ii) ∠ACB = ∠ADB [Angles in the same segment of a circle]
Hence, ∠ACB = 45°
3. Given O is the centre of the circle and ∠AOB = 70°.
Calculate the value of:
(i) ∠OCA,
(ii) ∠OAC.
Solution
Here, ∠AOB = 2∠ACB
[Angle at the center is double the angle at the circumference subtend by the same chord]
∠ACB = 70°/2 = 35°
Now, OC = OA [Radii of same circle]
Thus,
∠OCA = ∠OAC = 35°
4. In each of the following figures, O is the centre of the circle. Find the values of a, b and c.
Solution
Here, b = ½ ×130°
[Angle at the center is double the angle at the circumference subtend by the same chord]
Thus, b = 65°
Now,
a + b = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
⇒ a = 180° – 65° = 115°
Here, c = ½ × Reflex (112°)
[Angle at the center is double the angle at the circumference subtend by the same chord]
Thus, c = ½ ×(360o – 112o) = 124°
a = 115°, b = 65°, c = 124°
5. In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.
Solution
Here, ∠BAD = 90° [Angle in a semi-circle]
So, ∠BDA = 90° – 35° = 55°
And,
a = ∠ACB = ∠BDA = 55°
[Angles subtended by the same chord on the circle are equal]
Here, ∠DAC = ∠CBD = 25°
[Angles subtended by the same chord on the circle are equal]
And, we have
120° = b + 25°
[Exterior angle property of a triangle]
b = 95°
∠AOB = 2∠AOB = 2×50° = 100°
[Angle at the center is double the angle at the circumference subtend by the same chord]
Also, OA = OB
∠OBA = ∠OAB = c
c = (180°– 100°)/2 = 40°
We have, ∠APB = 90o [Angle in a semicircle]
∠BAP = 90° – 45° = 45°
Now, d = ∠BCP = ∠BAP = 45°
[Angles subtended by the same chord on the circle are equal]
a = 55°, b = 95°, c= 40°, d = 45°
6. In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centers of two circles.
Solution
It’s seen that,
∠DBA = ∠CBA = 90° [Angle in a semi-circle is a right angle]
So, adding both
∠DBA + ∠CBA = 180°
Thus, DBC is a straight line i.e. D, B and C form a straight line.
7. In the figure, given below, find:
(i) ∠BCD,
(ii) ∠ADC,
(iii) ∠ABC.
Show steps of your working.
Solution
From the given fig, it’s seen that
In cyclic quadrilateral ABCD, DC || AB
And given, ∠DAB = 105°
(i) So, ∠BCD = 180° – 105° = 75°
[Sum of opposite angles in a cyclic quadrilateral is 180o]
(ii) Now,
∠ADC and ∠DAB are corresponding angles.
So,
∠ADC + ∠DAB = 180°
∠ADC = 180° – 105°
Thus, ∠ADC = 75°
(iii) We know that, the sum of angles in a quadrilateral is 360°
So,
∠ADC + ∠DAB +∠BCD + ∠ABC = 360°
75° + 105° + 75° + ∠ABC = 360°
∠ABC = 360° – 255°
Thus, ∠ABC = 105°
8. In the figure, given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°;
find:
(i) ∠ACB,
(ii) ∠OBC,
(iii) ∠OAB,
(iv) ∠CBA.
Solution
Given, ∠AOB = 140° and ∠OAC = 50°
(i) Now,
∠ACB = ½ Reflex (∠AOB) = ½ (360° – 140°) = 110°
[Angle at the center is double the angle at the circumference subtend by the same chord]
(ii) In quadrilateral OBCA,
∠OBC + ∠ACB + ∠OCA + ∠AOB = 360° [Angle sum property of a quadrilateral]
∠OBC + 110° + 50° + 140° = 360°
Thus, ∠OBC = 360° – 300° = 60°
(iii) In ∆AOB, we have
OA = OB (radii)
So, ∠OBA = ∠OAB
Hence, by angle sum property of a triangle
∠OBA + ∠OAB + ∠AOB = 180°
⇒ 2∠OBA + 140° = 180°
⇒ 2∠OBA = 40°
⇒ ∠OBA = 20°
(iv) We already found, ∠OBC = 60°
And, ∠OBC = ∠CBA + ∠OBA
⇒ 60o = ∠CBA + 20°
∴ ∠CBA = 40°
9. Calculate:
(i) ∠CDB,
(ii) ∠ABC,
(iii) ∠ACB.
Solution
Here, we have
∠CDB = ∠BAC = 49°
∠ABC = ∠ADC = 43°
[Angles subtended by the same chord on the circle are equal]
Now, by angle sum property of a triangle we have
∠ACB = 180° – 49° – 43° = 88°
10. In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75o; ∠ABD = 58o and ∠ADC = 77o.
Find:
(i) ∠BDC,
(ii) ∠BCD,
(iii) ∠BCA.
Solution
(i) By angle sum property of triangle ABD,
∠ADB = 180o – 75o – 58o = 47o
Thus, ∠BDC = ∠ADC – ∠ADB = 77o – 47o = 30o
(ii) ∠BAD + ∠BCD = 180o
[Sum of opposite angles of a cyclic quadrilateral is 180o]
Thus, ∠BCD = 180o – 75o = 105o
(iii) ∠BCA = ∠ADB = 47o
[Angles subtended by the same chord on the circle are equal]
11. In the figure given below, O is the centre of the circle and triangle ABC is equilateral.
Find:
(i) ∠ADB,
(ii) ∠AEB
Solution
(i) As, it’s seen that ∠ACB and ∠ADB are in the same segment,
So, ∠ADB = ∠ACB = 60o
(ii) Now, join OA and OB.
And, we have
∠AEB = ½ Reflex (∠AOB) = ½ (360o – 120o) = 120o
[Angle at the center is double the angle at the circumference subtend by the same chord]
12. Given: ∠CAB = 75o and ∠CBA = 50o. Find the value of ∠DAB + ∠ABD.
Solution
Given, ∠CAB = 75o and ∠CBA = 50o
In ∆ABC, by angle sum property we have
∠ACB = 180o – (∠CBA + ∠CAB)
= 180o – (50o + 75o) = 180o – 125o
= 55o
And,
∠ADB = ∠ACB = 55o
[Angles subtended by the same chord on the circle are equal]
Now, taking ∆ABD
∠DAB + ∠ABD + ∠ADB = 180o
⇒ ∠DAB + ∠ABD + 55o = 180o
⇒ ∠DAB + ∠ABD = 180o – 55o
⇒ ∠DAB + ∠ABD = 125o
13. ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130o, find ∠BAC.
Solution
From the fig. its seem that,
∠ACB = 90o [Angle in a semi-circle is 90o]
Also,
∠ABC = 180o – ∠ADC = 180o – 130o = 50o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
By angle sum property of the right triangle ACB, we have
∠BAC = 90o – ∠ABC
= 90o – 50o
Thus, ∠BAC = 40o
14. In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110o, find ∠BDC.
Solution
Let’s join AD first.
So, we have
∠ADC = ½ ∠AOC = ½ x 110o = 55o
[Angle at the centre is double the angle at the circumference subtended by the same chord]
Also, we know that
∠ADB = 90o
[Angle in the semi-circle is a right angle]
∴ ∠BDC = 90o – ∠ADC = 90o – 55o
∠BDC = 35o
15. In the following figure, O is the centre of the circle; ∠AOB = 60o and ∠BDC = 100o, find ∠OBC.
Solution
Form the figure, we have
∠ACB = ½ ∠AOB = ½ x 60o = 30o
[Angle at the centre is double the angle at the circumference subtended by the same chord]
Now, by applying angle sum property in ∆BDC,
∠DBC = 180o – 100o – 30o = 50o
∴ ∠OBC = 50o
16. In ABCD is a cyclic quadrilateral in which ∠DAC = 27o, ∠DBA = 50o and ∠ADB = 33o. Calculate (i) ∠DBC, (ii) ∠DCB, (iii) ∠CAB.
Solution
(i) It’s seen that,
∠DBC = ∠DAC = 27o
[Angles subtended by the same chord on the circle are equal]
(ii) It’s seen that,
∠ACB = ∠ADB = 33o
And,
∠ACD = ∠ABD = 50o
[Angles subtended by the same chord on the circle are equal]
Thus,
∠DCB = ∠ACD + ∠ACB = 50o + 33o = 83o
(iii) In quad. ABCD,
∠DAB + ∠DCB = 180o
27o + ∠CAB + 83o = 180o
Thus,
∠CAB = 180o – 110o = 70o
17. In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80o and ∠CDE = 40o. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.
Solution
(i) Form the fig. its seen that,
∠DCE = 90o – ∠CDE = 90o – 40o = 50o
∴ ∠DEC = ∠OCB = 50o
(ii) In ∆BOC, we have
∠AOC = ∠OCB + ∠OBC [Exterior angle property of a triangle]
∠OBC = 80o – 50o = 30o [Given ∠AOC = 80o]
∴ ∠ABC = 30o
18. In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Solution
Firstly, join OB.
Then, ∠OBA = 90o [Angle in a semi-circle is a right angle]
That is, OB is perpendicular to AE.
Now, we know that the perpendicular draw from the centre to a chord bisects the chord.
∴ AB = BE
19. In the following figure,
(i) if ∠BAD = 96o, find ∠BCD and ∠BFE.
(ii) Prove that AD is parallel to FE.
Solution
(i) ABCD is a cyclic quadrilateral
So, ∠BAD + ∠BCD = 180o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
∠BCD = 180o – 96o = 84o
And, ∠BCE = 180o – 84o = 96o [Linear pair of angles]
Similarly, BCEF is a cyclic quadrilateral
So, ∠BCE + ∠BFE = 180o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
∠BFE = 180o – 96o = 84o
(ii) Now, ∠BAD + ∠BFE = 96o + 84o = 180o
But these two are interior angles on the same side of a pair of lines AD and FE.
∴ AD || FE.
20. Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.
Solution
(i) Let’s assume that ABCD is a parallelogram which is inscribed in a circle.
So, we have
∠BAD = ∠BCD [Opposite angles of a parallelogram are equal]
And ∠BAD + ∠BCD = 180o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
So, 2∠BAD = 180o
Thus, ∠BAD = ∠BCD = 90o
Similarly, the remaining two angles are 90o each and pair of opposite sides are equal.
∴ ABCD is a rectangle.
Hence, Proved
(ii) Let’s assume that ABCD is a rhombus which is inscribed in a circle.
So, we have
∠BAD = ∠BCD [Opposite angles of a rhombus are equal]
And ∠BAD + ∠BCD = 180o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
So, 2∠BAD = 180o
Thus, ∠BAD = ∠BCD = 90o
Similarly, the remaining two angles are 90o each and all the sides are equal.
∴ ABCD is a square.
Hence, Proved
21. In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.
Solution
Give, AB = AC
So, ∠B = ∠C … (1)
[Angles opposite to equal sides are equal]
And, DECB is a cyclic quadrilateral.
So, ∠B + ∠DEC = 180o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
∠C + ∠DEC = 180o …. (Using 1)
But this is the sum of interior angles on one side of a transversal.
DE || BC.
But, ∠ADE = ∠B and ∠AED = ∠C [Corresponding angles]
Thus, ∠ADE = ∠AED
AD = AE
⇒ AB – AD = AC = AE [As AB = AC]
⇒ BD = CE
Hence, we have DE || BC and BD = CE
∴ DECB is an isosceles trapezium.
22. Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Solution
Let O and O’ be the centres of two intersecting circles, where points of the intersection are P and Q and PA and PB are their diameters respectively.
Join PQ, AQ and QB.
Thus, ∠AQP = 90o and ∠BQP = 90o
[Angle in a semicircle is a right angle]
Now, adding both these angles we get
∠AQP + ∠BQP = 180o
∠AQB = 180o
∴ the points A, Q and B are collinear.
23. The figure given below, shows a circle with centre O.
(i) Find the relationship between a and b
(ii) Find the measure of angle OAB, if OABC is a parallelogram.
Solution
Given: ∠AOC = a and ∠ABC = b.
(i) It’s seen that,
∠ABC = ½ Reflex (∠COA)
[Angle at the centre is double the angle at the circumference subtended by the same chord]
So, b = ½ (360o - a)
a + 2b = 180o ….. (1)
(ii) As OABC is a parallelogram, the opposite angles are equal.
So, a = b
Now, using the above relationship in (1)
3a = 180o
⇒ a = 60o
Also, OC || BA
∠COA + ∠OAB = 180o
⇒ 60o + ∠OAB = 180o
∴ ∠OAB = 120o
24. Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the center O is equal to twice the angle APC
Solution
Required to prove: ∠AOC + ∠BOD = 2∠APC
OA, OB, OC and OD are joined.
Also, AD is joined.
Now, it’s seen that
∠AOC = 2∠ADC …. (1)
[Angle at the centre is double the angle at the circumference subtended by the same chord]
Similarly,
∠BOD = 2∠BAD …. (2)
Adding (1) and (2), we have
∠AOC + ∠BOD = 2∠ADC + 2∠BAD
= 2(∠ADC + ∠BAD) ….. (3)
And in ∆PAD,
Ext. ∠APC = ∠PAD + ∠ADC
= ∠BAD + ∠ADC …. (4)
So, from (3) and (4) we have
∠AOC + ∠BOD = 2∠APC
25. In the figure given RS is a diameter of the circle. NM is parallel to RS and 
MRS = 29o
Calculate: (i) ∠RNM;
(ii) ∠NRM.
Solution
(i) Join RN and MS
∠RMS = 90o [Angle in a semi-circle is a right angle]
So, by angle sum property of ∆RMS
∠RMS = 90o – 29o = 61o
And, ∠RNM = 180o – ∠RSM = 180o – 61o = 119o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
(ii) Now as RS || NM,
∠NMR = ∠MRS = 29o [Alternate angles]
⇒ ∠NMS = 90o + 29o = 119o
Also, we know that
∠NRS + ∠NMS = 180o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
∠NRM + 29o + 119o = 180o
⇒ ∠NRM = 180o – 148o
∴ ∠NRM = 32o
26. In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25o; find the angle AEB. Give reasons in support of your answer.
Solution
Join AC and BD.
So, we have
∠CAD = 90o and ∠CBD = 90o
[Angle is a semicircle is a right angle]
And, AB || CD
So, ∠BAD = ∠ADC = 25o [Alternate angles]
∠BAC = ∠BAD + ∠CAD = 25o + 90o = 115o
Thus,
∠ADB = 180o – 25o – ∠BAC = 180o – 25o – 115o = 40o
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
Finally,
∠AEB = ∠ADB = 40o
[Angles subtended by the same chord on the circle are equal]
27. Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.
Solution
Let’s join AC, PQ and BD.
As ACQP is a cyclic quadrilateral
∠CAP + ∠PQC = 180o … (i)
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
Similarly, as PQDB is a cyclic quadrilateral
∠PQD + ∠DBP = 180o …(ii)
Again, ∠PQC + ∠PQD = 180o … (iii) [Linear pair of angles]
Using (i), (ii) and (iii) we have
∠CAP + ∠DBP = 180o
Or ∠CAB + ∠DBA = 180o
We know that, if the sum of interior angles between two lines when intersected by a transversal are supplementary.
Then, AC || BD.
28. ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution
Let’s assume that ABCD be the given cyclic quadrilateral.
Also, PA = PD [Given]
So, ∠PAD = ∠PDA …(1)
[Angles opposite to equal sides are equal]
And,
∠BAD = 180o – ∠PAD [Linear pair of angles]
Similarly,
∠CDA = 180o – ∠PDA = 180o – ∠PAD [From (1)]
As the opposite angles of a cyclic quadrilateral are supplementary,
∠ABC = 180o – ∠CDA = 180o – (180o – ∠PAD) = ∠PAD
And, ∠DCB = 180o – ∠BAD = 180o – (180o – ∠PAD) = ∠PAD
Thus,
∠ABC = ∠DCB = ∠PAD = ∠PDA
Which is only possible when AD || BC.
Exercise 17(B)
1. In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
Prove it.
Solution
Let ABCD be the cyclic trapezium in which AB || DC, AC and BD are the diagonals.
Required to prove:
(i) AD = BC
(ii) AC = BD
Proof:
It’s seen that chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.
But, ∠ABD = ∠BDC [Alternate angles, as AB || DC with BD as the transversal]
So, Chord AD must be equal to chord BC
AD = BC
Now, in ∆ADC and ∆BCD
DC = DC [Common]
∠CAD = ∠CBD [Angles in the same segment are equal]
AD = BC [Proved above]
Hence, by SAS criterion of congruence
∆ADC ≅ ∆BCD
∴ by CPCT
AC = BD
2. In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110o, calculate:
(i) ∠AFE, (ii) ∠FAB.
Solution
Join AE, OB and OC.
(i) As AOD is the diameter
∠AED = 90o [Angle in a semi-circle is a right angle]
But, given ∠DEF = 110o
So, ∠AEF = ∠DEF – ∠AED = 110o – 90o = 20o
(ii) Also given, Chord AB = Chord BC = Chord CD
So,
∠AOB = ∠BOC = ∠COD [Equal chords subtends equal angles at the centre]
But,
∠AOB + ∠BOC + ∠COD = 180o [Since, AOD is a straight line]
Thus, ∠AOB = ∠BOC = ∠COD = 60o
Now, in ∆OAB we have
OA = OB [Radii of same circle]
So, ∠OAB = ∠OBA [Angles opposite to equal sides]
But, by angle sum property of ∆OAB
∠OAB + ∠OBA = 180o – ∠AOB
= 180o – 60o
= 120o
∴ ∠OAB = ∠OBA = 60o
Now, in cyclic quadrilateral ADEF
∠DEF + ∠DAF = 180o
∠DAF = 180o – ∠DEF
= 180o – 110o
= 70o
Thus,
∠FAB = ∠DAF + ∠OAB
= 70o + 60o = 130o
3. If two sides of a cycli-quadrilateral are parallel; prove that:
(i) its other two sides are equal.
(ii) its diagonals are equal.
Solution
Let ABCD is a cyclic quadrilateral in which AB || DC. AC and BD are its diagonals.
Required to prove:
(i) AD = BC
(ii) AC = BD
Proof:
(i) As AB || DC (given)
∠DCA = ∠CAB [Alternate angles]
Now, chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.
So,
∠DCA = ∠CAB
Hence, chord AD = chord BC or AD = BC.
(ii) Now, in ∆ABC and ∆ADB
AB = AB [Common]
∠ACB = ∠ADB [Angles in the same segment are equal]
BC = AD [Proved above]
Hence, by SAS criterion of congruence
∆ACB ≅ ∆ADB
∴ by CPCT
AC = BD
4. The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°.
Calculate:
(i) ∠POS,
(ii) ∠QOR,
(iii) ∠PQR.
Solution
Join OP, OQ, OR and OS.
Given, PQ = QR = RS
So, ∠POQ = ∠QOR = ∠ROS [Equal chords subtends equal angles at the centre]
Arc PQRS subtends ∠POS at the centre and ∠PTS at the remaining part of the circle.
Thus,
∠POS = 2×∠PTS = 2×75o = 150o
⇒ ∠POQ + ∠QOR + ∠ROS = 150o
⇒ ∠POQ = ∠QOR = ∠ROS = 150o/3 = 50o
In ∆OPQ we have,
OP = OQ [Radii of the same circle]
So, ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]
But, by angle sum property of ∆OPQ
∠OPQ + ∠OQP + ∠POQ = 180o
⇒ ∠OPQ + ∠OQP + 50o = 180o
⇒ ∠OPQ + ∠OQP = 130o
⇒ 2∠OPQ = 130o
⇒ ∠OPQ = ∠OPQ = 130o/2 = 65o
Similarly, we can prove that
In ∆OQR,
∠OQR = ∠ORQ = 65o
And in ∆ORS,
∠ORS = ∠OSR = 65o
Hence,
(i) ∠POS = 150o
(ii) ∠QOR = 50o and
(iii) ∠PQR = ∠PQO + ∠OQR = 65o + 65o = 130o
5. In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABC.
Solution
(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∠ACB = ½ ∠AOB
And as AB is the side of a regular hexagon, we have
∠AOB = 60o
(ii) Now,
∠ACB = ½ (60o) = 30o
(iii) Since AC is the side of a regular octagon,
∠AOC = 360o/ 8 = 45o
Again, arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.
∠ABC = ½ ∠AOC
⇒ ∠ABC = 45o/2 = 22.5o
Exercise 17(C)
1. In the given circle with diameter AB, find the value of x.
Solution
Now,
∠ABD = ∠ACD = 30o [Angles in the same segment]
In ∆ADB, by angle sum property we have
∠BAD + ∠ADB + ∠ABD = 180o
But, we know that angle in a semi-circle is 90o
∠ADB = 90o
So,
x + 90o + 30o = 180o
⇒ x = 180o – 120o
Hence, x = 60o
2. In the given figure, ABC is a triangle in which ∠BAC = 30o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.
Solution
Firstly, join OB and OC.
Proof:
∠BOC = 2∠BAC = 2×30o = 60o
Now, in ∆OBC
OB = OC [Radii of same circle]
So, ∠OBC = ∠OCB [Angles opposite to equal sides]
And in ∆OBC, by angle sum property we have
∠OBC + ∠OCB + ∠BOC = 180o
⇒ ∠OBC + ∠OBC + 60o = 180o
⇒ 2∠OBC = 180o – 60o = 120o
⇒ ∠OBC = 120o/2 = 60o
So, ∠OBC = ∠OCB = ∠BOC = 60o
Thus, ∆OBC is an equilateral triangle.
So, BC = OB = OC
But, OB and OC are the radii of the circum-circle.
∴ BC is also the radius of the circum-circle.
3. Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution
Let’s consider ∆ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.
And, join AD
Proof:
It’s seen that,
∠ADB = 90o [Angle in a semi-circle]
And,
∠ADC + ∠ADB = 180o [Linear pair]
Thus, ∠ADC = 90o
Now, in right ∆ABD and ∆ACD
AB = AC [Given]
AD = AD [Common]
∠ADB = ∠ADC = 90o
Hence, by R.H.S criterion of congruence.
∆ABD ≅ ∆ACD
Now, by CPCT
BD = DC
∴ D is the mid-point of BC.
4. In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65o, calculate ∠DEC.
Solution
Join OE.
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∠EOC = 2∠EBC = 2 x 65o = 130o
Now, in ∆OEC
OE = OC [Radii of the same circle]
So, ∠OEC = ∠OCE
But, in ∆EOC by angle sum property
∠OEC + ∠OCE + ∠EOC = 180o [Angles of a triangle]
⇒ ∠OCE + ∠OCE + ∠EOC = 180o
⇒ 2 ∠OCE + 130o = 180o
⇒ 2 ∠OCE = 180o – 130o
⇒ ∠OCE = 50o/2 = 25o
And, AC || ED [Given]
∠DEC = ∠OCE [Alternate angles]
Thus, ∠DEC = 25o
5. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution
Let ABCD be a cyclic quadrilateral and PQRS be the quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.
Required to prove: PQRS is a cyclic quadrilateral.
Proof:
By angle sum property of a triangle
In ∆APD,
∠PAD + ∠ADP + ∠APD = 180o …. (i)
And, in ∆BQC
∠QBC + ∠BCQ + ∠BQC = 180o …. (ii)
Adding (i) and (ii), we get
∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180o + 180o = 360o … (iii)
But,
∠PAD + ∠ADP + ∠QBC + ∠BCQ = ½ [∠A + ∠B + ∠C + ∠D]
= ½ ×360o = 180o
∴ ∠APD + ∠BQC = 360o – 180o = 180o [From (iii)]
But, these are the sum of opposite angles of quadrilateral PRQS.
∴ Quadrilateral PQRS is also a cyclic quadrilateral.
6. In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Solution
(i) Given that BD is a diameter of the circle.
And, the angle in a semicircle is a right angle.
So, ∠BCD = 90°
Also given that,
∠DBC = 58°
In ∆BDC,
∠DBC + ∠BCD + ∠BDC = 180o
⇒ 58° + 90° + ∠BDC = 180o
⇒ 148o + ∠BDC = 180o
⇒ ∠BDC = 180o – 148o
Thus, ∠BDC = 32o
(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.
So, in cyclic quadrilateral BECD
∠BEC + ∠BDC = 180o
⇒ ∠BEC + 32o = 180o
⇒ ∠BEC = 148o
(iii) In cyclic quadrilateral ABEC,
∠BAC + ∠BEC = 180o [Opposite angles of a cyclic quadrilateral are supplementary]
⇒ ∠BAC + 148o = 180o
⇒ ∠BAC = 180o – 148o
Thus, ∠BAC = 32o
7. D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Solution
Given,
∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE.
And, DE is joined.
Required to prove: Points B, C, E and D are concyclic
Proof:
In ∆ABC,
AB = AC [Given]
So, ∠B = ∠C [Angles opposite to equal sides]
Similarly,
In ∆ADE,
AD = AE [Given]
So, ∠ADE = ∠AED [Angles opposite to equal sides]
Now, in ∆ABC we have
AD/AB = AE/AC
Hence, DE || BC [Converse of BPT]
So,
∠ADE = ∠B [Corresponding angles]
⇒ (180o – ∠EDB) = ∠B
⇒ ∠B + ∠EDB = 180o
But, it’s proved above that
∠B = ∠C
So, ∠C + ∠EDB = 180o
Thus, opposite angles are supplementary.
Similarly,
∠B + ∠CED = 180o
Hence, B, C, E and D are concyclic.
8. In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92o, ∠FAE = 20o; determine ∠BCD. Given reason in support of your answer.
Solution
Given,
In cyclic quad. ABCD
AF || CB and DA is produced to E such that ∠ADC = 92o and ∠FAE = 20o
So, ∠B + ∠D = 180o
∠B + 92o = 180o
⇒ ∠B = 88o
As AF || CB, ∠FAB = ∠B = 88o
But, ∠FAD = 20o [Given]
Ext. ∠BAE = ∠BAF + ∠FAE
= 88o + 22o = 108o
But, Ext. ∠BAE = ∠BCD
∴ ∠BCD = 108o
9. If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66o and ∠ABC = 80o. Calculate:
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC
Solution
Join DB and DC, IB and IC.
Given, if ∠BAC = 66o and ∠ABC = 80o, I is the incentre of the ∆ABC.
(i) As it’s seen that ∠DBC and ∠DAC are in the same segment,
So, ∠DBC = ∠DAC
But, ∠DAC = ½ ∠BAC = ½ x 66o = 33o
Thus, ∠DBC = 33o
(ii) And, as I is the incentre of ∆ABC, IB bisects ∠ABC.
∴ ∠IBC = ½ ∠ABC = ½ x 80o = 40o
(iii) In ∆ABC, by angle sum property
∠ACB = 180o – (∠ABC + ∠BAC)
⇒ ∠ACB = 180o – (80o + 66o)
⇒ ∠ACB = 180o – 156o
⇒ ∠ACB = 34o
And since, IC bisects ∠C
Thus, ∠ICB = ½ ∠C = ½ x 34o = 17o
Now, in ∆IBC
∠IBC + ∠ICB + ∠BIC = 180o
⇒ 40o + 17o + ∠BIC = 180o
⇒ 57o + ∠BIC = 180o
⇒ ∠BIC = 180o – 57o
∴ ∠BIC = 123o
10. In the given figure, AB = AD = DC = PB and ∠DBC = xo. Determine, in terms of x:
(i) ∠ABD, (ii) ∠APB.
Hence or otherwise, prove that AP is parallel to DB.
Solution
Given, AB = AD = DC = PB and ∠DBC = xo
Join AC and BD.
Proof:
∠DAC = ∠DBC = xo [Angles in the same segment]
And, ∠DCA = ∠DAC = xo [As AD = DC]
Also, we have
∠ABD = ∠DAC [Angles in the same segment]
And, in ∆ABP
Ext. ∠ABD = ∠BAP + ∠APB
But, ∠BAP = ∠APB [Since, AB = BP]
⇒ 2 xo = ∠APB + ∠APB = 2∠APB
⇒ 2∠APB = 2xo
So, ∠APB = xo
Thus, ∠APB = ∠DBC = xo
But these are corresponding angles,
∴ AP || DB.
11. In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.
Solution
Join EB.
Then, in cyclic quad.ABEP
∠APE + ∠ABE = 180o ….. (i) [Opposite angles of a cyclic quad. are supplementary]
Similarly, in cyclic quad.BCQE
∠CQE + ∠CBE = 180o ….. (ii) [Opposite angles of a cyclic quad. are supplementary]
Adding (i) and (ii), we have
∠APE + ∠ABE + ∠CQE + ∠CBE = 180o + 180o = 360o
⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 360o
But, ∠ABE + ∠CBE = 180o [Linear pair]
⇒ ∠APE + ∠CQE + 180o = 360o
⇒ ∠APE + ∠CQE = 180o
∴ ∠APE and ∠CQE are supplementary.
12. In the given, AB is the diameter of the circle with centre O.
If ∠ADC = 32o, find angle BOC.
Solution
Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.
Thus, ∠AOC = 2∠ADC
∠AOC = 2×32o = 64o
As ∠AOC and ∠BOC are linear pair, we have
∠AOC + ∠BOC = 180o
⇒ 64o + ∠BOC = 180o
⇒ ∠BOC = 180o – 64o
∴ ∠BOC = 116o