# Selina Solutions Concise Maths Class 10 Chapter 17 Circles

A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point in the same plane always remains constant. In this chapter, students will learn the basic types and parts of a circle, cyclic properties and some important theorems and results. Students wanting to get a clear picture of concepts and problem solving techniques can access the Selina Solutions for Class 10 Mathematics prepared by expert faculty at BYJUâ€™S. The solutions will also boost confidence among students to take up their Class 10 ICSE. The Selina Solutions for Class 10 Mathematics Chapter 17 Circles PDFs are available exercise-wise in the link below.

## Selina Solutions Concise Maths Class 10 Chapter 17 Circles Download PDF

### Exercises of Concise Selina Solutions Class 10 Maths Chapter 17 Circles

Exercise 17(A) Solutions

Exercise 17(B) Solutions

Exercise 17(C) Solutions

## Access Selina Solutions Concise Maths Class 10 Chapter 17 Circles

Exercise 17(A) Page No: 257

1. In the given figure, O is the center of the circle.Â âˆ OAB and âˆ OCB are 30o and 40o respectively. Find âˆ AOC Show your steps of working.

Solution:

Firstly, letâ€™s join AC.

And, let âˆ OAC = âˆ OCA = x [Angles opposite to equal sides are equal]

So, âˆ AOC = 180o â€“ 2x

Also,

âˆ BAC = 30o + x

âˆ BCA = 40o + x

Now, in âˆ†ABC

âˆ ABC = 180o – âˆ BAC – âˆ BCA [Angles sum property of a triangle]

= 180o â€“ (30o + x) â€“ (40o + x)

= 110o â€“ 2x

And, âˆ AOC = 2âˆ ABC

[Angle at the center is double the angle at the circumference subtend by the same chord]

180o â€“ 2x = 2(110o â€“ 2x)

2x = 40o

x = 20o

Thus, âˆ AOC = 180o â€“ 2×20o = 140o

2. In the given figure,Â âˆ BAD = 65Â°,Â âˆ ABD = 70Â°,Â âˆ BDC = 45Â°

(i) Prove that AC is a diameter of the circle.

(ii) FindÂ âˆ ACB.

Solution:

(i) In âˆ†ABD,

âˆ DAB + âˆ ABD + âˆ ADB = 180o

65o + 70o + âˆ ADB = 180o

135o + âˆ ADB = 180o

âˆ ADB = 180o – 135o = 45o

Now,

âˆ ADC = âˆ ADB + âˆ BDC = 45o + 45o = 90o

As âˆ ADC is the angle of semi-circle for AC as the diameter of the circle.

(ii) âˆ ACB = âˆ ADB [Angles in the same segment of a circle]

Hence, âˆ ACB = 45o

3. Given O is the centre of the circle andÂ âˆ AOB = 70o.

Calculate the value of:

(i)Â âˆ OCA,

(ii)Â âˆ OAC.

Solution:

Here, âˆ AOB = 2âˆ ACB

[Angle at the center is double the angle at the circumference subtend by the same chord]

âˆ ACB = 70o/ 2 = 35o

Now, OC = OA [Radii of same circle]

Thus,

âˆ OCA = âˆ OAC = 35o

4. In each of the following figures, O is the centre of the circle. Find the values of a, b and c.

Solution:

(i) (ii)

(i) Here, b = Â½ x 130o

[Angle at the center is double the angle at the circumference subtend by the same chord]

Thus, b = 65o

Now,

a + b = 180o [Opposite angles of a cyclic quadrilateral are supplementary]

a = 180o â€“ 65o = 115o

(ii) Here, c = Â½ x Reflex (112o)

[Angle at the center is double the angle at the circumference subtend by the same chord]

Thus, c = Â½ x (360o â€“ 112o) = 124o

5. In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

Solution:

(i) Here, âˆ BAD = 90o [Angle in a semi-circle]

So, âˆ BDA = 90o â€“ 35o = 55o

And,

a = âˆ ACB = âˆ BDA = 55o

[Angles subtended by the same chord on the circle are equal]

(ii) Here, âˆ DAC = âˆ CBD = 25o

[Angles subtended by the same chord on the circle are equal]

And, we have

120o = b + 25o

[Exterior angle property of a triangle]

b = 95o

(iii) âˆ AOB = 2âˆ AOB = 2 x 50o = 100o

[Angle at the center is double the angle at the circumference subtend by the same chord]

Also, OA = OB

âˆ OBA = âˆ OAB = c

c = (180o– 100o)/ 2 = 40o

(iv) We have, âˆ APB = 90o [Angle in a semicircle]

âˆ BAP = 90o â€“ 45o = 45o

Now, d = âˆ BCP = âˆ BAP = 45o

[Angles subtended by the same chord on the circle are equal]

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. O1Â and O2Â are the centers of two circles.

Solution:

Itâ€™s seen that,

âˆ DBA = âˆ CBA = 90o [Angle in a semi-circle is a right angle]

âˆ DBA + âˆ CBA = 180o

Thus, DBC is a straight line i.e. D, B and C form a straight line.

7. In the figure, given below, find:

(i) âˆ BCD,

(iii) âˆ ABC.

Solution:

From the given fig, itâ€™s seen that

In cyclic quadrilateral ABCD, DC || AB

And given, âˆ DAB = 105o

(i) So,

âˆ BCD = 180o â€“ 105o = 75o

[Sum of opposite angles in a cyclic quadrilateral is 180o]

(ii) Now,

âˆ ADC and âˆ DAB are corresponding angles.

So,

âˆ ADC + âˆ DAB = 180o

âˆ ADC = 180o â€“ 105o

Thus,

(iii) We know that, the sum of angles in a quadrilateral is 360o

So,

âˆ ADC + âˆ DAB +âˆ BCD + âˆ ABC = 360o

75o + 105o + 75o + âˆ ABC = 360o

âˆ ABC = 360o â€“ 255o

Thus,

âˆ ABC = 105o

8. In the figure, given below, O is the centre of the circle. If âˆ AOB = 140o and âˆ OAC = 50o;

find:

(i) âˆ ACB,

(ii) âˆ OBC,

(iii) âˆ OAB,

(iv) âˆ CBA.

Solution:

Given, âˆ AOB = 140o and âˆ OAC = 50o

(i) Now,

âˆ ACB = Â½ Reflex (âˆ AOB) = Â½ (360o â€“ 140o) = 110o

[Angle at the center is double the angle at the circumference subtend by the same chord]

âˆ OBC + âˆ ACB + âˆ OCA + âˆ AOB = 360o [Angle sum property of a quadrilateral]

âˆ OBC + 110o + 50o + 140o = 360o

Thus, âˆ OBC = 360o â€“ 300o = 60o

(iii) In âˆ†AOB, we have

So, âˆ OBA = âˆ OAB

Hence, by angle sum property of a triangle

âˆ OBA + âˆ OAB + âˆ AOB = 180o

2âˆ OBA + 140o = 180o

2âˆ OBA = 40o

âˆ OBA = 20o

(iv) We already found, âˆ OBC = 60o

And, âˆ OBC = âˆ CBA + âˆ OBA

60o = âˆ CBA + 20o

Therefore,

âˆ CBA = 40o

9. Calculate:

(i) âˆ CDB,

(ii) âˆ ABC,

(iii) âˆ ACB.

Solution:

Here, we have

âˆ CDB = âˆ BAC = 49o

âˆ ABC = âˆ ADC = 43o

[Angles subtended by the same chord on the circle are equal]

Now, by angle sum property of a triangle we have

âˆ ACB = 180o â€“ 49o â€“ 43o = 88o

10. In the figure given below, ABCD is a cyclic quadrilateral in whichÂ âˆ BAD = 75o;Â âˆ ABD = 58oÂ andÂ âˆ ADC = 77o.

Find:

(i) âˆ BDC,

(ii) âˆ BCD,

(iii) âˆ BCA.

Solution:

(i) By angle sum property of triangle ABD,

âˆ ADB = 180o â€“ 75o â€“ 58o = 47o

Thus, âˆ BDC = âˆ ADC – âˆ ADB = 77o â€“ 47o = 30o

(ii) âˆ BAD + âˆ BCD = 180o

[Sum of opposite angles of a cyclic quadrilateral is 180o]

Thus, âˆ BCD = 180o – 75o = 105o

(iii) âˆ BCA = âˆ 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) âˆ ADB, (ii) âˆ AEB

Solution:

(i) As, itâ€™s seen that âˆ ACB and âˆ ADB are in the same segment,

So,

âˆ ADB = âˆ ACB = 60o

(ii) Now, join OA and OB.

And, we have

âˆ AEB = Â½ Reflex (âˆ AOB) = Â½ (360o â€“ 120o) = 120o

[Angle at the center is double the angle at the circumference subtend by the same chord]

12. Given: âˆ CAB = 75o and âˆ CBA = 50o. Find the value of âˆ DAB + âˆ ABD.

Solution:

Given, âˆ CAB = 75o and âˆ CBA = 50o

In âˆ†ABC, by angle sum property we have

âˆ ACB = 180o â€“ (âˆ CBA + âˆ CAB)

= 180o â€“ (50o + 75o) = 180o â€“ 125o

= 55o

And,

âˆ ADB = âˆ ACB = 55o

[Angles subtended by the same chord on the circle are equal]

Now, taking âˆ†ABD

âˆ DAB + âˆ ABD + âˆ ADB = 180o

âˆ DAB + âˆ ABD + 55o = 180o

âˆ DAB + âˆ ABD = 180o – 55o

âˆ DAB + âˆ ABD = 125o

13. ABCD is a cyclic quadrilateral in a circle with centre O. If âˆ ADC = 130o, findÂ âˆ BAC.

Solution:

From the fig. its seem that,

âˆ ACB = 90o [Angle in a semi-circle is 90o]

Also,

âˆ ABC = 180o – âˆ 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

âˆ BAC = 90o – âˆ ABC

= 90o â€“ 50o

Thus, âˆ BAC = 40o

14. In the figure given alongside, AOB is a diameter of the circle andÂ âˆ AOC = 110o, findÂ âˆ BDC.

Solution:

So, we have

âˆ ADC = Â½ âˆ AOC = Â½ x 110o = 55o

[Angle at the centre is double the angle at the circumference subtended by the same chord]

Also, we know that

[Angle in the semi-circle is a right angle]

Therefore,

âˆ BDC = 90o – âˆ ADC = 90o â€“ 55o

âˆ BDC = 35o

15. In the following figure, O is the centre of the circle; âˆ AOB = 60oÂ andÂ âˆ BDC = 100o, findÂ âˆ OBC.

Solution:

Form the figure, we have

âˆ ACB = Â½ âˆ AOB = Â½ 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 âˆ†BDC,

âˆ DBC = 180o – 100o â€“ 30o = 50o

Therefore,

âˆ OBC = 50o

16. In ABCD is a cyclic quadrilateral in which âˆ DAC = 27o, âˆ DBA = 50oÂ andÂ âˆ ADB = 33o. Calculate (i)Â âˆ DBC, (ii)Â âˆ DCB, (iii) âˆ CAB.

Solution:

(i) Itâ€™s seen that,

âˆ DBC = âˆ DAC = 27o

[Angles subtended by the same chord on the circle are equal]

(ii) Itâ€™s seen that,

âˆ ACB = âˆ ADB = 33o

And,

âˆ ACD = âˆ ABD = 50o

[Angles subtended by the same chord on the circle are equal]

Thus,

âˆ DCB = âˆ ACD + âˆ ACB = 50o + 33o = 83o

âˆ DAB + âˆ DCB = 180o

27o + âˆ CAB + 83o = 180o

Thus,

âˆ CAB = 180o â€“ 110o = 70o

17. In the figure given alongside, AB and CD are straight lines through the centre O of a circle. IfÂ âˆ AOC = 80oÂ andÂ âˆ CDE = 40o. Find the number of degrees in: (i)Â âˆ DCE; (ii)Â âˆ ABC.

Solution:

(i) Form the fig. its seen that,

âˆ DCE = 90o – âˆ CDE = 90o â€“ 40o = 50o

Therefore,

âˆ DEC = âˆ OCB = 50o

(ii) In âˆ†BOC, we have

âˆ AOC = âˆ OCB + âˆ OBC [Exterior angle property of a triangle]

âˆ OBC = 80oÂ – 50oÂ = 30oÂ  [Given âˆ AOC = 80o]

Therefore, âˆ 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, âˆ 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.

Therefore,

AB = BE

19. (a) In the following figure,

(i) ifÂ âˆ BAD = 96o, findÂ âˆ BCD andÂ âˆ BFE.

(ii) Prove that AD is parallel to FE.

(b) ABCD is a parallelogram. A circle

Solution:

(i) ABCD is a cyclic quadrilateral

So, âˆ BAD + âˆ BCD = 180o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

âˆ BCD = 180o – 96o = 84o

And, âˆ BCE = 180o – 84o = 96o [Linear pair of angles]

Similarly, BCEF is a cyclic quadrilateral

So, âˆ BCE + âˆ BFE = 180o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

âˆ BFE = 180o – 96o = 84o

(ii) Now, âˆ BAD + âˆ BFE = 96o + 84o = 180o

But these two are interior angles on the same side of a pair of lines AD and 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) Letâ€™s assume that ABCD is a parallelogram which is inscribed in a circle.

So, we have

âˆ BAD = âˆ BCD [Opposite angles of a parallelogram are equal]

And âˆ BAD + âˆ BCD = 180o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

Thus, âˆ BAD = âˆ BCD = 90o

Similarly, the remaining two angles are 90o each and pair of opposite sides are equal.

Therefore,

ABCD is a rectangle.

– Hence Proved

(ii) Letâ€™s assume that ABCD is a rhombus which is inscribed in a circle.

So, we have

âˆ BAD = âˆ BCD [Opposite angles of a rhombus are equal]

And âˆ BAD + âˆ BCD = 180o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

Thus, âˆ BAD = âˆ BCD = 90o

Similarly, the remaining two angles are 90o each and all the sides are equal.

Therefore,

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, âˆ B = âˆ C â€¦ (1)

[Angles opposite to equal sides are equal]

And, DECB is a cyclic quadrilateral.

So, âˆ B + âˆ DEC = 180o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

âˆ C + âˆ DEC = 180o â€¦. (Using 1)

But this is the sum of interior angles on one side of a transversal.

DE || BC.

But, âˆ ADE = âˆ B and âˆ AED = âˆ C [Corresponding angles]

Thus, âˆ ADE = âˆ AED

AB â€“ AD = AC = AE [As AB = AC]

BD = CE

Hence, we have DE || BC and BD = CE

Therefore,

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 Oâ€™ 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, âˆ AQP = 90o and âˆ BQP = 90o

[Angle in a semicircle is a right angle]

Now, adding both these angles we get

âˆ AQP + âˆ BQP = 180o

âˆ AQB = 180o

Therefore, the points A, Q and B are collinear.

23. The figure given below, shows a circle with centre O. Given:Â âˆ AOC = a andÂ âˆ ABC = b.

(i) Find the relationship between a and b

(ii) Find the measure of angle OAB, if OABC is a parallelogram.

Solution:

(i) Itâ€™s seen that,

âˆ ABC = Â½ Reflex (âˆ COA)

[Angle at the centre is double the angle at the circumference subtended by the same chord]

So, b = Â½ (360o – a)

a + 2b = 180o â€¦.. (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

âˆ COA + âˆ OAB = 180o

60o + âˆ OAB = 180o

Therefore,

âˆ 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: âˆ AOC + âˆ BOD = 2âˆ APC

OA, OB, OC and OD are joined.

Now, itâ€™s seen that

âˆ AOC = 2âˆ ADC â€¦. (1)

[Angle at the centre is double the angle at the circumference subtended by the same chord]

Similarly,

âˆ BOD = 2âˆ BAD â€¦. (2)

Adding (1) and (2), we have

So, from (3) and (4) we have

âˆ AOC + âˆ BOD = 2âˆ APC

25. In the figure given RS is a diameter of the circle. NM is parallel to RS andÂ MRS = 29o

Calculate: (i)Â âˆ RNM; (ii)Â âˆ NRM.

Solution:

(i) Join RN and MS

âˆ RMS = 90o [Angle in a semi-circle is a right angle]

So, by angle sum property of âˆ†RMS

âˆ RMS = 90o â€“ 29o = 61o

And,

âˆ RNM = 180o – âˆ RSM = 180oÂ â€“ 61o = 119o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

(ii) Now as RS || NM,

âˆ NMR = âˆ MRS = 29o [Alternate angles]

âˆ NMS = 90o + 29o = 119o

Also, we know that

âˆ NRS + âˆ NMS = 180o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

âˆ NRM + 29o + 119o = 180o

âˆ NRM = 180o â€“ 148o

Therefore,

âˆ NRM = 32o

26. In the figure given alongside, AB || CD and O is the center of the circle. IfÂ âˆ ADC = 25o; find the angle AEB. Give reasons in support of your answer.

Solution:

Join AC and BD.

So, we have

âˆ CAD = 90oÂ and âˆ CBD = 90o

[Angle is a semicircle is a right angle]

And, AB || CD

âˆ BAC = âˆ BAD + âˆ CAD = 25o + 90o = 115o

Thus,

âˆ ADB = 180o â€“ 25o – âˆ BAC = 180o â€“ 25o â€“ 115o = 40o

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

Finally,

âˆ AEB = âˆ 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:

Letâ€™s join AC, PQ and BD.

As ACQP is a cyclic quadrilateral

âˆ CAP + âˆ PQC = 180o â€¦â€¦. (i)

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

Similarly, as PQDB is a cyclic quadrilateral

âˆ PQD + âˆ DBP = 180o â€¦â€¦. (ii)

Again, âˆ PQC + âˆ PQD = 180o â€¦â€¦ (iii) [Linear pair of angles]

Using (i), (ii) and (iii) we have

âˆ CAP + âˆ DBP = 180o

Or âˆ CAB + âˆ 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:

Letâ€™s assume that ABCD be the given cyclic quadrilateral.

Also, PA = PD [Given]

So, âˆ PAD = âˆ PDA â€¦â€¦ (1)

[Angles opposite to equal sides are equal]

And,

Similarly,

âˆ CDA = 180o – âˆ PDA = 180o – âˆ PAD [From (1)]

As the opposite angles of a cyclic quadrilateral are supplementary,

âˆ ABC = 180o – âˆ CDA = 180o â€“ (180o – âˆ PAD) = âˆ PAD

And, âˆ DCB = 180o – âˆ BAD = 180o â€“ (180o – âˆ PAD) = âˆ PAD

Thus,

âˆ ABC = âˆ DCB = âˆ PAD = âˆ PDA

Which is only possible when AD || BC.

Exercise 17(B) Page No: 265

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:

(ii) AC = BD

Proof:

Itâ€™s seen that chord AD subtends âˆ ABD and chord BC subtends âˆ BDC at the circumference of the circle.

But, âˆ ABD = âˆ BDC [Alternate angles, as AB || DC with BD as the transversal]

So, Chord AD must be equal to chord BC

DC = DC [Common]

âˆ CAD = âˆ CBD [Angles in the same segment are equal]

Hence, by SAS criterion of congruence

Therefore, by CPCT

AC = BD

2. In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CDÂ areÂ equal. IfÂ âˆ DEF = 110o, calculate:

(i)Â âˆ AFE, (ii)Â âˆ FAB.

Solution:

Join AE, OB and OC.

(i) As AOD is the diameter

âˆ AED = 90o [Angle in a semi-circle is a right angle]

But, given âˆ DEF = 110o

So,

âˆ AEF = âˆ DEF – âˆ AED = 110o â€“ 90o = 20o

(ii) Also given, Chord AB = Chord BC = Chord CD

So,

âˆ AOB = âˆ BOC = âˆ COD [Equal chords subtends equal angles at the centre]

But,

âˆ AOB + âˆ BOC + âˆ COD = 180o [Since, AOD is a straight line]

Thus,

âˆ AOB = âˆ BOC = âˆ COD = 60o

Now, in âˆ†OAB we have

OA = OB [Radii of same circle]

So, âˆ OAB = âˆ OBA [Angles opposite to equal sides]

But, by angle sum property of âˆ†OAB

âˆ OAB + âˆ OBA = 180o – âˆ AOB

= 180o â€“ 60o

= 120o

Therefore, âˆ OAB = âˆ OBA = 60o

âˆ DEF + âˆ DAF = 180o

âˆ DAF = 180o – âˆ DEF

= 180o â€“ 110o

= 70o

Thus,

âˆ FAB = âˆ DAF + âˆ OAB

= 70o + 60o = 130o

3. If two sides of aÂ cycli-quadrilateral are parallel; proveÂ that:

(i)Â itsÂ other two sides are equal.

(ii)Â itsÂ diagonals are equal.

Solution:

Let ABCD is a cyclic quadrilateral in which AB || DC. AC and BD are its diagonals.

Required to prove:

(ii) AC = BD

Proof:

(i) As AB || DC (given)

âˆ DCA = âˆ CAB [Alternate angles]

Now, chord AD subtends âˆ DCA and chord BC subtends âˆ CAB at the circumference of the circle.

So,

âˆ DCA = âˆ CAB

(ii) Now, in âˆ†ABC and âˆ†ADB

AB = AB [Common]

âˆ ACB = âˆ ADB [Angles in the same segment are equal]

Hence, by SAS criterion of congruence

Therefore, by CPCT

AC = BD

4. The given figure show a circle with centre O. Also, PQ = QR = RS andÂ âˆ PTS = 75Â°.

Calculate:

(i)Â âˆ POS,

(ii)Â âˆ QOR,

(iii)Â âˆ PQR.

Solution:

Join OP, OQ, OR and OS.

Given, PQ = QR = RS

So, âˆ POQ = âˆ QOR = âˆ ROS [Equal chords subtends equal angles at the centre]

Arc PQRS subtends âˆ POS at the centre and âˆ PTS at the remaining part of the circle.

Thus,

âˆ POS = 2 x âˆ PTS = 2 x 75o = 150o

âˆ POQ + âˆ QOR + âˆ ROS = 150o

âˆ POQ = âˆ QOR = âˆ ROS = 150o/ 3 = 50o

In âˆ†OPQ we have,

OP = OQ [Radii of the same circle]

So, âˆ OPQ = âˆ OQP [Angles opposite to equal sides are equal]

But, by angle sum property of âˆ†OPQ

âˆ OPQ + âˆ OQP + âˆ POQ = 180o

âˆ OPQ + âˆ OQP + 50o = 180o

âˆ OPQ + âˆ OQP = 130o

2 âˆ OPQ = 130o

âˆ OPQ = âˆ OPQ = 130o/ 2 = 65o

Similarly, we can prove that

In âˆ†OQR,

âˆ OQR = âˆ ORQ = 65o

And in âˆ†ORS,

âˆ ORS = âˆ OSR = 65o

Hence,

(i) âˆ POS = 150o

(ii) âˆ QOR = 50o and

(iii) âˆ PQR = âˆ PQO + âˆ 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)Â âˆ AOB,

(ii)Â âˆ ACB,

(iii)Â âˆ ABC.

Solution:

(i) Arc AB subtends âˆ AOB at the centre and âˆ ACB at the remaining part of the circle.

âˆ ACB = Â½ âˆ AOB

And as AB is the side of a regular hexagon, we have

âˆ AOB = 60o

(ii) Now,

âˆ ACB = Â½ (60o) = 30o

(iii) Since AC is the side of a regular octagon,

âˆ AOC = 360o/ 8 = 45o

Again, arc AC subtends âˆ AOC at the centre and âˆ ABC at the remaining part of the circle.

âˆ ABC = Â½ âˆ AOC

âˆ ABC = 45o/ 2 = 22.5o

Exercise 17(C) Page No: 265

1. In the given circle with diameter AB, find the value of x.

Solution:

Now,

âˆ ABD = âˆ ACD = 30o [Angles in the same segment]

In âˆ†ADB, by angle sum property we have

But, we know that angle in a semi-circle is 90o

So,

x + 90o + 30o = 180o

x = 180o – 120o

Hence, x = 60o

2. In the given figure, ABC is a triangle in whichÂ âˆ 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:

âˆ BOC = 2âˆ BAC = 2 x 30o = 60o

Now, in âˆ†OBC

OB = OC [Radii of same circle]

So, âˆ OBC = âˆ OCB [Angles opposite to equal sides]

And in âˆ†OBC, by angle sum property we have

âˆ OBC + âˆ OCB + âˆ BOC = 180o

âˆ OBC + âˆ OBC + 60o = 180o

2 âˆ OBC = 180o â€“ 60oÂ = 120o

âˆ OBC = 120o/ 2 = 60o

So, âˆ OBC = âˆ OCB = âˆ BOC = 60o

Thus, âˆ†OBC is an equilateral triangle.

So,

BC = OB = OC

But, OB and OC are the radii of the circum-circle.

Therefore, 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:

Letâ€™s consider âˆ†ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.

Proof:

Itâ€™s seen that,

âˆ ADB = 90o [Angle in a semi-circle]

And,

Now, in right âˆ†ABD and âˆ†ACD

AB = AC [Given]

Hence, by R.H.S criterion of congruence.

âˆ†ABD â‰… âˆ†ACD

Now, by CPCT

BD = DC

Therefore, D is the mid-point of BC.

4. In the given figure, chord ED is parallel to diameter AC of the circle. GivenÂ âˆ CBE = 65o, calculate âˆ DEC.

Solution:

Join OE.

Arc EC subtends âˆ EOC at the centre and âˆ EBC at the remaining part of the circle.

âˆ EOC = 2âˆ EBC = 2 x 65o = 130o

Now, in âˆ†OEC

OE = OC [Radii of the same circle]

So, âˆ OEC = âˆ OCE

But, in âˆ†EOC by angle sum property

âˆ OEC + âˆ OCE + âˆ EOC = 180o [Angles of a triangle]

âˆ OCE + âˆ OCE + âˆ EOC = 180o

2 âˆ OCE + 130o = 180o

2 âˆ OCE = 180o – 130o

âˆ OCE = 50o/ 2 = 25o

And, AC || ED [Given]

âˆ DEC = âˆ OCE [Alternate angles]

Thus,

âˆ 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 âˆ A, âˆ B, âˆ C and âˆ D.

Required to prove: PQRS is a cyclic quadrilateral.

Proof:

By angle sum property of a triangle

In âˆ†APD,

âˆ PAD + âˆ ADP + âˆ APD = 180o â€¦. (i)

And, in âˆ†BQC

âˆ QBC + âˆ BCQ + âˆ BQC = 180o â€¦. (ii)

Adding (i) and (ii), we get

âˆ PAD + âˆ ADP + âˆ APD + âˆ QBC + âˆ BCQ + âˆ BQC = 180o + 180o = 360o â€¦â€¦ (iii)

But,

âˆ PAD + âˆ ADP + âˆ QBC + âˆ BCQ = Â½ [âˆ A + âˆ B + âˆ C + âˆ D]

= Â½ x 360o = 180o

Therefore,

âˆ APD + âˆ BQC = 360o – 180o = 180o [From (iii)]

But, these are the sum of opposite angles of quadrilateral PRQS.

Therefore,

6. In the figure,Â âˆ DBC = 58Â°. BD is a diameter of the circle. Calculate:

(i)Â âˆ BDC

(ii)Â âˆ BEC

(iii)Â âˆ BAC

Solution:

(i) Given that BD is a diameter of the circle.

And, the angle in a semicircle is a right angle.

So, âˆ BCD = 90Â°

Also given that,

âˆ DBC = 58Â°

In âˆ†BDC,

âˆ DBC + âˆ BCD + âˆ BDC = 180o

58Â° + 90Â° + âˆ BDC = 180o

148o + âˆ BDC = 180o

âˆ BDC = 180o – 148o

Thus, âˆ BDC = 32o

(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.

âˆ BEC + âˆ BDC = 180o

âˆ BEC + 32o = 180o

âˆ BEC = 148o

âˆ BAC + âˆ BEC = 180o [Opposite angles of a cyclic quadrilateral are supplementary]

âˆ BAC + 148o = 180o

âˆ BAC = 180o – 148o

Thus, âˆ 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,

âˆ†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 âˆ†ABC,

AB = AC [Given]

So, âˆ B = âˆ C [Angles opposite to equal sides]

Similarly,

So, âˆ ADE = âˆ AED [Angles opposite to equal sides]

Now, in âˆ†ABC we have

Hence, DE || BC [Converse of BPT]

So,

âˆ ADE = âˆ B [Corresponding angles]

(180o – âˆ EDB) = âˆ B

âˆ B + âˆ EDB = 180o

But, itâ€™s proved above that

âˆ B = âˆ C

So,

âˆ C + âˆ EDB = 180o

Thus, opposite angles are supplementary.

Similarly,

âˆ B + âˆ 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. IfÂ âˆ ADC = 92o, âˆ FAE = 20o; determine âˆ BCD. Given reason in support of your answer.

Solution:

Given,

AF || CB and DA is produced to E such that âˆ ADC = 92o and âˆ FAE = 20o

So,

âˆ B + âˆ D = 180o

âˆ B + 92o = 180o

âˆ B = 88o

As AF || CB, âˆ FAB = âˆ B = 88o

But, âˆ FAD = 20o [Given]

Ext. âˆ BAE = âˆ BAF + âˆ FAE

= 88o + 22o = 108o

But, Ext. âˆ BAE = âˆ BCD

Therefore,

âˆ 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 âˆ BAC = 66o andÂ âˆ ABC =Â 80o. Calculate:

(i) âˆ DBC,

(ii) âˆ IBC,

(iii) âˆ BIC

Solution:

Join DB and DC, IB and IC.

Given, if âˆ BAC = 66o andÂ âˆ ABC =Â 80o, I is the incentre of the âˆ†ABC.

(i) As itâ€™s seen that âˆ DBC and âˆ DAC are in the same segment,

So, âˆ DBC = âˆ DAC

But, âˆ DAC = Â½ âˆ BAC = Â½ x 66o = 33o

Thus, âˆ DBC = 33o

(ii) And, as I is the incentre of âˆ†ABC, IB bisects âˆ ABC.

Therefore,

âˆ IBC = Â½ âˆ ABC = Â½ x 80o = 40o

(iii) In âˆ†ABC, by angle sum property

âˆ ACB = 180o â€“ (âˆ ABC + âˆ BAC)

âˆ ACB = 180o â€“ (80o + 66o)

âˆ ACB = 180o â€“ 156o

âˆ ACB = 34o

And since, IC bisects âˆ C

Thus, âˆ ICB = Â½ âˆ C = Â½ x 34o = 17o

Now, in âˆ†IBC

âˆ IBC + âˆ ICB + âˆ BIC = 180o

40o + 17o + âˆ BIC = 180o

57o + âˆ BIC = 180o

âˆ BIC = 180o â€“ 57o

Therefore, âˆ BIC = 123o

10. In the given figure, AB = AD = DC = PB andÂ âˆ DBC = xo. Determine, in terms of x:

(i) âˆ ABD, (ii) âˆ APB.

Hence or otherwise, prove that AP is parallel to DB.

Solution:

Given, AB = AD = DC = PB andÂ âˆ DBC = xo

Join AC and BD.

Proof:

âˆ DAC = âˆ DBC = xo [Angles in the same segment]

And, âˆ DCA = âˆ DAC = xo [As AD = DC]

Also, we have

âˆ ABD = âˆ DAC [Angles in the same segment]

And, in âˆ†ABP

Ext. âˆ ABD = âˆ BAP + âˆ APB

But, âˆ BAP = âˆ APB [Since, AB = BP]

2 xo = âˆ APB + âˆ APB = 2âˆ APB

2âˆ APB = 2xo

So, âˆ APB = xo

Thus, âˆ APB = âˆ DBC = xo

But these are corresponding angles,

Therefore, AP || DB.

11. In the given figure; ABC, AEQ and CEP are straight lines. Show that âˆ APE and âˆ CQE are supplementary.

Solution:

Join EB.

âˆ APE + âˆ ABE = 180o â€¦.. (i) [Opposite angles of a cyclic quad. are supplementary]

âˆ CQE + âˆ CBE = 180o â€¦.. (ii) [Opposite angles of a cyclic quad. are supplementary]

Adding (i) and (ii), we have

âˆ APE + âˆ ABE + âˆ CQE + âˆ CBE = 180o + 180o = 360o

âˆ APE + âˆ ABE + âˆ CQE + âˆ CBE = 360o

But, âˆ ABE + âˆ CBE = 180o [Linear pair]

âˆ APE + âˆ CQE + 180o = 360o

âˆ APE + âˆ CQE = 180o

Therefore, âˆ APE and âˆ CQE are supplementary.

12. In the given, AB is the diameter of the circle with centre O.

If âˆ ADC =Â 32o, find angle BOC.

Solution:

Arc AC subtends âˆ AOC at the centre and âˆ ADC at the remaining part of the circle.

Thus, âˆ AOC = 2âˆ ADC

âˆ AOC = 2 x 32o = 64o

As âˆ AOC and âˆ BOC are linear pair, we have

âˆ AOC + âˆ BOC = 180o

64o + âˆ BOC = 180o

âˆ BOC = 180o â€“ 64o

Therefore, âˆ BOC = 116o

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2023-24 will be updated soon.