Unit Test (Solutions): Quadrilaterals | Mathematics Class 8- New NCERT (Ganita Prakash) 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 number 12 & 13 carry 5 marks each

Q1: The quadrilateral whose all its sides are equal and angles are equal to 90 degrees, it is called:  (1 Mark)

a. Rectangle

b. Square

c. Kite

d. Parallelogram

Answer: b

Q2: The sum of all the angles of a quadrilateral is equal to:  (1 Mark)

a. 180°

b. 270°

c. 360°

d. 90°

Answer: c

Q3: A trapezium has:  (1 Mark)

a. One pair of opposite sides parallel

b. Two pairs of opposite sides parallel to each other

c. All its sides are equal

d. All angles are equal

Answer: a

Explanation: A trapezium has only one pair of opposite sides parallel to each other, and the other two sides are non-parallel.

Q4: A rhombus can be a:  (1 Mark)

a. Parallelogram

b. Trapezium

c. Kite

d. Square

Answer: d

Q5: A diagonal of a parallelogram divides it into two congruent:  (1 Mark)

a. Square

b. Parallelogram

c. Triangles

d. Rectangle

Answer: c

​Q6: Write true and false against each of the given statements.  (2 Marks)
(a) Diagonals of a rhombus are equal.
(b) Diagonals of rectangles are equal.
(c) Kite is a parallelogram.
(d) Sum of the interior angles of a triangle is 180°.

Ans: 

(a) False

(b) True

(c) False

(d) True

Q7: Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is (2 Marks)

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 

x = 120º.

Q8: The angles of a quadrilateral are in the ratio 4: 5: 10: 11. The angles are: (2 Marks)

Sol: Let x be the common angle among all the four angles of a quadrilateral.

As per angle sum property, we know:

4x+5x+10x+11x = 360°

30x = 360°

x = 12°

Hence, angles are

4x = 4 (12) = 48°

5x = 5 (12) = 60°

10x = 10 (12) = 120°

11x = 11 (12) = 132°

Q9: A diagonal of a rectangle is inclined to one side of the rectangle at 25º. Then find the acute angle between the diagonals. (3 Marks)

Sol: Consider the rectangle ABCD

​In a triangle BOC, 

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

Therefore, ∠OBC + ∠OCB+∠BOC = 180º

25º+25º + ∠BOC = 180º

∠BOC = 180º- 50º

∠BOC = 130º.

By using the linear pair, 

∠AOB + ∠BOC = 180º

∠AOB = 180º – 130º

∠AOB= 50º

Hence, the acute angle between the diagonals is 50º.

Q10: ABCD is a rhombus such that ∠ACB = 40º. Then find ∠ADB.  (3 Marks)

Sol: ​We know that the diagonals of the rhombus bisect each other perpendicularly.
​​​

By using the alternate interior angles, and angle sum property of triangle, we can say:

From the triangle, BOC,

∠BOC + ∠OCB + ∠OBC = 180º

(where  ∠BOC= 90º, ∠OCB = 40º)

90º+40º+ ∠OBC = 180º

∠OBC = 180º – 130º

∠OBC = 50º

∠OBC =∠DBC

Now, by using alternate angles, we can say

∠ADB = 50º

Q11: In the given parallelogram ABCD, find the value of x and y.  (3 Marks)
Sol:
∠A + ∠B = 180° (adjacent angles of a parallelogram are supplementary)
3y + 2y – 5 = 180°
⇒ 5y – 5 = 180°
⇒ 5y = 180 + 5°
⇒ 5y = 185°
⇒ y = 37°
Now ∠A = ∠C [Opposite angles of a parallelogram]
3y = 3x + 3
⇒ 3 × 37 = 3x + 3
⇒ 111 = 3x + 3
⇒ 111 – 3 = 3x
⇒ 108 = 3x
⇒ x = 36°
Hence, x = 36° and y – 37°.

Q12: Find the values of x and y in the following parallelogram.  (5 Marks)
Sol: Since, the diagonals of a parallelogram bisect each other.
OA = OC
x + 8 = 16 – x
⇒ x + x = 16 – 8
⇒ 2x = 8
x = 4
Similarly, OB = OD
5y + 4 = 2y + 13
⇒ 3y = 9
⇒ y = 3
Hence, x = 4 and y = 3

Q13: ABCD is a rhombus with ∠ABC = 126°, find the measure of ∠ACD.  (5 Marks)
Sol: ∠ABC = ∠ADC (Opposite angles of a rhombus)
∠ADC = 126°

 (Diagonal of rhombus bisects the respective angles)

⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)
In ΔOCD,
∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)
⇒ ∠OCD + 63° + 90° = 180°
⇒ ∠OCD + 153° = 180°
⇒ ∠OCD = 180° – 153° = 27°
Hence ∠OCD or ∠ACD = 27°