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Elementary Mathematics

31). If the diagonals of a quadrilateral are equal and bisect each other at right angles, then the quadrilateral is a

A). rectangle

B). square

C). rhombus

D). trapezium

View Answer

32). In the figure given below, PQRS is a parallelogram. If AP, AQ, CR and CS are bisectors of \( \Large \angle P \), \( \Large \angle Q \), \( \Large \angle R \) and \( \Large \angle S \) respectively, then ABCD is a

A). square

B). rhombus

C). rectangle

D). None of these

View Answer

33).

In the figure given above, ABCD is a trapezium. EF is parallel to AD and BC Then, \( \Large \angle y \) is equal to

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

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

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

D). \( \Large 65 ^{\circ} \)

View Answer

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

From figure,

\( \Large x ^{\circ} = z ^{\circ} = 50 ^{\circ} \)

\( \Large \theta + z ^{\circ} = 180 ^{\circ} \)

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

Now, in quadrilateral AQFD,

\( \Large x ^{\circ} +y ^{\circ} +120 ^{\circ} + \theta = 360 ^{\circ} \)

=> \( \Large 50 ^{\circ} +y ^{\circ} + 120 ^{\circ} + 130 ^{\circ} = 360 ^{\circ} \)

\( \Large \therefore y = 360 ^{\circ} - 300 ^{\circ} = 60 ^{\circ} \)

34). A quadrilateral ABCD is inscribed in a circle. If AB is parallel to CD and AC = BD, then the quadrilateral must be a

A). parallelogram

B). rhombus

C). trapezium

D). None of these

View Answer

35). Let ABCD be a parallelogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2mn and \( \Large BD = m^{2} - n^{2} \). Let \( \Large AB = \left(m^{2} + n^{2}\right)/2 \) Statement I AC > BD
Statement II ABCD is a rhombus.
Which one of the following is correct in respect of the above statements?
(a) Both statements I and ii are true and statement ll is the correct explanation of statement I
(b) Both statements I and II are true but statement H is not the correct 50' explanation of statement I
(c) Statement | is true but statement II is false
(d) Statement II is true but statement | is false

A). Both statements I and ii are true and statement ll is the correct explanation of statement |

B). Both statements I and II are true but statement H is not the correct 50' explanation of statement l

C). Statement | is true but statement II is false

D). Statement II is true but statement | is false

View Answer

Correct Answer: Both statements I and II are true but statement H is not the correct 50' explanation of statement l

Given that, \( \Large AC = 2mn, BD = m^{2}-n^{2} and AB = \frac{m^{2}+n^{2}}{2} \)

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

=> \( \Large \left(4m^{2}n^{2}+m^{4}+n^{4}-2m^{2}n^{2}\right) \)

= \( \Large 2 \{ \frac{1}{4} \left(m^{2}+n^{2}\right)^{2}+BC^{2} \} \)

=> \( \Large \left(m^{2}+n^{2}\right)^{2} = \frac{1}{2} \left(m^{2}+n^{2}\right)^{2}+2BC^{2} \)

=> \( \Large 2BC^{2} = \frac{1}{2} \left(m^{2}+n^{2}\right)^{2} \)

=> \( \Large BC^{2} = \frac{ \left(m^{2}+n^{2}\right)^{2} }{4} \)

=> \( \Large BC = \frac{m^{2}+n^{2}}{2} \)

\( \Large \therefore ABCD is\ a\ rhombus. \)

Let \( \Large AC > BD => 2mn > m^{2}-n^{2} \)

\( \Large (m + n)^{2} > 2m^{2}\)

which is always true for every positive integers m, n, where n < m < 2n.

36). If ABCD be a rectangle and P,Q,R,S be the mid-points of \( \Large \bar{AB} , \bar{BC}, \bar{CD} \ and \ \bar{DA}\)respectively, then the area of the quadrilateral PQRS is equal to

A). \( \Large \frac{1}{3} area \left(ABCD\right) \)

B). \( \Large \frac{3}{4} area \left(ABCD\right) \)

C). \( \Large \frac{1}{2} area \left(ABCD\right) \)

D). \( \Large area \left(ABCD\right) \)

View Answer

Correct Answer: \( \Large \frac{1}{2} area \left(ABCD\right) \)

Let length and breadth of rectangle ABCD are 2x and 2y, respectively.

Then, \( \Large Area\ of ABCD = 2x \times 2y = 4xy \)

Area of all the four triangles

=\( \Large 4 \times \frac{1}{2} \times x \times y = 2xy \)

Area of PQRS = \( \Large 4xy - 2xy = 2xy \)

= \( \Large \frac{1}{2} \left(4xy\right)=\frac{1}{2} area \left(ABCD\right) \)

37). Let X be any point within a square4 ABCD. On AX, a square AXYZ is described such that D is within it. Which one of the following is correct?

A). AX = DZ

B). \( \Large \angle ADZ \angle BAX \)

C). AD = DZ

D). BX = DZ

View Answer

Correct Answer: BX = DZ

In \( \Large \triangle ABX \ and \triangle ACZ, \)

\( \Large AB = AD [side\ of\ square\ ABCD] \)

and \( \Large AX = AZ [side\ of\ a\ square\ AXYZ] \)

\( \Large Let \angle BAX = \theta \)

\( \Large \therefore \angle XAD = 90 ^{\circ} - \theta \)

Also, AXYZ is a square.

\( \Large \therefore \angle ZAX = 90 ^{\circ} \)

=> \( \Large \angle ZAD + \angle XAD = 90 ^{\circ} \)

=> \( \Large \angle ZAD = 90 ^{\circ} - \left(90 ^{\circ} -\theta\right) = \theta \)

i.e., \( \Large \angle BAX = \angle ZAD \)

\( \Large \therefore \triangle ABX = \triangle ADZ \)

\( \Large \therefore BX = DZ \)

38). AB is the diameter of a circle with centre O and P is a point on it. If \( \Large \angle POA = 120^{\circ}\), then the value of \( \Large \angle \)PBO is

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

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

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

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

View Answer

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

The angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of the circle.

\( \Large \therefore \angle PBO = \frac{1}{2} \times \angle POA \)

\( \Large = \frac{1}{2} \times 120 ^{\circ} \)

\( \Large = 60 ^{\circ} \)

39). Two circles of same radius 5 cm, intersect each other at A and B. If AB = 8 cm, then the distance between the centres is

A). 10 cm

B). 4 cm

C). 6 cm

D). 8 cm

View Answer

Correct Answer: 6 cm

From the figure,

on joining AO and AP.

In \( \Large \triangle AOX, \ \)

\( \Large AO^{2} = AX^{2}+OX^{2} => \left(5\right)^{2}= \left(\frac{AB}{2}\right)^{2}+ \left(OX\right)^{2} \)

=> \( \Large 25 = \left(\frac{8}{2}\right)^{2}+OX^{2} \)\

=> \( \Large 25 = \left(4\right)^{2}+OX^{2} \)

=> \( \Large \left(OX\right)^{2} = 25 - 16 = 9 \)

\( \Large \therefore OX = \sqrt{9} = 3 cm \)

Similarly, in \( \Large \triangle APX, PX = 3 cm \)

Therefore, Distance between the centre

= \( \Large OX + PX = 3 + 3 = 6 cm \)

40). O is the centre of a circle. AC and BD are two chords of the circle intersecting each other at P. If \( \Large \angle \)AOB = \( \Large 15 ^{\circ} \) and \( \Large \angle \)APB = \( \Large 30 ^{\circ} \), then ( \Large \angle COD \) is equal to

A). \( \Large 35 ^{\circ} \)

B). \( \Large 25 ^{\circ} \)

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

D). \( \Large 55 ^{\circ} \)

View Answer

Correct Answer: \( \Large 45 ^{\circ} \)

From figure,

\( \Large \angle AOB + \angle COD = 2 \angle ACB + 2 \angle DBC \)

=> \( \Large 15 ^{\circ} + \angle COD = 2 \left( \angle ACB + \angle DBC\right) \)

=> \( \Large 15 ^{\circ} + \angle COD = 2 \angle APB \)

=> \( \Large \angle COD = 2 \times 30 ^{\circ} - 15 ^{\circ} \)

\( \Large \therefore \angle COD = 60 ^{\circ} - 15 ^{\circ} = 45 ^{\circ} \)

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