Selina Solutions Concise Maths Class 10 Chapter 17 Circles
Access Selina Solutions Concise Maths Class 10 Chapter 17 Circles 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
Therefore, 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
Therefore, ∠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
Therefore, 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 x ∠PTS = 2 x 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
0 Comments
Please Comment