Unit Test (Solutions): Quadrilaterals | Mathematics (Maths) Class 9 PDF Download

Time: 1 hour
M.M. 30 
Attempt all questions. 
Question numbers 1 to 5 carry 1 mark each. 
Question numbers 6 to 8 carry 2 marks each. 
Question numbers 9 to 11 carry 3 marks each. 
Question numbers 12 & 13 carry 5 marks each.

Q1. Which of the following is not a quadrilateral? (1 Mark)
(a) Kite
(b) Square
(c) Triangle
(d) Rhombus

Ans: (c)
Sol: A quadrilateral has 4 sides. Kite, Square, and Rhombus have 4 sides each. A triangle has only 3 sides, so it is not a quadrilateral.

Q2. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if (1 Mark)
(a) PQRS is a rectangle
(b) PQRS is a parallelogram
(c) Diagonals of PQRS are perpendicular
(d) Diagonals of PQRS are equal

Ans: (c)
Sol: If the diagonals of PQRS are perpendicular, then the mid-point quadrilateral becomes a rectangle. This follows from the mid-point theorem in quadrilaterals.

Q3. Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is (1 Mark)
(a) 90º
(b) 95º
(c) 105º
(d) 120º

Ans: (d)
Sol: We know that the sum of angles of a quadrilateral is 360º.
Let the unknown angle be x.
Therefore, 75º + 90º + 75º + x = 360º
x = 360º – 240º = 120º.

Q4. Which of the following is not true for a parallelogram? (1 Mark)
(a) Opposite sides are equal
(b) Opposite angles are equal
(c) Opposite angles are bisected by the diagonals
(d) Diagonals bisect each other.

Ans: (c)
Sol: Opposite angles are bisected by the diagonals is not true for a parallelogram. Whereas opposite sides are equal, opposite angles are equals, diagonals bisect each other are the properties of a parallelogram.

Q5. A trapezium has: (1 Mark)
(a) One pair of opposite sides is parallel
(b) Two pairs of opposite sides parallel to each other
(c) All its sides are equal
(d) All angles are equal

Ans: (a)
Sol: A trapezium has only one pair of opposite sides parallel to each other, and the other two sides are non-parallel.

Q6. In a parallelogram, one angle measures 60°. Find the measures of all other angles. (2 Marks)

Ans: 

Let the parallelogram be ABCD.
Given: ∠A = 60°
Since opposite angles of a parallelogram are equal,
∠C = ∠A = 60°
Also, adjacent angles of a parallelogram are supplementary.
So, ∠B = 180° – ∠A = 180° – 60° = 120°
∠D = ∠B = 120°
Hence, the four angles are: 60°, 120°, 60°, and 120°.

Q7. Find the perimeter of the quadrilateral with sides 5 cm, 7 cm, 9 cm and 11 cm. (2 Marks)

Ans: Given, sides of a quadrilateral are 5 cm, 7 cm, 9 cm and 11 cm.
Therefore, the perimeter of the quadrilateral is: P = 5 cm + 7 cm + 9 cm + 11 cm = 32 cm

Q8. Determine the area of a parallelogram with a base of 5 cm and a height of 3 cm. (2 Marks)

Ans: Given that the base length is 5 cm and the height is 3 cm,
Area = 5 x 3 = 15 sq.cm, according to the formula.

Q9. The perimeter of the quadrilateral is 50 cm and the lengths of the three sides are 9 cm, 13 cm and 17 cm. Find the missing side of the quadrilateral. (3 Marks)

Ans: Let the unknown side of the quadrilateral = x
Given, Perimeter of the quadrilateral = 50 cm
The lengths of the other three sides are 9 cm, 13 cm and 17 cm
As we know,
Perimeter = sum of all four sides.
50 = 9 cm + 13 cm + 17 cm + x
50 = 39 + x
x = 50 – 39
x = 11
Therefore, the fourth side of the quadrilateral = 11 cm

Q10. In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals. (3 Marks)

Ans: 

In rectangle ABCD, diagonal AC makes an angle of 25° with side AB.
So, ∠CAB = 25° (Given)

Since ∠A = 90° in a rectangle,
∠ABC = 90°, and triangle ABC is right-angled at B.

In triangle ABC:
∠CAB + ∠ACB = 90°
⇒ ∠ACB = 90° – 25° = 65°

Now, diagonals AC and BD intersect at point O.
In triangle AOB, the angles at A and B are both 65°.

So, angle between diagonals = ∠AOB
∠AOB = 180° – ∠CAB – ∠ACB
= 180° – 65° – 65° = 50°

Q11. Calculate all the angles of a quadrilateral if they are in the ratio 2 : 5 : 4 : 1.  (3 Marks)

Ans: As the angles are in the ratio 2 : 5 : 4 : 1, they can be written as-
2x, 5x, 4x, and x
Now, as the sum of the angles of a quadrilateral is 360°,
2x + 5x + 4x + x = 360°
Or, x = 30°
Now, all the angles will be,
2x =2 × 30° = 60°
5x = 5 × 30° = 150°
4x = 4 × 30° = 120°, and
x = 30°
Therefore, the angles are 60°, 150°, 120°, and 30°.

Q12.  In an isosceles trapezium ABCD, AB ∥ CD and AD = BC. Show that ∠A = ∠B and ∠C = ∠D. (5 Marks)

Ans: 

Given:

ABCD is an isosceles trapezium
AB ∥ CD (AB and CD are the parallel sides)
AD = BC (non-parallel sides are equal)

To Prove:

∠A = ∠B

∠C = ∠D

Proof:

Since AB ∥ CD and AD = BC, triangle ABD ≅ triangle CBA by SSS or ASA rule (depending on the approach).

Alternatively, consider ∆ABD and ∆CDB:

AB ∥ CD, so ∠A and ∠B are alternate interior angles

AD = BC (Given)

So, ∠A = ∠B and ∠C = ∠D

Conclusion:

Base angles of an isosceles trapezium are equal

Hence, ∠A = ∠B and ∠C = ∠D

Q13. Show that the diagonals of a rhombus bisect each other at right angles. (5 Marks)

Ans: 

Let ABCD be a rhombus with diagonals AC and BD intersecting at O.
Since diagonals of a rhombus bisect each other,
⇒ AO = OC and BO = OD
Also, all sides of a rhombus are equal: AB = BC = CD = DA
In triangles △AOB and △COB:

• AO = OC (Given)

• BO = BO (Common)

• AB = BC (Sides of rhombus)
  So, △AOB ≅ △COB (SSS)
  ⇒ ∠AOB = ∠COB (CPCT)

But ∠AOB and ∠COB form a linear pair.
So, ∠AOB + ∠COB = 180°
⇒ 2∠AOB = 180° ⇒ ∠AOB = 90°
Therefore, diagonals of a rhombus bisect each other at right angles.