A is a point outside a circle of radius 5 cm and two tangents AP and AQ are drawn. Line joining point A with center of the circle O cuts the circle at D and is 13 cm long. The tangent drawn from D intersects the AP and AQ at points B and C respectively. Find the length of BC.
(a)
10/3
(b)
20/3
(c)
3
(d)
None of these
Answer: Option
B
Explanation :
Tangents make a right angle with the radius ⇒ ∠APO = ∠AQO = 90°.
In right ∆APO
AP2 + PO2 = AO2
⇒ AP2 + 52 = 132
⇒ AP = 12 cm.
Similarly, AQ = AP = 12 cm.
Now AD = AO - radius = 13 - 5 = 8 cm.
Let BP = x, hence, BD = x [Tangents drawn from a point are equal in length]
and, AB = 12 - x
In right ∆BDA, ∠BDA = 90°
AD2 + DB2 = AB2
⇒ 82 + x2 = (12 - x)2
64 + x2 = 144 - 24x + x2
⇒ 24x = 80
⇒ x = 10/3 cm
In the figure below, AB = 32 cm and is tangent to the two smaller circles. Find the area of the shaded part (in cm2). All three circles touch each other.
(a)
128
(b)
32π
(c)
64π
(d)
128π
(e)
Cannot be determined
Answer: Option
D
Explanation :
Let the radius of the largest circle be R and that of the two smaller circles be r1 and r2.
When two chords of a circle intersect, product of two parts of both these chords is same.
⇒ PC × CQ = AC × BC
⇒ 2r1 × 2r2 = 16 × 16
⇒ r1 × r2 = 64
In the figure given below, radius of the circle is √3 cm. What is the length of CE given that CE || OA, DE || BO and ∠AOB = 60°.
(a)
1.5 cm
(b)
3√3/2 cm
(c)
√3 cm
(d)
2 cm
Answer: Option
D
Explanation :
∠AOB = 60° and AO = BO = radius of the circle.
⇒ Triangle AOB is an equilateral triangle.
Since CE || OA, DE || BO and ∠AOB = 60°, it implies that ∠CED is also equal to 60°.
Also, CE = ED, hence, triangle CED is also an equilateral triangle.
Height of triangle CED = radius of the circle = √3
Height of an equilateral triangle = √3 × side/2 = √3
⇒ Side of triangle CED = 2 cm
In the given figure (not drawn to scale), a semicircle with its diameter on the hypotenuse of a right-angled triangle is drawn touching the remaining sides of the triangle. The two parts of the hypotenuse made by the centre of the semicircle have lengths 24 cm and 32 cm respectively. Find the radius of the semicircle (in cm).
In the figure below, AB is a diameter of the circle and C and D are such points that CD = BD. AB and CD intersect at O. If angle AOD = 35°, find angle ADC (in degrees).
(a)
30
(b)
40
(c)
50
(d)
60
Answer: Option
B
Explanation :
Let ∠ACD = x
Now, ∠ACD = ∠ABD = x (angle subtended by same chord AD)
∠ACB = 90° (angle subtended by diameter AB on circle will be right angle)
⇒ ∠DCB = 90 - x
Now, In triangle DCB, DC = DB
⇒ ∠DCB = ∠DBC = 90 - x
⇒ ∠ABC = ∠DBC - ∠DBO = 90 - x - x = 90 - 2x
In the figure given below, PR and ST are perpendicular to tangent QR. PQ passes through center O of the circle whose diameter is 10 cm. If PR = 9 cm, then what is the length (in cm) of ST?
(a)
1
(b)
1.25
(c)
1.2
(d)
2
Answer: Option
A
Explanation :
In the figure drawn above, XR = OM = 5 cm (radius)
∴ PX = 9 - 5 = 4 cm.
Now, In right triangle PXO, PX = 4, PO = 5
⇒ OX = 3 cm (pythagorean triplet)
Now, PR || OM || ST and RQ and RQ are two common transversals, hence
RM : MT = PO : OS = 1 : 1
⇒ RM = MT
⇒ OX = SY = 3 cm.
Now, In right triangle OYS, SY = 3, SO = 5
⇒ OY = 4 cm (pythagorean triplet)
In the figure given below, a smaller circle touches a larger circle at P and passes through its center O. PR is a chord of length 34 cm, then what is the legth (in cm) of PS?
(a)
9
(b)
17
(c)
21
(d)
25
Answer: Option
B
Explanation :
Let us join QR and SO
OP is the diameter of smaller circle, hence ∠OSP = 90°
QP is the diameter of larger circle, hence ∠QRP = 90°
Now, ∆PQR ≈ ∆POS [∠OSP = ∠QRP = 90° and ∠OPS is common angle]
In the figure given below, ABC is a triangle in which AB = 10 cm, AC = 6 cm and altitude AE = 4 cm. If AD is the diameter of the circum-circle, then what is the legth (in cm) of circum-radius of triangle ABC?
(a)
3
(b)
7.5
(c)
12
(d)
15
Answer: Option
B
Explanation :
Let us join BD.
∠ABD = 90° (angle subtended by diameter on cirlce is right angle)
In ∆ABD and ∆AEC
∠ABD = ∠AEC = 90°
∠BDA = ∠BCA (angle subtened by same chord AB)
∴ ∆ABD ≈ ∆AEC (by AA rule)
⇒ AB : AE = AD : AC
⇒ 10 : 4 = AD : 6
∴ AD = 15 cm
In the figure, ACB is a right triangle. CD is the altitude. The circles are inscribed within the triangles ACD and BCD. P and Q are the centres of the circles. Find the distance PQ.
(a)
5
(b)
6
(c)
5√2
(d)
8
Answer: Option
C
Explanation :
∆ABC is a right triangle, hence AB = 25 cm (15, 20 and 25 is a pythagorean triplet)
∴ CD = = = 12 cm
In right ∆ACD
AD = 9 cm (9, 12 and 15 is a pythagorean triplet)
In the figure given below, smaller circle touches bigger circle at point C. A is the center of bigger circle while B is the center of smaller circle. ∠ABE = 45°. If BC = 3 cm, find DE
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