Olympiad Mathematics - Circles
Exam Duration: 45 Mins Total Questions : 30
48 cm long chord of a circle is at a distance of 7 cm from the centre. Find the radius of the circle.
- (a)
5 cm
- (b)
17 cm
- (c)
25 cm
- (d)
16 cm
In the given figure, ΔABC is inscribed in a circle with centre O. lf ㄥACB = 65°, find ㄥABC.
- (a)
25°
- (b)
35°
- (c)
90°
- (d)
65°
In the given figure, O is the centre of a circle. If ㄥDAC = 54° and ㄥACB = 63°, find ㄥOAB.
- (a)
72°
- (b)
54°
- (c)
81°
- (d)
27°
What is the least number of noncollinear points required to draw a circle passing through them?
- (a)
Two
- (b)
Three
- (c)
Four
- (d)
Nine
In the given figure, POQ is a diameter of a circle with centre 0 and PQRS is a cyclic quadrilateral. SQ is drawn. If ㄥR = 138°, find ㄥPQS.
- (a)
90°
- (b)
42°
- (c)
48°
- (d)
38°
PQRS is a cyclic quadrilateral.
⇒ ㄥP + ㄥR =180°
⇒ ㄥP = 180° - 138° = 42°
In ΔPQS, ㄥP + ㄥQ + ㄥS = 180°
∴ ㄥPQS = 48°.
Two circles intersect in A and B. Quadrilaterals PCBA and ABDE are inscribed in these circles such that PAE and CBD are line segments. If ㄥP is 95° find the value of z.
- (a)
650
- (b)
1050
- (c)
950
- (d)
850
In a circle, the major arc is twice the minor arc. What are the corresponding central angles and the degree measures of the two arcs?
- (a)
90° and 270°
- (b)
60° and 300°
- (c)
240° and 120°
- (d)
140° and 220°
In the given figure, ABCD is a quadrilateral inscribed in a circle. Diagonals AC and BD are drawn. If ㄥCAD = 40° and ㄥBDC = 25°, find ㄥBCD.
- (a)
85°
- (b)
120°
- (c)
115°
- (d)
95°
O is the centre of the circle as shown in the figure. ㄥORP = 35° and the distance between P and Q through 'O' is 4 cm. What is the measure of ㄥROQ?
- (a)
55°
- (b)
35°
- (c)
105°
- (d)
70°
Since PQ = 4 cm = 2 x OQ
= 2 x radius, PQ is the diameter of circle Join RQ. ㄥPRQ = 90°.
(Angle in a semicircle.)
∴ ㄥORQ = 90° - 35° = 55°
But OR=OQ.
∴ ㄥORQ = ㄥOQR = 55°
∴ y = 180° - (55° + 55°) = 70°
In the fi1ure given, \(\bar{DC}\) is produced to E and if \(\bar{AC}\) is the bisector of \(\angle
\)A, find the measure of \(\angle
\)BCD.
- (a)
120°
- (b)
160°
- (c)
60°
- (d)
240°
Since ABCD is a cyclic quadrilateral,
ㄥDAB = ㄥBCE = 120°.
Since AC is the bisector of ㄥDAB,
ㄥDAC = \(\frac { 120^{ 0 } }{ 2 } \) = 600.
Since angles in the same segment are equal, ㄥDBC = ㄥDAC = 600.
Two parallel chords on the same side of the centre of a circle are 10 cm and 24 cm long and their distance apart is 7 cm. Determine the radius of the circle.
- (a)
13 cm
- (b)
8 cm
- (c)
5 cm
- (d)
7 cm
A chord of a circle divides the circular region into two parts. What is the region that contains the centre?
- (a)
Minor Arc
- (b)
Major Arc
- (c)
Minor Segment
- (d)
Major Segment
In the given figure, \(\bar { PQR } =\bar { SRQ } \).
Which of the following is correct?
- (a)
PQ≠SR
- (b)
PS=QR
- (c)
PQ=SR
- (d)
PS≠QR
Since, \(\overline { PQR } =\overline { SRQ } \)
⇒ \(\overline { PQR } -\overline { QR } =\overline { SRQ } -\overline { QR } \)
⇒ \(\overline { PQ } -\overline { SR } \)
∴ PQ=SR
In the given figure, 0 is the centre of the circle and OM丄PO. If PQ = 6 cm and OM = 4 cm, find the radius of the circle.
- (a)
6 cm
- (b)
3 cm
- (c)
2 cm
- (d)
5 cm
Refer the question figure, OM丄PQ
⇒ M is the mid-point of PQ
MQ= \(\frac{1}{2}\) x 6 = 3 cm
Join OQ, In rt. ΔOMQ, we have
OQ2= OM2+ MQ2= 42+ 32= 25
⇒ OQ = \(\sqrt{25}\) = 5 cm.
AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm. find the distance of AB from the centre of the circle.
- (a)
17 cm
- (b)
15 cm
- (c)
4 cm
- (d)
8 cm
The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. lf PR= RQ= 8 cm and RB=4, cm find the radius of the circle.
- (a)
10 cm
- (b)
8 cm
- (c)
16 cm
- (d)
6 cm
In the given figure, 0 is the centre of the circle. A is any point on minor arc BC. Find the value of ㄥBAC - ㄥOBC.
- (a)
900
- (b)
1200
- (c)
600
- (d)
450
In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cm. What is the distance of the centre of the circle from BC?
- (a)
1.4 cm
- (b)
2.1 cm
- (c)
2.4 cm
- (d)
2.7 cm
AB is a chord of a circle of radius 4.3 cm and P is a point on AB which divides it into two parts in the ratio 7:10. If P is 2.7 cm away from the centre O, find the length of AB.
- (a)
5 cm
- (b)
6.8 cm
- (c)
6.4 cm
- (d)
6.1 cm
The radius of a circle is 2.5 cm. AB and CD are two parallel chords 2.7 cm apart. If AB = 4.8 cm, find the measure of CD.
- (a)
4.8 cm
- (b)
2.4 cm
- (c)
3 cm
- (d)
4 cm
ABCD is a quadrilateral whose vertices are on a semicircle such that AB = BC = CD = 10 cm and AD is diameter of the circle with centre 0. Find the perimeter of the quadrilateral ABCD.
- (a)
80 cm
- (b)
70 cm
- (c)
60 cm
- (d)
50 cm
In the given figure, if \(\angle\)DAB = 62° and \(\angle\)ABD=58°, then \(\angle\)ACB is equal to _______
- (a)
60°
- (b)
58°
- (c)
62°
- (d)
None of these
Given, \(\angle\)DAB = 62°, \(\angle\)ABD = 58°
In \(\Delta\)ADB, \(\angle\)DAB + \(\angle\)ABD + \(\angle\)ADB = 180°
\(\Rightarrow\)62° + 58° + \(\angle\)ADB = 180°
\(\Rightarrow\) \(\angle\)ADB = 180° - 120° = 60°
Now, \(\angle\)ACB = \(\angle\)ADB = 60° (Angles in same segment are equal)
In the given figure, \(\angle\)PQR = 120°, where P, Q and R are points on a circle with centre O. Then \(\angle\)OPR is
- (a)
20°
- (b)
10°
- (c)
30°
- (d)
40°
Given, \(\angle\)PQR = 120°
\(\therefore\)Reflex \(\angle\)POR = 2\(\angle\)PQR = 2(120°) = 240°
Now, \(\angle\)POR = 360° - Reflex \(\angle\)POR
= 360° - 240° = 120° ...... (i)
Also, OP = OR \(\Rightarrow\) \(\angle\)OPR = \(\angle\)ORP ...... (ii)
(Angles opposite to equal sides of a triangle are equal)
In \(\Delta\)OPR, \(\angle\)OPR + \(\angle\)ORP + \(\angle\)POR = 180°
\(\Rightarrow\)2\(\angle\)PR + 120° = 180° [From (i) & (ii)]
\(\Rightarrow\)2\(\angle\)OPR = 60° \(\Rightarrow\) \(\angle\)OPR = 30°
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If \(\angle\)QPR = 67o and \(\angle\)SPR = 72°, then \(\angle\)QRS =
- (a)
41°
- (b)
23°
- (c)
67o
- (d)
18°
\(\angle\)QPS = \(\angle\)QPR + \(\angle\)SPR
=67o+72°=139°
PQRS is a cyclic quadrilateral.
\(\Rightarrow\)\(\angle\)QRS + \(\angle\)QPS = 180
\(\Rightarrow\)\(\angle\)QRS = 180° - 139° = 41°
In the given figure, ABeD is a cyclic quadrilateral whose side AB is a diameter of the circle passing through A, B, C and D. If \(\angle\)ADC = 130°, find \(\angle\)BAC
- (a)
45°
- (b)
58°
- (c)
60°
- (d)
40°
Given, ABCD is a cyclic quadrilateral, \(\angle\)ADC = 130°
\(\therefore\)\(\angle\)ADC + \(\angle\)ABC = 180°
\(\Rightarrow\)\(\angle\)ABC = 50°
Since, AB is a diameter of circle
\(\therefore\)\(\angle\)ACB = 90°
Now, In \(\Delta\)ACB,
\(\angle\)ABC + \(\angle\)ACB + \(\angle\)BAC = 180°
\(\Rightarrow\)50° + 90° + \(\angle\)BAC = 180° \(\Rightarrow\)\(\angle\)BAC = 40°
In the given figure, AEDF is a cyclic quadrilateral. The values of x and y respectively are
- (a)
79°, 47o
- (b)
89°, 37o
- (c)
89o, 47o
- (d)
79o, 37o
In \(\Delta\)AEB, \(\angle\)ABE + \(\angle\)BEA + \(\angle\)BAE = 180°
\(\Rightarrow\)35° + \(\angle\)BEA + 54° = 180°
\(\Rightarrow\)\(\angle\)BEA=91°
Now, \(\angle\)AFD + \(\angle\)DEA = 180°
(opposite angles of cyclic quadrilateral)
\(\Rightarrow\)x+91°=180°\(\Rightarrow\) x=89°
In \(\Delta\)AFC, \(\angle\)AFC + \(\angle\)FCA + \(\angle\)CAF = 180°
\(\Rightarrow\)89° + y+ 54° = 180° \(\Rightarrow\) y= 3r
o is the centre of the circle having radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. If OA meets BC at P then OP = ______
- (a)
3.6 cm
- (b)
1.4 cm
- (c)
2 cm
- (d)
3 cm
We have, AB = AC = 6 cm
OB = OC = 5 cm
\(\therefore\) ABOC is a kite
Since, diagonals of kite are perpendicular to one another.
\(\angle\)OPB = \(\angle\)OPC = 90°
Let OP = x \(\therefore\)AP = OA - OP = 5 - x ... (i)
In \(\Delta\)OPB, (OB)2 = (OP)2 + (PB)2
\(\Rightarrow\)(5)2 = x2 + (PB)2 \(\Rightarrow\) (PB)2 = 25 - X2 ... (ii)
In \(\Delta\)APB, (AB)2 = (AP)2 + (PB)2
\(\Rightarrow\)(6)2 = (5 - x)2 + (PB)2
\(\Rightarrow\)(PB)2 = (6)2 - (5 - x)2 ... (iii)
From (ii) and (iii), we have
25 - x2 = (6)2 - (5 - x)2
\(\Rightarrow\)25 - x2= 36 - 25 - x2 + 10x
\(\Rightarrow\)14 = 10x \(\Rightarrow\) x = 1.4 cm
C1 is a circle of radius 6 cm, C2 is a circle of radius 8 cm. Jyoti wants the two circles to touch tangentially. She knows that there are two possibilities for the distance between their centres. What are these two distances?
- (a)
3 cm and 4 cm
- (b)
2 cm and 8 cm
- (c)
2 cm and 14 cm
- (d)
6 cm and 8 cm
Two circles can touch each other as shown below
For case I, distance between centres is (8 - 6) = 2 cm
For case II, distance between centres is (8 + 6) = 14 cm
In the given figure, AOB is the diamete of a circle and CD || AB. If \(\angle\)BAD = 30°, then \(\angle\)CAD = _______
- (a)
30°
- (b)
60°
- (c)
45°
- (d)
50°
Given that, CD II AB and \(\angle\)BAD = 30°.
\(\therefore\) AOB is the diameter.
\(\therefore\)\(\angle\)ADB = 90° (Angle in semicircle)
Now, In \(\Delta\)AOB, \(\angle\)DAB + \(\angle\)ADB + \(\angle\)ABD = 180°
30° + 90° + \(\angle\)ABD = 180° \(\Rightarrow\)\(\angle\)ABD = 60°
In cyclic quadrilateral ABDC, we have
\(\angle\)ABD + \(\angle\)ACD = 180°
\(\Rightarrow\)60° + \(\angle\)ACD = 180° \(\Rightarrow\)\(\angle\)ACD = 120°
Also, \(\angle\)CDA = 30° = \(\angle\)DAB (Alternate angles)
In \(\Delta\)CAD, \(\angle\)CAD + \(\angle\)CDA + \(\angle\)ACD = 180°
\(\Rightarrow\)\(\angle\)CAD + 30° + 120° = 180° \(\Rightarrow\) \(\angle\)CAD = 30°
Two circles intersect at two points A and B. If AD and AC are diameters of the circles, then which of the following step is INCORRECT in order to prove that B lies on the line segment DC?
(P) Join AB.
(Q) \(\angle\)ABD = 90° and \(\angle\)ABC = 90° (Angle in semicircle)
(R) \(\angle\)ABD + \(\angle\)ABC = 360°
(S) DBC is a straight line segment. Hence B lies on the line segment DC.
- (a)
P
- (b)
Q
- (c)
R
- (d)
S
(R) is incorrect. The correct step should be \(\angle\)ABD + \(\angle\)ABC = 180°, for DBC to be a straight line segment.