QUADRILATERAL AND POLYGON (CDS)

Total Questions: 41

31. Consider the following statements [2014 (II) Evening Shift]

1. ‘Area of ∆ADM − Area of ∆ ABM’ is always equal to area of ∆DCM, if AB = CD.
2. Half of area of ∆ABM is equal to one-eighth of area of trapezium ABCD, if AB = CD.

Which of the above statement(s) is/are correct?

Correct Answer: (c) Both 1 and 2
Solution:


32. ABCD is a parallelogram. P and R are the mid-points of DC and BC, respectively. The line PR intersects the diagonal AC at Q. The distance CQ will be [2014 (II) Evening Shift]

Correct Answer: (a) AC/4
Solution:Given, ABCD is a parallelogram. Join AC and BDwhich intersect each other at O.

33. Bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect each other at a point P. Which one of the following is correct? [2014 (II) Evening Shift]

Correct Answer: (a)
Solution:Given, a quadrilateral ABCD, AP and BP are bisectors of ∠A and ∠B, respectively.

34. Two light rods AB = a + b and CD = a −b symmetrically lying on a horizontal AB. There are kept intact by two strings AC and BD. The perpendicular distance between rods is a. The length of AC is given by [2014 (I) Morning Shift]

Correct Answer: (d)
Solution:Since, they are symmetrically lying on horizontal plane.

35. If PQRS is a rectangle such PQ = √3 QR. Then, what is ∠PRS equal to? [2014 (I) Morning Shift]

Correct Answer: (c) 30°
Solution:In rectangle PQRS, PQ || RS

36. In a trapezium, the two nonparallel sides are equal in length, each being of 5 cm. The parallel sides are at a distance of 3 cm apart. If the smaller side of the parallel sides is of length 2 cm, then the sum of the diagonals of the trapezium is [2014 (I) Morning Shift]

Correct Answer: (b) 6√5 cm
Solution:In ∆BCF, by Pythagoras theorem, (5²) = (3²) + (BF²)

37. The area of a rectangle lies between 40 cm² and 45 cm². If one of the sides is 5 cm, then its diagonal lies between [2014 (I) Morning Shift]

Correct Answer: (b) 9 cm and 11 cm
Solution:Area of rectangle lies between 40 cm² and 45 cm² .

38. Let ABCD be a parallelogram. Let P, Q, R and S be the mid-points of sides AB, BC, CD and DA, respectively. Consider the following statements. [2014 (I) Morning Shift]

I. Area of ∆APS < Area of ∆DSR, if BD < AC.
II. Area of ∆ABC = 4 (Area of ∆BPQ).

Select the correct answer using the codes given below.

Correct Answer: (b) Only II
Solution:

39. Consider the following statements [2014 (I) Morning Shift]

I. Let ABCD be a parallelogram which is not a rectangle. Then, (AB² + BC²) ≠ AC² + BD²
II. If ABCD is a rhombus with AB = 4 cm, then AC² + BD² = n³ for some positive integer n.

Which of the above statement(s) is/are correct?

Correct Answer: (b) Only II
Solution:

40. ABCD is a parallelogram. E is a point on BC such that BE : EC = m : n. If AE and DB intersect in F, then what is the ratio of the area of ∆FEB to the area of ∆AFD? [2014 (I) Morning Shift]

Correct Answer: (d) [m/(m + n)]²
Solution:In ∆AFD and ∆BFE,