Quadrilaterals Class 9 MCQ Maths

Quadrilaterals Class 9 MCQ with Solution

1. The figure formed by joining the consecutive mid-points of any rhombus is always-

(A) a square

(B) a rhombus

(D) None of the above

Ans. (C)

Sol. The figure formed by joining the consecutive mid-points of any rhombus is always a parallelogram (by using mid point theorem.)

2. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order is a rhombus if :

(A) PQRS is rhombus

(B) PQRS is a parallelogram

(D) Diagonals of PQRS are equal

Ans. (D)

Sol. PQ = SR = 1/2 AC

SP = QR = 1/2 BD

For Rhombus PQ = SP

so 1/2 AC = 1/2 BD

AC = BD

so diagonals should be equal.

3. Choose the correct statement :

(A) The diagonals of a parallelogram are equal

(B) The diagonals of a rectangle are perpendicular to each other.

(D) Every quadrilateral is either a trapezium or a parallelogram or a kite

Ans. (C)

Sol. If the diagonals of a quadrilateral intersect at right angles.

It is not necessarily a rhombus.

4. ABCD is a rhombus, EABF is a straight line such that EA = AB = BF and ED and FC when produced meet at G, then∠DGC is :

(A) An acute angle

(B) An obtuse angle

(D) A reflex angle

Ans. (C)

Sol. Let $\angle DAB$ = θ

So $\angle CBA$ = $180^\circ - \theta$

here DA = EA

\angle EDA

\angle AED

so $\angle AED$ = $\theta /2$

similarly $\angle BFC$ = $\frac{{180 - \theta }}{2}$

In $\Delta EGF$ $\angle DGC$ = $180 - \left[ {\frac{\theta }{2} + \frac{{180 - \theta }}{2}} \right]$

= 90°

5. ABCD is a ||gm. AB is produced to E so that BE = AB. If DE and BC intersect at O and OB = 5 cm, then AD =

(A) 5 cm

(B) 10 cm

(D) Can’t be determined

Ans. (B)

Sol. AB = BE

& BC || AD

so O will be mid point of BC

so BC = CD

& AD = BC = 10

6. In figure, ABCD is a ||gm then ∠A = :

(A) 60°

(B) 75°

(D) None of these

Ans. (C)

Sol. $\angle A$ = $\angle C = 5a$

so 2a + 3a + 5a = 180°

10a = 180°

a = 18°

so $\angle A$ = 5a = 90°

7. Which one is not true in the following :

(A) A square, rectangle and rhombus are parallelograms

(B) A parallelogram is a trapezium, and a trapezium is also a parallelogram

(D) A kite is not a paralleogram

Ans. (B)

Sol. A parallelogram is a trapezium but trapezium is also a parallelogram is not true.

8. One angle of a cyclic trapezium is double the other. The measure of the larger angle is :

(A) 60º

(B) 75º

(D) 120º

Ans. (D)

Sol. $\alpha + 2\alpha$ = 180°

$\alpha$ = 60°

⇒ 2 $\alpha$ = 120°

9. The adjacent sides of a parallelogram are in the ratio of a : b. The ratio of the corresponding altitude is

(A) a : b

(B) b : a

(D) b² : a²

Ans. (B)

Sol. Area of parallelogram will be constant

a.h₁ = b.h₂

\frac{{{h_1}}}{{{h_2}}}

\frac{b}{a}

10. A trapezium has its non-parallel sides congruent, then its opposite angles are :

(A) Congruent

(B) Supplementary

(D) None of these

Ans. (B)

Sol. It will be a cyclic trapezium

So opposite angles will be supplementary.

11. The sum of the interior angles of a 12 sided polygon is :

(A) 1860^o

(B) 1980^o

(D) 1800^o

Ans. (D)

Sol. (n –2) × 180º = 10 × 180º = 1800°

12. The sum of the interior angles of a polygon is seven times the sum of its exterior angles. Then number of sides in polygon is :

(A) 12

(B) 15

(D) None of these

Ans. (C)

Sol. (n – 2) × 180º = 7 × 360º

\Rightarrow

n = 16

13. The interior angles of a pentagon are ao, (a+10)^o, (a + 20)^o, (a+30)^o and (a+40)^o. Then which of the following is one of the angles of the pentagon.

(A) 68^o

(B) 96^o

(D) 108^o

Ans. (D)

Sol. a + a + 10º + a + 20º + a + 30º + a + 40º = (5 – 2) × 180º

\Rightarrow

5a = 440º

\Rightarrow

a = 88º

angles are 88º, 98º, 108º, 118º and 128º.

14. In the given figure, ABCD is an isosceles trapezium in which ∠CDA = 2x° and ∠BAD = 3x° then x =:

(A) 18°

(B) 27°

(D) 45°

Ans. (C)

Sol. 2xº + 3xº = 180º

\Rightarrow

x =

= 36º

15. The sum of interior angles of a polygon is three times the sum of its exterior angles. Then number of sides in the polygon is :

(A) 6

(B) 7

(D) 9

Ans (C)

Sol. (n – 2) × 180° = 3 × 360°

⇒ n = 8

16. The sides BA and DC of the quadrilateral ABCD are produced as shown in the figure then :

(A) a + x = b + y

(B) a + y = b + x

(D) None of these

Ans. (C)

Sol. x + 180º – b + y + 180º – a = 360º

\Rightarrow

a + b = x + y

17. Four angles of an 8-sided polygon are each of 154°. If the remaining four angles are equal, the measure of each remaining angle is :

(A) 116°

(B) 120°

(D) 128°

Ans. (A)

Sol. (4×154°) (4 × x) = (8 – 2) × 180°

4x = 1080º – 616º = 464º

x = 116°

18. ABCD is a || gm. If two diagonals are equal, then $\angle A{\bf{B}}C$ =

(A) 60^o

(B) 45^o

(D) Can’t be determined.

Ans. (C)

Sol. [ABCD will be a rectangle, so ∠ABC = 90°]

19. The bisectors of the angles of a || gm enclose a :

(A) Parallelogram

(B) Rhombus

(D) Square

Ans. (C)

Sol.

ABCD be a parallelogram and AP, BP, CQ, DQ are angle bisectors

∠P = ∠Q = ∠R = ∠S = 90°

So, PQRS is a rectangle.

20. In a quadrilateral ABCD, if AO and BO are the bisectors $\angle A$ of $\angle B$ and respectively, $\angle C = {70^o}$ and $\angle D = {130^o}$ , then $\angle AOB$ is :

(A) 40^o

(B) 80^o

(D) 100^o

Ans. (D)

Sol.

∠A + ∠B = 360° – (70° + 130°)

= 160°

2(x+y) = 160°

⇒ x+y = 80°

∠AOB = 180° – (x+y) = 100°

21. If one angle of a parallelogram is 24^o less than twice the smallest angle, then the largest angle of the parallelogram is :

(A) 176°

(B) 68°

(D) 102°

Ans. (C)

Sol. x + x+(2x–24°) +(2x–24°) = 360°

2x + 4x – 48° = 360°

6x = 408°

x = 68°

Largest Angle = 2 × 68° – 24

= 136° – 24° = 112°

22. Which one of the following statements is correct for a square :

(A) Diagonals are equal and bisect each other at right angles.

(B) Diagonals are unequal & do not bisect each other.

(D) Diagonals are unequal and bisect each other at right angle.

Ans. (A)

Sol. For a square, diagonals are equal and bisect each other at right angles.

23. The adjacent interior angles of a parallelogram are (2x – 15)^o and (7x – 75)^o then x =

(A) 25^o

(B) 30^o

(D) 40^o

Ans. (B)

Sol. (2x –15)° +(7x –75)° = 180°

9x = 180° + 90° = 270°

x = 30°

24. In which of the following is the lengths of diagonals equal ?

(A) Rhombus

(B) Parallelogram

(D) Rectangle

Ans. (D)

Sol. Rectangle

25. The length of a side of a rhombus is 5 m and one of its diagonals is of length 8 m. The length of the other diagonal is :

(A) 5 m

(B) 7 m

(D) 8 m

Ans. (C)

Sol.

ABCD is a rhombus

AO = $\sqrt {{5^2}\, - \,{4^2}}$ = 3 m

AC = 2(AO) = 6 m

26. If the lengths of two diagonals of a rhombus are 12 cm and 16 cm, then the length of each side of the rhombus is:

(A) 10 cm

(B) 14 cm

(D) None of these

Ans. (A)

Sol. Each side of rhombus = $\sqrt {{6^2} + {8^2}} = 10\,cm$

27. ABCD is a rectangle with $\angle ABD = {40^o}$ . Then DBC is :

(A) 40^o

(B) 90^o

(D) None of these

Ans. (C)

Sol. ∠DBC = 90° – 40° = 50°

28. If ABCD is a quadrilateral such that its diagonals AC and BD intersect at O into four triangles of equal area. Then ABCD must be a :

(A) Parallelogram

(B) Rhombus

(D) Rectangle

Ans. (A)

Sol. Parallelogram

29. In figure ABCD and AEFG are both || gm.

If ∠C = 60° then ∠F = :

(A) 60°

(B) 120°

(D) None of these

Ans. (A)

Sol. ∠C = ∠F = 60°

30. ABCD is a rhombus with $\angle$ ABC = 58^o. Then $\angle$ ACD is :

(A) 60^o

(B) 62^o

(D) 61^o

Ans. (D)

Sol. ∠ACD = $\frac{{180^\circ \, - \,58^\circ }}{2}\, = \,61^\circ$

31. In a parallelogram ABCD, AB = 10cm. The altitudes corresponding to the sides AB and AD are 7 cm and 8 cm respectively. Length of the AD is :

(A) 5 cm

(B) 8.75 cm

(D) 11.75 cm

Ans. (B)

Sol. AD × 8 = 10 × 7 $\Rightarrow$ AD = $\frac{{10 \times 7}}{8}$ = 8.75 cm

32. If O is a point within a quadrilateral then,

(A) OA < OB

(B) 2.OB = AO.OD

(D) OA + OB + OC + OD > AC + BD

Ans. (D)

Sol. In ΔAOC and ΔBOD, OA + OC > AC and OB + OD > BD

Hence, OA + OB + OC + OD > AC + BD

33. In the figure, ABCD is a parallelogram and the bisector of ∠ A bisects BC at x. Then,

(A) AD = BD

(B) $AC = \frac{1}{2}AX$

(D) None of these

Ans. (C)

Sol. ∠DAX = ∠XAB = ∠BXA

So ∠1 = ∠2 and AB = BX = XC

$\Rightarrow$ 2AB = AD

34. If each interior angle of a regular polygon is 178^o, then the number of sides of that polygon is :

(A) 180 sides

(B) 360 sides

(D) 178 sides

Ans. (A)

Sol. exterior angle = (180º – 178º)

= 2º

number of sides = = 180

35. Which of the following is not a polygon :

(A) Triangle

(B) Square

(D) Heptagon

Ans. (C)

Sol. Circle

36. In figure ABCD is a rhombus. If ∠CAB = 2a and ∠BCA = 3a and ∠DAC = a + 40, find the value of ∠ABC :

(A) 70°

(B) 80°

(D) 60°

Ans. (B)

Sol. a + 40° = 3a

⇒ a = 20°

∠ABC = 180° – (2a + 3a) = 180° – 5a = 80°

37. The perimeter of a rhombus is 20 cm. One of its diagonal is 8 cm. Then its area:

(A) 24 sq. cm

(B) 48 sq. cm

(D) 92 sq. cm

Ans. (A)

Sol. Side of rhombus = $\frac{{20}}{4}$ = 5 cm

Other diagonal = $2 \times \,\sqrt {{5^2}\, - \,{4^2}} \,\, = \,\,6\,cm$

Area = = 24 sq.cm

38. In figure, ABCD is a rhombus with ∠BAD = 50^o , then (x,y) =

(A) (25^o,65^o)

(B) (25^o,50^o)

(D) None of these

Ans. (D)

Sol. yº = ∠BAC = = 25º

∠ABC = 2xº

= 180º – 50º

= 130º

\Rightarrow

x = 65º

Hence, (x, y) = (65º, 25º)

39. The angles of a quadrilateral are x°, x – 10°, x + 30° and 2x°. Find the greatest angle:

(A) 136°

(B) 180°

(D) None of these

Ans. (A)

Sol. x + x – 10º + x + 30º + 2xº = 360º

5xº = 360º + 10º – 30º

= 340º

\Rightarrow

xº =

greatest angle = 2xº = 68º × 2

= 136º

40. Two adjacent angles of a parallelogram are in the ratio 4 : 5. The angles are:

(A) 180°, 180°

(B) 36°, 144°

(D) None of these

Ans. (C)

Sol. 4x + 5x = 180º

\Rightarrow

x = 20º

angle are 4x, 5x

\Rightarrow

80º, 100º

41. If one of the interior angles of a regular polygon is found to be equal to $\frac{9}{8}$ times of one of the interior angles of a regular hexagon, then the number of sides of the polygon is

(A) 4

(B) 5

(D) 8

Ans. (D)

Sol. $\frac{{(n\, - \,2) \times \,180^\circ }}{n}$ = $\frac{9}{8}\, \times \,\frac{{(6\, - 2) \times \,180^\circ }}{6}$

⇒ n = 8

42. The diagonals of a rectangle ABCD meet at O. If BOC = 44°, then OAD equals :

(A) 44°

(B) 136°

(D) 68°

Ans. (D)

Sol.

∠OAD = $\left( {\frac{{180^\circ \, - \,44^\circ }}{2}} \right)$

= 68°

43. In the given figure, ABCD is a || gm and angle bisectors of ∠A and ∠B meet at P. Then ∠APB equals :

(A) 90°

(B) 45°

(D) None of these

Ans. (A)

Sol. ∠A + ∠B = 180º

∠APB = 180º – $\left( {\frac{{\angle A}}{2} + \frac{{\angle B}}{2}} \right)$

= 180º –

= 90º

44. ABCD is a parallelogram. If AB = 2AD and P mid point of AB, then CPD is equal to :

(A) 90°

(B) 60°

(D) 135°

Ans. (A)

Sol.

∠A = 180 –2y

∠B = 180 –2x

∠A + ∠B = 180°

180° – 2y + 180 – 2x = 180°

x + y = 90°

x + y + ∠CPD = 180°

∠CPD = 90°

45. If PQ and RS are two perpendicular diameters of a circle, then PRQS is a :

(A) Rectangle

(B) Trapezium

(D) Rhombus but not square

Ans. (C)

Sol. Square

46. In a rhombus ABCD, ∠A = 60° and AB = 6cm. Then the diagonal BD is :

(A) $2\sqrt 3$ cm

(B) 6 cm

(D) Insufficient data

Ans. (B)

Sol. ∠A = 60º = ∠B = ∠D

So, ΔABD is an equilateral triangle.

Hence, BD = 6 cm.

47. In a || gm ABCD, ∠A = 60°. If the bisectors of ∠A and ∠B meet DC at P then ∠APB is :

(A) Acute angle

(B) Obtuse angle

(D) None of these

Ans. (C)

Sol. ∠A + ∠B = 180º

∠APB = 180º – $\left( {\frac{{\angle A}}{2} + \frac{{\angle B}}{2}} \right)$

= 180º – 90º

= 90º

48. In a || gm ABCD, ∠A = 60°. If the bisectors of ∠A and ∠B meet DC at P then ΔBCP is :

(A) An isosceles triangle

(B) A right triangle

(D) An equilateral triangle

Ans. (D)

Sol. ∠A = ∠C = 60°

∠CBP = ∠BPC = 60° (∠BPC = 180° – 120° = 60°)

Hence ΔBCP is an equilateral triangle

49. In a parallelogram, which one is not correct :

(A) Its opposite sides are equal

(B) Its opposite angles are equal

(D) None is false

Ans. (C)

Sol. In a parallelogram, diagonals do not bisect at 90º.

50. If PQ and RS are two unequal perpendicular diagonals of a quadrilateral and these are bisecting each other then the quadrilateral is :

(A) Square

(B) Rhombus

(D) Both (B) and (C)

Ans. (B)

Sol. Quadrilateral can be rhombus.

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