The problems of this exercise strengthen the main concepts of the chapter. The Selina Solutions for Class 10 Maths is a one-stop solution to all the doubts and clarifications pertaining to solving problems of the book. The solutions PDF of the Concise Selina Solutions for Class 10 Maths Chapter 17 Circles Exercise 17(C) is available in the link given below.
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Access Selina Solutions Concise Maths Class 10 Chapter 17 Circles Exercise 17(C)
1. In the given circle with diameter AB, find the value of x.
∠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
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.
Firstly, join OB and OC.
∠BOC = 2∠BAC = 2 x 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.
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.
Let’s consider ∆ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.
And, join AD
It’s seen that,
∠ADB = 90o [Angle in a semi-circle]
∠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
Therefore, 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.
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]
∠DEC = 25o
5. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
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.
By angle sum property of a triangle
∠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)
∠PAD + ∠ADP + ∠QBC + ∠BCQ = ½ [∠A + ∠B + ∠C + ∠D]
= ½ x 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) 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°
∠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.
∆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
AB = AC [Given]
So, ∠B = ∠C [Angles opposite to equal sides]
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]
∠ADE = ∠B [Corresponding angles]
(180o – ∠EDB) = ∠B
∠B + ∠EDB = 180o
But, it’s proved above that
∠B = ∠C
∠C + ∠EDB = 180o
Thus, opposite angles are supplementary.
∠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.
In cyclic quad. ABCD
AF || CB and DA is produced to E such that ∠ADC = 92o and ∠FAE = 20o
∠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:
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
Therefore, ∠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.
Given, AB = AD = DC = PB and ∠DBC = xo
Join AC and BD.
∠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,
Therefore, AP || DB.
11. In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.
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
Therefore, ∠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.
Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.
Thus, ∠AOC = 2∠ADC
∠AOC = 2 x 32o = 64o
As ∠AOC and ∠BOC are linear pair, we have
∠AOC + ∠BOC = 180o
64o + ∠BOC = 180o
∠BOC = 180o – 64o
Therefore, ∠BOC = 116o