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Elementary Mathematics

61). Consider the following statements in respect of two chords XY and ZT of a circle intersecting at P.
I. PX-PY = PZ-PT
II. PXZ and PTY are similar triangles.
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

Correct Answer: Both I and II

When two chords of a circle intersects internally, then they are divided in a proportion.

i.e., PX.PY = PZ.PT

In \( \Large \triangle PXZ and \triangle PTY, \)

\( \Large \angle ZPX = \angle YPT \)

[vertically opposite angles]

\( \Large \angle PZX = \angle PYT \)

[angles in same segment]

\( \Large \angle PXZ = \angle PTY \)

[angles in same segment]

\( \Large \therefore \triangle PXZ \cong \triangle PTY \)

Hence, the both statements are correct.

62). A regular hexagon is inscribed in a circle of radius 5 cm. If x is the area inside the circle but outside the regular hexagon, then which one of the following is correct?

A). \( \Large 13 cm^{2} < x < 15 cm^{2} \)

B). \( \Large 15 cm^{2} < x < 17 cm^{2} \)

C). \( \Large 17 cm^{2} < x < 19 cm^{2} \)

D). \( \Large 19 cm^{2} < x < 21 cm^{2} \)

View Answer

Correct Answer: \( \Large 13 cm^{2} < x < 15 cm^{2} \)

OB = 0A = radius

Also, \( \Large \angle AOB = 60 ^{\circ} \left(\frac{360 ^{\circ} }{6}\right) = 60 ^{\circ} \)

and \( \Large \angle OAB = \angle OBA = 60 ^{\circ} \)

\( \Large \therefore \triangle AOB\ is\ an\ equilateral\ triangle \)

Then, AB = 5 cm

So, Area \( \Large x = Area\ of\ circle - Area\ of\ hexagon \)

= \( \Large \pi r^{2} - \frac{3\sqrt{3}a^{2}}{2} \)

= \( \Large \frac{22}{7} \times \left(5\right)^{2} - \frac{3\sqrt{3}}{2} \times \left(5\right)^{2} \)

\([ \Large \because r = a = 5] \)

= \( \Large 78.57 - 64.95 = 13.62 cm^{2} \)

63). The diameter of a circle with centre at C is 50 cm. CP is a radial segment of the circle. AB is a chord perpendicular to CP and passes through P. CP produced intersects the circle at D. If DP = 18 cm, then what is the length of AB?

A). 24 cm

B). 32 cm

C). 40 cm

D). 48 cm

View Answer

Correct Answer: 48 cm

In \( \Large \triangle ACP, \)

\( \Large CP = CD - PD = 25 - 18 = 7 \)

Now, \( \Large AC^{2} = CP^{2} + AP^{2} \)

\( \Large \therefore AP = \sqrt{AC^{2}-CP^{2}} = \sqrt{ \left(25\right)^{2} - \left(7\right)^{2} } \)

= \( \Large \sqrt{625-49} = \sqrt{576} = 24 cm \)

Similarly, PB = 24 cm

Therefore, AB= AP + PB = 24 + 24 = 48 cm

64). ABC is an equilateral triangle inscribed in a circle. D is any point on the arc BC. What is \( \Large \angle \) ADB equal to?

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

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

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

D). None of these

View Answer

65). AB and CD are two chords of a circle meeting externally at P. Then, which of the following is/are correct?
I. PA x PD = PC x PB
II. \( \Large \triangle \)PAC and \( \Large \triangle \)PDB are similar.
Select the correct answer using the codes given below.

A). Only I

B). Only II

C). Both I and II

D). Neither I nor ll

View Answer

66). ABC is triangle right angled at B. If AB = 6 cm and BC = 8 cm, then what is the length of the circumradius of the \( \Large \triangle \)ABC?

A). 10 cm

B). 7 cm

C). 6 cm

D). 5 cm

View Answer

Correct Answer: 5 cm

Since, \( \Large \triangle ABC \) is right angled a B.

Using Pythagoras theorem,

\( \Large AC^{2} = AB^{2} + BC^{2} \)

=> \( \Large AC^{2} = 6^{2} + 8^{2} \)

=> \( \Large AC^{2} = 36 + 64 = 100 \)

Therefore, AC = 10 cm

and in case of right angled triangle, radius lies of h hypotenuse and is the circumcircle of \( \Large \angle ABC \).

Therefore, Radius of circumcircle = \( \Large \frac{10}{2} = 5 cm \)

67). PQ is chord of length 6 cm of a circle of radius 5 cm. Tangents to the circle at P and Q meet at T. Length of TP is

A). 4.75 cm

B). 2.75 cm

C). 3.75 cm

D). 4.25 cm

View Answer

Correct Answer: 3.75 cm

Let distance from M point to T point = x cm.

In \( \Large \triangle OPM \),

\( \Large OM = \sqrt{5^{2}-3^{2}} = 4 \)

\( \Large \therefore MN = ON - OM = 5 - 4 = 1 cm \)

In \( \Large \triangle POT, \)

\( \Large PT^{2} = OT^{2} - OP^{2} = \left(5 + x\right)^{2} - 5^{2} \)

= \( \Large x^{2} + 10x \) ...(i)

In \( \Large \triangle PMT, \)

\( \Large PT^{2} = PM^{2} + TM^{2} \)

= \( \Large 3^{2} + \left(1 + x\right)^{2} \)

= \( \Large x^{2} + 2x + 10 \) ...(ii)

\( \Large x^{2} + 2x + 10 = x^{2} + 10x \)

=> \( \Large x = \frac{10}{8} = \frac{5}{4} \)

From Eq. (i).

\( \Large PT = \sqrt{x^{2}+10x} \)

= \( \Large \sqrt{ \left(\frac{5}{4}\right)^{2} + 10 \left(\frac{5}{4}\right) } \)

= 3.75 cm

68). ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB = AE = BF, then

A). \( \Large ED^{2} + CF^{2} = EF^{2} \)

B). \( \Large ED || CF \)

C). \( \Large ED > CF \)

D). \( \Large ED \perp CF \)

View Answer

Correct Answer: \( \Large ED \perp CF \)

Produce EC and FC to meet at a point M.

Let each side of the rhombus be x. Then,

AB = BC = CD = AD = BF = AE = x

\( \Large \because \angle DAB + \angle CBA = 180 ^{\circ} \)

\( \Large \therefore \angle DAE + \angle CBF = 180 ^{\circ} \)

In, \( \Large \triangle ADF, \)

\( \Large \angle EAD + \angle AED + \angle ADE = 180 ^{\circ} \) ...(i)

\( \Large \triangle BCF, \)

\( \Large \angle CBE + \angle CFB + \angle BCF = 180 ^{\circ} \) ...(ii)

On adding Eqs. (i) and (ii), we get

\( \Large \left( \angle EAD + \angle CBF\right) + \left( \angle AED + \angle ADE \right) + \left( \angle CFB + \angle BCF\right) \)

=>\( \Large 180 ^{\circ} + 2\angle AED + 2 \angle CFB = 360 ^{\circ} \)

[Therefore, AE = AD and BC = BF]

=> \( \Large \angle AED + \angle CFB = 90 ^{\circ} \)

=> \( \Large \angle EMF = 90 ^{\circ} \)

=> \( \Large EM \perp FM \)

\( \Large \therefore ED \perp CF \)

69). AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are of the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is

A). 12 cm

B). 13 cm

C). 10 cm

D). 11cm

View Answer

Correct Answer: 13 cm

Let radius of the circle be r.

Given, MN = 17 cm

Let ON = x

Then, OM = 17 - x

\( \Large \triangle AOM, \)

\( \Large OA^{2} = OM^{2} + AM^{2} \)

=> \( \Large r^{2} = \left(17-x\right)^{2} + 5^{2} \)

= \( \Large 289 + x^{2} - 34x + 25 \)

\( \Large r^{2} = 314 + x^{2} - 34x \) ...(i)

\( \Large OC^{2} = ON^{2} + CN^{2} \)

=> \( \Large r^{2} = x^{2} + 12^{2} = x^{2} + 144 \) ...(ii)

From Eqs. (i) and (ii), we get

\( \Large 314 + x^{2} - 34x = x^{2} + 144 \)

=> 34x = 170

=> x = 5

From Eq. (ii), \( \Large r^{2} = 25 + 144 = 169 \)

r = 13 cm

70). Consider the following statements
I. Let ABCD be a parallelogram which is not a rectangle. Then, \( \Large 2 \left(AB^{2}+ BC^{2}\right) \Large \ne AC^{2} + BD^{2} \)
II. If ABCD is a rhombus with AB =4 cm, then \( \Large AC^{2} + BD^{2} = n^{3} \) for some positive integer n. Which of the above statements is/are correct?

A). Only I

B). Only II

C). Both I and II

D). Neither I nor ll

View Answer

Correct Answer: Only II

Statement I ABCD is a parallelogram,

then

In \( \Large \triangle ABC, \)

\( \Large AC^{2} = AB^{2} + BC^{2} \) ...(i)

and in \( \Large \triangle CBD, \)

\( \Large BD^{2} = BC^{2} + CD^{2} \) ...(ii)

From Eqs. (i) and (ii), we get

\( \Large AC^{2} + BD^{2} = AB^{2}+BC^{2}+BC^{2}+CD^{2} \)

= \( \Large 2AB^{2} + 2BC^{2} \)

\( \Large AC^{2} + BD^{2} = 2 \left(AB^{2}+BC^{2}\right) \)

Statement II ABCD is a rhombus and diagonals AC and BD bisect each other.

Therefore, AO = OC and OB = OD

\( \Large AB^{2} = AO^{2} + OB^{2} \)

=> \( \Large \left(4\right)^{2} = \left(\frac{AC}{2}\right)^{2} + \left(\frac{BD}{2}\right)^{2} \)

\( \Large \therefore AC^{2} + BD^{2} = 64 = \left(4\right)^{3} \)

i.e., \( \Large n^{3} \)

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