ML Aggarwal Solution Class 10 Chapter 15 Circles MCQs
MCQs
Question 1
In the given figure, O is the centre of the circle. If ∠ABC = 20°, then ∠AOC is equal to
(a) 20°
(b) 40°
(c) 60°
(d) 10°
Question 2
In the given figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is. equal to
(a) 30°
(b) 60°
(c) 90°
(d) 45°
But ∠A+∠B=90∘
∴∠A=∠B=90∘2=45∘
∴∠CAB=45∘
Ans (d)
Question 3
In the given figure, if ∠DAB = 60° and ∠ABD = 50° then ∠ACB is equal to
(a) 60°
(b) 50°
(c) 70°
(d) 80°
Question 4
In the given figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to
(a) 50°
(b) 40°
(c) 60°
(d) 70°
∴∠AOB=180∘−(40∘+40∘)
=180∘−80∘=100∘
But arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴∠ACB=12∠AOB=12×100∘=50∘
Ans (a)
Question 5
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to
(a) 80°
(b) 50°
(c) 40°
(d) 30°
Sol :
ABCD is a cyclic quadrilateral,
AB is the diameter of the circle circumscribing it
∠ADC = 140°, ∠BAC = Join AC
Question 6
In the given figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 120°
∴∠ABO=∠BAO=60∘
Ext. ∠AOC=∠BAO+∠ABO
=60∘+60∘=120∘
Now arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle
∴∠AOC=2∠ADC⇒2∠ADC=120∘
⇒∠ADC=120∘2=60∘
Ans (c)
Question 7
In the given figure, O is the centre of the circle. If ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to
(a) 30°
(b) 45°
(c) 90°
(d) 60°
In ΔAOB
Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴∠ACB=12∠AOB=12×90∘=45∘
Now in ΔACB,∠ABC=30∘,∠ACB=45∘
∴∠BAC=180∘−(30∘+45∘)
=180∘−75∘=105∘
But ∠OAB=45∘
∠CAO = 105° – 45° = 60°
Ans (d)
Question 8
In the given figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then ∠PRQ is
(a) 60°
(b) 45°
(c) 30°
(d) 15°
Question 9
In the given figure, if O is the centre of the circle then the value of x is
(a) 18°
(b) 20°
(c) 24°
(d) 36°
(Angles in the same segment)
arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle
∴∠AOB=2∠ADB=2×2x=4x
InΔOAB
∠OAB=∠OBA=3x(OA=OB)
Sum of angles of ΔOAB=180∘ ⇒3x+3x+4x=180∘
⇒10x=180∘
⇒x=180∘10=18∘
∴x=18∘
Ans (a)
Question 10
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Sol :
From Q, length of tangent PQ to the circle = 24 cm
and QO = 25 cm
∴Radius of the circle=7 cm
Ans (a)
Question 11
From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Sol :
Let point P is 13 cm from O, the centre of the circle
Radius of the circle (OQ) = 5 cm
PQ and PR are tangents from P to the circle
Join OQ and OR
Now area of ΔOPQ=12PQ×OQ
(12 base × height )
=12×12×5=30 cm2
∴ area of quad. PQOR=2×30=60 cm2
(a)
Question 12
If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Sol :
Angles between two radii OA and OB = 130°
From A and B, tangents are drawn which meet at P
∴∠AOB+∠APB=180∘
⇒130∘+∠APB=180∘
⇒∠APB=180∘−130∘=50∘
Ans (b)
Question 13
In the given figure, PQ and PR are tangents from P to a circle with centre O. If ∠POR = 55°, then ∠QPR is
(a) 35°
(b) 55°
(c) 70°
(d) 80°
Question 14
If tangents PA and PB from an exterior point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 100°
Sol :
Length of tangents PA and PB to the circle from a point P
outside the circle with centre O, and inclined an angle of 80°
∴∠OAP=90∘ and ∠OPA=12∠APB
=12×80∘=40∘
=12×80∘=40∘
Ans (a)
Question 15
In the given figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to
(a) 5 cm
(b) 10 cm
(c) 7.5 cm
(d) 5√2 cm
∴∠APO=90∘×12=45∘
∴∠AOP=90∘−45∘=45∘
i.e. OA=AP=5 cm
∴OP=√OA2+PA2=√52+52
=√25+25=√50=√2×25
=√2×5 cm=5√2 cm
Ans (d)
Question 16
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Sol :
AB is the diameter of a circle with radius 5 cm
At A, XAY is a tangent to the circle
CD || XAY at a distance of 8 cm from A
Join OC
Ans (d)
Question 17
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other is
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Sol :
Radii of two concentric circles are 4 cm and 5 cm
AB is a chord of the bigger circle
which is tangent to the smaller circle at C.
Join OA, OC
OC=4 cm,OA=5 cm
and OC⊥ACB
∴ In right ΔOAC
OA2=OC2+AC2⇒52=42+AC2
⇒25=16+AC2⇒AC2=25−16=9=(3)2
∴AC=3 cm
∴ Length of chord AB=2×AC
=2×3=6 cm
Ans (b)
Question 18
In the given figure, AB is a chord of the circle such that ∠ACB = 50°. If AT is tangent to the circle at the point A, then ∠BAT is equal to
(a) 65°
(b) 60°
(c) 50°
(d) 40°
Question 19
∠RPQ=50∘
∵ OP is radius and PR is tangent to the circle
∴ OP ⊥PR
But ∠OPQ+∠RPQ=90∘
⇒∠OPQ+50∘=90∘
⇒∠OPQ=90∘−50∘=40∘
∵OP=OQ (radii of the same circle)
∴∠OQP=∠OPQ=40∘
and ∠POQ=180∘−(∠OPQ+∠OQP)
=180∘−(40∘+40∘)
=180∘−80∘=100∘
Ans (a)
Question 20
In the given figure, PA and PB are tangents to a circle with centre O. If ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Ans (a)
Question 21
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CA = 8 cm, then the length of side AB is
(a) 12 cm
(b) 11 cm
(c) 10 cm
(d) 9 cm
∴BF=BD=4 cm
Similarly, CD=CE=3 cm
∴AE=AC−CE=8−3=5 cm
and AF=AE=5 cm
Now AB=AF+BF=5+4=9 cm
Ans (d)
Question 22
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at the points P, Q and R respectively. If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of ∆ABC is
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
and BP=BR=6 cm
Now AB=AR+BR=4+6=10 cm
BC=BP+CP=6+5=11 cm
AC=AQ+CQ=4+5=9 cm
∴ Perimeter of the ΔABC=AB+BC+CA
=10+11+9=30 cm
Ans (c)
Question 23
PQ is a tangent to a circle at point P. Centre of circle is O. If ∆OPQ is an isosceles triangle, then ∠QOP is equal to
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Sol :
PQ is tangent to the circle at point P centre of the circle is O.
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