Chapter 3 Understanding Quadrilaterals Ex 3.3 NCERT Solutions for Class 8 Maths

Latest Updated : October 2023

If you are in search of NCERT solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals, then you are at the right place. When it comes to Mathematics, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed NCERT solutions for Class 8 Maths of all the chapters, exercise wise with updated syllabus by NCERT. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark ncertforclass8.com to boost your preparation.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

NCERT solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

Exercise 3.3

1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.

(i) AD = ……………. (ii) ∠ DCB = …………………
(iii) OC = ………..… (iv) m ∠DAB + m ∠CDA = …………….

Solution:

(i) AD = BC [Since opposite sides of a parallelogram are equal]
(ii) ∠ DCB= ∠ DAB [Since opposite angles of a parallelogram are equal]
(iii) OC = OA [Since diagonals of a parallelogram bisect each other]
(iv) m∠DAB+ m∠CDA = 180° [Adjacent angles in a parallelogram are supplementary]

2. Consider the following parallelograms. Find the values of the unknowns x, y, z.

Solution:

(i) ∠B+∠C 180°
[Adjacent angles in a parallelogram are supplementary]
⇒100°+x=180°
⇒x=180°-100°=80°
And, z=x=80°
[Since opposite angles of a parallelogram are equal]
Also, y=100°
[Since opposite angles of a parallelogram are equal]

Solution:

(ii) x+50° 180°
[Adjacent angles in a parallelogram is supplementary]
⇒x=180°-50°=130°
⇒z=x=130°
[Corresponding angles]

Solution:

(iii) x=90° [Vertically opposite angles]
⇒y+x+30=180° [Angle sum property of a triangle]
⇒y+90° +30° =180°
⇒y+120° =180°
⇒y=180°-120° =60°
⇒z=y=60° [Alternative interior angles]

Solution:

(iv) z=80° [Corresponding angles]
⇒x+80°=180° [Adjacent angles in a ||gm is supplementary]
⇒x=180°-80° = 100°
And, y=80° [Opposite angles are equal in a ||gm]

Solution:

(v) y=112° [Opposite angles are equal in a parallelogram]
⇒ 40°+y+x=180° [Angle sum property of a triangle]
⇒40°+112°+x=180°
⇒ 152°+x=180°
⇒x=180°-152° = 28°
And, z=x=28° [Alternate angles]

3. Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?

Solution:

(i) It can be, but not necessarily always.

(ii) No, in this case only one pair of opposite sides are equal and another pair of opposite sides are unequal. So, it is not a parallelogram.

(iii) No. ∠A ≠ ∠C
Since opposite angles are equal in parallelogram and here opposite angles are not equal in quadrilateral ABCD. Therefore, it is not a parallelogram.

4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Solution:

ABCD is a quadrilateral. ∠B = ∠D, it can be a kite.

5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.

Solution:

Let two adjacent angles be 3x and 2x.
Since the adjacent angles in a parallelogram are supplementary
Therefore, 3x+2x=180°
⇒5x=180°
⇒x= 180°/5
⇒x= 36°

One angle ⇒ 3x=3×36°=108°
And, another angle ⇒ 2x=2×36°=72°

6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Solution:

Let each adjacent angle be x.
Since the adjacent angles in a parallelogram are supplementary.
Therefore, x+x=180°
⇒ 2x=180°
⇒ x=180°/2
⇒ x=90°

Hence, each adjacent angle is 90°
Therefore,
⇒x+x+x=180°
⇒3x=180°
⇒ x = 180°/3
⇒x=60°

7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:

⇒∠HOP +70°=180° [Angles of linear pair]
⇒∠HOP= 180°-70°=110°
And, ∠E =HOP [Opposite angles of a ||gm are equal]
⇒x=110°
⇒∠PHE = ∠HPO [Alternate angles]
Therefore, y=40°
Now ∠EHO = ∠O = 70° [Corresponding angles]
⇒40°+z=70°
⇒z=70°-40°=30°
Hence, x=110°, y=40° and z=30°

8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:

(i) In parallelogram GUNS,
GS = UN [Opposite sides of parallelogram are equal]
⇒3x=18
⇒x=18/3 = 6 cm
Also, GU = SN [Opposite sides of parallelogram are equal]
⇒3y-1=26
⇒3y=26+1
⇒3y=27
⇒y=27/2
⇒y= 9 cm

Hence, x = 6 cm and y = 9 cm.

(ii) In parallelogram RUNS,
⇒y+7=20 [Diagonals of ||gm bisects each other]
⇒y=20-7=13 cm
And, x+y=16
⇒x+13=16
⇒x=16-13 = 3cm

Hence, x=3 cm and y=13 cm.

9. In the figure both RISK and CLUE are parallelograms. Find the value of x.

Solution:

In parallelogram RISK,
∠RIS = ∠K = 120° [Opposite angles of a ||gm is equal]
∠m+120°=180° [Linear pair]
⇒∠m=180°-120°=60°

And, ∠ECI =∠L = 70° [Corresponding angles]
⇒∠m+∠n+∠ECI = 180° [Angle sum property of a triangle]
⇒60°+n+70° = 180°
⇒130°+n=180°
⇒n=180°-130° = 50°
Also, x=n=50° [Vertically opposite angles]

10. Explain how this figure is a trapezium. Which of its two sides are parallel?

Solution:

Here, ∠M + ∠L = 100° + 80° = 180° [Sum of interior opposite angles is 180°]
Therefore, NM and KL are parallel.
Hence, KLMN is a trapezium.

11. Find m∠C in Fig if AB II DC .

Solution:

Here, ∠B + ∠C = 180°
⇒Therefore, 120° + m∠C = 180° [ Since AB II DC]
⇒m∠C = 180° – 120° = 60°
⇒m∠C =60°

12. Find the measure of ∠P and ∠S if SP II RQ in Figure below
(If you find m∠R, is there more than one method to find m∠P?)

Solution:

Here, ∠P + ∠Q = 180° [Sum of co-interior angles is 180°]
⇒ ∠P+130°=180°
⇒ ∠P= 180°-130°
⇒∠P= 50°

⇒∠R = 90° [Given]
⇒∠S+90°= 180°
⇒∠S = 180°-90°
⇒ ∠S = 90°
Yes, one more method is there to find ∠P
⇒S+∠R+∠Q+ ∠P = 360° [Angle sum property of quadrilateral]
⇒ 90°+90°+130°+∠P=360°
⇒310°+∠P=360°
⇒ ∠P= 360°-310°
⇒ ∠P= 50°

Using Class 8 NCERT solutions offers several key advantages for students:

  1. Accurate Answers: NCERT solutions for Class 8 Maths are prepared by expert educators and thoroughly reviewed to ensure quality and accurate answers.
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