Quadrilaterals MCQs

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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°

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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°

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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°

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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.

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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°

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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°

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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°

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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°

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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°

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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

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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°

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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.

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them

Monthly consumption 
(in units)

 Number of consumers

65 - 85 

4

85 - 105

5

105 - 125

13

125 - 145

20

145 - 165

14

165 - 185

8

185 - 205

4

      2.

      A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

          3.
          Find the sums given below :
          1. \(7 + 10\frac 12+ 14 + ....... + 84\)
          2. \(34 + 32 + 30 + ....... + 10\)
          3. \(–5 + (–8) + (–11) + ....... + (–230)\)

              4.

              Form the pair of linear equations for the following problems and find their solution by substitution method.

              (i) The difference between two numbers is 26 and one number is three times the other. Find them.

              (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

              (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

              (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

              (v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

              (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

                  5.

                  Solve the following pair of linear equations by the substitution method. 
                  (i) x + y = 14 
                      x – y = 4   

                  (ii) s – t = 3 
                      \(\frac{s}{3} + \frac{t}{2}\) =6 

                  (iii) 3x – y = 3 
                        9x – 3y = 9

                  (iv) 0.2x + 0.3y = 1.3 
                       0.4x + 0.5y = 2.3 

                  (v)\(\sqrt2x\) + \(\sqrt3y\)=0
                      \(\sqrt3x\) - \(\sqrt8y\) = 0

                  (vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
                      \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

                      6.
                      The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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