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Elementary Mathematics

41). P is a point outside a circle and is 13 cm away from its centre. A secant drawn from the point P intersects the circle at points A and B in such a way that PA = 9 cm and AB = 7 cm. The radius of the circle is

A). 5.5 cm

B). 5 cm

C). 4 cm

D). 4.5 cm

View Answer

Correct Answer: 5 cm

Let the radius of circle4 = r

From figure,

=>\( \Large PA \times PB = PQ \times PR \)

=> \( \Large 9 \times \left(9+7\right)= \left(13-r\right) \left(13+r\right) \)

=> \( \Large 169 - r^{2} = 144 => r^{2} = 25 \)

=> r = 5 cm

42). Consider the following statements
I. The perpendicular bisector of a chord of a circle does not pass through the centre of the circle.
II. The angle in a semicircle is a right angle.
Which of the statements given above is/are correct?

A). Only I

B). Only II

C). Both I and II

D). Neither I nor II

View Answer

43). From the circumcentre I of the \( \Large \triangle \) ABC, \( \Large \perp \) ID is drawn on BC. If \( \Large \angle \) BAC = \( \Large 60 ^{\circ} \), then the value of \( \Large \angle \) BID is

A). \( \Large 75 ^{\circ} \)

B). \( \Large 60 ^{\circ} \)

C). \( \Large 45 ^{\circ} \)

D). \( \Large 80 ^{\circ} \)

View Answer

Correct Answer: \( \Large 60 ^{\circ} \)

From figure

\( \Large \angle BIC = 2 \times \angle BAC = 120 ^{\circ} and\ IB = IC \)

\( \Large \therefore \angle IBD = \angle ICD \)

= \( \Large \frac{180 ^{\circ} - 120 ^{\circ} }{2} = 30 ^{\circ} \)

Now, \( \Large \angle BID = 90 ^{\circ} - 30 ^{\circ} = 60 ^{\circ} \)

44). In a \( \Large \triangle \) ABC, O is its circumcentre and \( \Large \angle BAC = 50 ^{\circ} \). The measure of \( \Large \angle OBC \) is

A). \( \Large 60 ^{\circ} \)

B). \( \Large 30 ^{\circ} \)

C). \( \Large 40 ^{\circ} \)

D). \( \Large 50 ^{\circ} \)

View Answer

Correct Answer: \( \Large 40 ^{\circ} \)

The angle subtended by an arc at the centre of the circle is twice the angle subtended by the arc at any point on the remaining part of the circle.

\( \Large \therefore \angle BOC = 2 \angle BAC = 2 \times 50 ^{\circ} = 100 ^{\circ} \)

Now, in \( \Large \triangle BOC \)

\( \Large OB = OC \) [radii of circumcentre]

\( \Large \therefore \angle OBC = \angle OCB = x \)

\( \Large \therefore x + x + 100 ^{\circ} = 180 ^{\circ} \)

=> \( \Large 2x = 80 ^{\circ} \)

=> \( \Large x = 40 ^{\circ} \)

45). The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at the point P. Then, it is always true that

A). AP.CP = BP.DP

B). AP. BP = CP.DP

C). AP.CD = AB.CP

D). BP. AB = CD.CP

View Answer

46). A, B, C and D are four points on a circle. AC and BD intersect at a point E such that \( \Large \angle \) BEC = \( \Large 130 ^{\circ} \) and \( \Large \angle \) ECD = \( \Large 20 ^{\circ} \). Then, \( \Large \angle \) BAC is equal to

A). \( \Large 90 ^{\circ} \)

B). \( \Large 100 ^{\circ} \)

C). \( \Large 110 ^{\circ} \)

D). \( \Large 120 ^{\circ} \)

View Answer

Correct Answer: \( \Large 110 ^{\circ} \)

\( \Large \angle CED = 180 ^{\circ} - 130 ^{\circ} = 50 ^{\circ} \)

Now, in \( \Large \triangle CED \)

\( \Large \angle ECD + \angle CED + \angle CDE = 180 ^{\circ} \)

=> \( \Large \angle CDE = 180 ^{\circ} - 50 ^{\circ} - 20 ^{\circ} = 110 ^{\circ} \)

\( \Large \therefore \angle BAC = \angle CDE = 110 ^{\circ} \)

[angles in same segment are equal]

47). From the circumcentre I of the \( \Large \triangle \) ABC, \( \Large \perp \) ID is drawn on BC. If \( \Large \angle \) BAC = \( \Large 60 ^{\circ} \), then the value of \( \Large \angle \) BID is. Find the value of X.

A). \( \Large 75 ^{\circ} \)

B). \( \Large 60 ^{\circ} \)

C). \( \Large 45 ^{\circ} \)

D). \( \Large 80 ^{\circ} \)

View Answer

Correct Answer: \( \Large 60 ^{\circ} \)

From figure,

\( \Large \angle BIC = 2 \times \angle BAC = 120 ^{\circ} and\ IB = IC \)

\( \Large \therefore \angle IBD = \angle ICD = \frac{180 ^{\circ} - 120 ^{\circ} }{2} = 30 ^{\circ} \)

Now, \( \Large \angle BID = 90 ^{\circ} - 30 ^{\circ} = 60 ^{\circ} \)

48). An equilateral \( \Large \triangle \) TQR is drawn inside a square PQRS. The value of the \( \Large \angle \) PTS (in degrees) is

A). 75

B). 90

C). 120

D). 150

View Answer

Correct Answer: 150

In \( \Large \triangle SRT \)

\( \Large \angle SRT = 90 ^{\circ} - 60 ^{\circ} = 30 ^{\circ} \)

\( \Large \therefore \angle RTS = \frac{1}{2} \left(180 ^{\circ} - 30 ^{\circ} \right) = 75 ^{\circ} \)

\( \Large \therefore \angle PTQ = \angle RTS = 75 ^{\circ} \)

\( \Large \because \angle PTS + \angle

PTQ + \angle QTR + \angle RTS = 360 ^{\circ} \)

=> \( \Large \angle PTS + 75 ^{\circ} + 60 ^{\circ} + 75 ^{\circ} = 360 ^{\circ} \)

=> \( \Large \angle PTS = 360 ^{\circ} - 210 ^{\circ} = 150 ^{\circ} \)

49). If the sides of a quadrilateral ABCD touch a circle and AB = 6 cm, CD = 5 cm BC = 7 cm, then the length of AD (in cm) is

A). 4

B). 6

C). 8

D). 9

View Answer

50). Two concentric circles having common centre O and chord AB of the outer circle intersect the inner circle at points C and D. If distance of chord from the centre is 3 cm, outer radius is 13 cm and inner radius is 7 cm, then length of AC (in cm) is

A). \( \Large 8\sqrt{10} \)

B). \( \Large 6\sqrt{10} \)

C). \( \Large 4\sqrt{10} \)

D). \( \Large 2\sqrt{10} \)

View Answer

Correct Answer: \( \Large 2\sqrt{10} \)

From figure,

In \( \Large \triangle OCM, CM^{2} = 7^{2} - 3^{2} = 40 \)

\( \Large \therefore CM = 2\sqrt{10} cm \)

In \( \Large \triangle AOM, \)

\( \Large AM = \sqrt{13^{2} - 3^{2}} = 4\sqrt{10} \)

Now, \( \Large AC = AM - CM =4\sqrt{10} - 2\sqrt{10} \)

= \( \Large 2\sqrt{10} cm \)

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