Quadrilaterals MCQs

Collegedunia Team logo

Collegedunia Team

Content Curator

Quadrilateral is a closed form that has four sides, four vertices, and four angles. It's made by connecting four non-collinear points. The total of quadrilateral inner angles is always equal to 360 degrees.

The term quadrilateral is derived from the Latin terms 'Quadra' (four) and 'Latus' (sides). It is not required for all four sides of a quadrilateral to be the same length. As a result, we can have different kinds of quadrilaterals dependent on their sides and angles. In this post, we'll look at some more fascinating facts regarding quadrilaterals.

Question 1. Three angles of a quadrilateral are 75°, 90°and 75°, the fourth angle is

(a) 90°

(b) 95°

(c) 105°

(d) 120°

Click Here for Answer

Ans. d

Explanation: According to the information provided

A quadrilateral's three angles are 75°, 90°, and 75°

Assume that the fourth angle is x.

The total of all four internal angles is 360 degrees, according to the quadrilateral angle sum property.

The sum of all quadrilateral angles equals 360°.

75° + 90° + 75° + x = 360°

⇒ 240° + x = 360°

x = 360° – 240°

x = 120°

Hence, the fourth angle is 120°.

Question 2. A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is

(a) 55°

(b) 50°

(c) 40°

(d) 25°

Click Here for Answer

Ans. b

Explanation:

In °Â³BOC,

∠OBC=∠OCB (Opposite angle of isosceles triangle)

∠OBC+∠OCB+∠BOC=180°ÂÂ

25°ÂÂ+25°ÂÂ+∠BOC=180°ÂÂ

Therefore, ∠BOC=130°ÂÂ

So, ∠AOB+∠BOC=180 (Linear pair)

∠AOB=180°ÂÂ−130°ÂÂ

∠AOB=50°ÂÂ

So, the acute angle between the diagonal is 50°ÂÂ.

Question 3. ABCD is a rhombus such that ∠ACB = 40°, then ∠ADB is

(a) 40°

(b) 45°

(c) 50°

(d) 60°

Click Here for Answer

Ans. c

Explanation:

ABCD is a rhombus in the diagram. We know that rhombus diagonals bisect each other perpendicularly.

∠BOC= 90°ÂÂ

∠OCB = 40°ÂÂ

AD||BC and BD is the transversal (The opposite sides of a rhombus are parallel to one another.)

∴ ∠ADB = ∠DBC ---- (Alternate angles)

In OBC,

∠BOC + ∠OCB + ∠OBC = 180°ÂÂ

⇒ 90°ÂÂ+ 40°ÂÂ + ∠OBC = 180°ÂÂ

⇒ ∠OBC =180°ÂÂ - 130°ÂÂ

∴ ∠OBC = 50°ÂÂ

But ∠OBC =∠DBC

∴ ∠ADB = 50° ---( Alternate angle)

Question 4. If angles A, B, C and D of a quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a

(a) rhombus

(b) parallelogram

(c) trapezium

(d) kite.

Click Here for Answer

Ans. c

Explanation: Because the angles A, B, C, and D of the quadrilateral ABCD are in the ratio 3: 7: 6: 4, we obtain angles A, B, C, and D = 3x, 7x, 6x, and 4x. Now, the sum of a quadrilateral's angles equals 360o. As a result, ABCD is a trapezium.

Question 5. The diagonals AC and BD of a || gm ABCD intersect each other at the point O. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC is equal to

(a) 24°

(b) 86°

(c) 38°

(d) 32°

Click Here for Answer

Ans. c

Explanation: So, AD °Â£°Â£ BC

∴ ∠DAC = ∠ACB --- ( Alternate angle)

∴ ∠ACB = 32°ÂÂ

∠AOB + ∠BOC = 180°ÂÂ --- (straight angle)

⇒70°ÂÂ + ∠BOC = 180°ÂÂ

∴ ∠BOC = 110°ÂÂ

In °Â³BOC,

∠OBC + ∠BOC + ∠OCB = 180°ÂÂ

⇒∠OBC + 110°ÂÂ + 32°ÂÂ = 180°ÂÂ

⇒ ∠OBC = 38°ÂÂ

∴ ∠DBC = 38°ÂÂ

Question 6. ABCD is a rhombus such that ∠ABC = 40°, then ∠ADC is equal to

(a) 40°

(b) 45°

(c) 50°

(d) 20°

Click Here for Answer

Ans. a

Explanation: ∠BCD=2×∠ACB=2×40°ÂÂ=80°ÂÂBut∠BCD+∠ADC=180°ÂÂ (Sum of consecutive angle of||gm)⇒80°+∠ADC=180°⇒∠ADC=180°−80°=100°∴∠ADB=12∠ADC=12×100°=50°

Question 7. In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 60°, then ∠GFE is

(a) 30°

(b) 60°

(c) 90°

(d) 120°

Click Here for Answer

Ans. b

Explanation: The letters ABCD and AEFG form a parallelogram.

Consider the parallelogram ABCD first.

Since it is assumed that C = 60°

As a result, A or DAB = 60°. (angles opposite in a parallelogram are equal)

However, GAE is a component of DAB.

As a result, GAE = 60° - (i)

Take a look at the parallelogram AEFG.

GAE = 60° according to equation I

As a result, GFE = 60°. (angles opposite in a parallelogram are equal).

As a result, the value of GFE is 60°.

Question 8. The bisectors of any two adjacent angles of a || gm intersect at

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Click Here for Answer

Ans. d

Explanation: The bisectors of any two adjacent parallelogram angles meet at 90 degrees.

Question 9. If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of a parallelogram is

(a) 176°

(b) 68°

(c) 112°

(d) 102°

Click Here for Answer

Ans. c

Explanation: Let x be the smallest angle.

The maximum angle will then be =180x.

However, the same equals [2x24].

As a result, [2x24]=180x

3x=204

x=68

As a result, the maximum angle is 180-68=112 degrees.

Question 10. If the diagonal of a rhombus are 18 cm and 24 cm respectively, then its side is equal to

(a) 16 cm

(b) 15 cm

(c) 20 cm

(d) 17 cm

Click Here for Answer

Ans. b

Explanation: 

AC = 24 cm, BD = 18 cm, henceAB=BC=CD=DA [side of rhombus] We know that rhombus diagonals intersect each other at 90°. In the appropriate

 AOB, AB2 = BO2 + AO2 AB2

 = 122 + 92 = 144 + 81 = 225 AB

 = √ 225 225 = 15 cm 

Side of rhombus = 15 cm

Question 11. In the given figure, ABCD is a parallelogram. Find the value of x.

(a) 25°

(b) 60°

(c) 75°

(d) 45°

Click Here for Answer

Ans. d

Explanation: The opposite angles of a parallelogram are of equal size. A and C are opposing angles in the parallelogram ABCD, while B and D are opposite angles. The measures of A and C are given in the question, and they must be equal.

3x-50=x+40

2x=90

X=45

Also Read:

CBSE X Related Questions

  • 1.
    Determine the ratio in which the line \(2x + y = 6\) divides the line segment joining the points (1, 3) and (2, 5).


      • 2.
        In the given figure, \(\Delta ABC\) is a right-angled triangle with \(\angle A = 90^\circ\). AD is perpendicular to BC.

        35(a)(i) Prove that \(\Delta DBA \sim \Delta DAC\)


          • 3.
            Prove that : \(\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1}\).


              • 4.
                If 14th term of an A.P. is 4 and its 15th term is zero, then its first term is

                  • –48
                  • –56
                  • 56
                  • 48

                • 5.
                  Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories. Based on above information, answer the following questions:

                  38(i) Represent the above situation in terms of a pair of linear equations in two variables.


                    • 6.
                      In the given figure, two triangles ABC and PQR are shown such that \(\angle A = \angle P\) and \(\angle C = \angle R\). If \(AD \perp BC\) and \(PS \perp QR\), then prove that (i) \(\Delta ADB \sim \Delta PSQ\) (ii) \(AD \times QS = BD \times PS\).

                        Comments


                        No Comments To Show