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

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

Chapter 3 of Class 8 Maths NCERT, "Understanding Quadrilaterals," delves into the properties and types of four-sided figures, focusing on the fundamentals of quadrilaterals. Exercise 3.3 is particularly important as it helps students to apply these concepts in solving various problems. This chapter covers different kinds of quadrilaterals like parallelograms, trapeziums, and rhombuses, emphasizing their unique properties and theorems related to their angles and sides.

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In Exercise 3.3, students should focus on understanding the properties and characteristics of specific quadrilaterals. It is crucial to grasp the relationships between angles and sides to solve problems accurately. This exercise enhances analytical skills and provides a solid foundation for advanced geometric concepts. By mastering this chapter, students will be well-prepared for higher-level geometry studies.

Access NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.3 Solutions

Exercise 3.3

1. Give a parallelogram \[\text{ABCD}\]. Complete each statement along with the definition of property used

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\[\left( \text{i} \right)\text{AD=}\]

Ans: As the given figure is a parallelogram and we know that that in parallelogram, the opposite sides are equal to each other, so \[\text{AD=BC}\]

\[\left( \text{ii} \right)\angle \text{DAB}\]

Ans: As the given figure is a parallelogram and we know that that in parallelogram, the opposite angles are equal to each other, so \[\angle \text{DCB=}\angle \text{DAB}\]

\[\left( \text{iii} \right)\text{OC=}\]

Ans: As the given figure is a parallelogram and we know that that in parallelogram, diagonals bisect each other, so \[\text{OC=OA}\]

\[\left( \text{iv} \right)\text{m}\angle \text{DAB+m}\angle \text{CDA=}\]

Ans: As the given figure is a parallelogram and we know that that in parallelogram, the adjacent angles are supplementary to each other, so

\[\text{m}\angle \text{DAB+m}\angle \text{CDA=18}{{\text{0}}^{\text{o}}}\]


2. Complete the following parallelograms

\[\left( \text{i} \right)\]

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Ans: We know that in parallelograms adjacent angles are supplementary, we got $\text{x+10}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x=18}{{\text{0}}^{\text{o}}}\text{-10}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x=8}{{\text{0}}^{\text{o}}}  $


\[\left( \text{ii} \right)\]


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Ans: We know that in parallelograms adjacent angles are supplementary, we got

 $\text{y+5}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{y=18}{{\text{0}}^{\text{o}}}\text{-5}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x=13}{{\text{0}}^{\text{o}}}  $


Now, we got

\[\text{y=x=13}{{\text{0}}^{\text{o}}}\], as the opposite angles are equal

And 

\[\text{z=x=13}{{\text{0}}^{\text{o}}}\], as they are corresponding angles

\[\left( \text{iii} \right)\]

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Ans: From observing the figure, we got

\[\text{x=9}{{\text{0}}^{\text{o}}}\], as vertically opposite angle

Now, applying angle sum property of triangle, we got


$\text{x+y+3}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{9}{{\text{0}}^{\text{o}}}\text{+y+3}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

$\Rightarrow \text{y+12}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

$\Rightarrow \text{y=6}{{\text{0}}^{\text{o}}}  $

\[\text{z=y=6}{{\text{0}}^{\text{o}}}\], as they are alternate interior angles

\[\left( \text{iv} \right)\]


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Ans: From observing the figure, we get

\[\text{z=8}{{\text{0}}^{\text{o}}}\], as they are corresponding angles

\[\text{y=8}{{\text{0}}^{\text{o}}}\], as the opposites angles are equal

As we know adjacent angles are supplementary, we get

 $\text{x+y=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x=18}{{\text{0}}^{\text{o}}}\text{-8}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x=10}{{\text{0}}^{\text{o}}}  $


\[\left( \text{v} \right)\]


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From the figure, we got that

\[\text{y=11}{{\text{2}}^{\text{o}}}\], as opposite angles are equal

Now applying angle sum property of triangle, we got,


$\text{x+y+4}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x+11}{{\text{2}}^{\text{o}}}\text{+4}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

$\Rightarrow \text{x=18}{{\text{0}}^{\text{o}}}\text{-15}{{\text{2}}^{\text{o}}} $ 

$\Rightarrow \text{x=2}{{\text{8}}^{\text{o}}}  $

And

\[\text{z=x=2}{{\text{8}}^{\text{o}}}\], as they are alternate interior angles.

3. Can a quadrilateral \[\text{ABCD}\] be a parallelogram if,

\[\left( \text{i} \right)\angle \text{D+}\angle \text{B=18}{{\text{0}}^{\text{o}}}\]

Ans: According to the condition given, the quadrilateral may or may not be a parallelogram.

For a quadrilateral to be a parallelogram, there are some more conditions are present that is,

The sum of the adjacent angles should be \[\text{18}{{\text{0}}^{\text{o}}}\] and opposite angles should be of same length.

If these conditions are also fulfilled, then we can say that quadrilateral is also a parallelogram.

\[\left( \text{ii} \right)\text{AB=DC=8}\] cm, \[\text{AD=4}\] cm and \[\text{BC=4}\text{.4}\] cm

Ans: As we know that in parallelogram, the opposite sides are equal

So, according to the given condition, the opposite sides aren’t equal

Therefore, we can’t say that given quadrilateral is a parallelogram.

\[\left( \text{iii} \right)\angle \text{A=7}{{\text{0}}^{\text{o}}}\] and \[\angle \text{C=6}{{\text{5}}^{\text{o}}}\]

Ans: As we know that in parallelogram, the opposite angles are equal

So, according to the given condition, the opposite angles aren’t equal

Therefore, we can’t say that given quadrilateral is a parallelogram.


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

Ans: 


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As we can see, we have drawn a quadrilateral \[\text{ABCD}\] with \[\angle \text{B}\] and \[\angle \text{D}\] as opposite angles which are also equal to each other.

Now, we can also see that the quadrilateral drawn isn’t a parallelogram as in parallelogram, opposite sides are also equal to each other and we can also see that the \[\angle \text{A}\] and \[\angle \text{C}\] aren’t equal to each other.


5. The measures of two adjacent angles of a parallelogram are in the ratio \[\text{3:2}\]. Find the measure of each of the angles of the parallelogram.

Ans: Let us consider the two angles be \[\angle \text{A}\] and \[\angle \text{B}\]

According to the question, angles are in a ratio \[\text{3:2}\]

So, \[\angle \text{A=3x}\] and \[\angle \text{B=2x}\]

As we know that the sum of the adjacent angles in a parallelogram is \[\text{18}{{\text{0}}^{\text{o}}}\], we got

 $\angle \text{A+}\angle \text{B=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{3x+2x=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{5x=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{x=3}{{\text{6}}^{\text{o}}} $


So, we got,

\[\angle \text{A=}\angle \text{C}\Rightarrow \text{3x=10}{{\text{8}}^{\text{o}}}\] and \[\angle \text{B=}\angle \text{D}\Rightarrow \text{2x=7}{{\text{2}}^{\text{o}}}\], as they are opposite angles.

Therefore, the measure of each angles are,

 $\angle \text{A=10}{{\text{8}}^{\text{o}}} $ 

 $\angle \text{B=7}{{\text{2}}^{\text{o}}} $ 

 $\angle \text{C=10}{{\text{8}}^{\text{o}}} $ 

 $\angle \text{D=7}{{\text{2}}^{\text{o}}}  $



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

Ans: As we know that the sum of adjacent angles of parallelogram is \[\text{18}{{\text{0}}^{\text{o}}}\], we got,

\[\angle \text{A+}\angle \text{B=18}{{\text{0}}^{\text{o}}}\] 

According to the question, adjacent angles are of equal measures, we got

 $\angle \text{A+}\angle \text{A=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{2}\angle \text{A=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \angle \text{A=9}{{\text{0}}^{\text{o}}}  $

We know that the adjacent angles are of equal measures

Therefore, we got \[\angle \text{A=}\angle \text{B=9}{{\text{0}}^{\text{o}}}\], \[\angle \text{A=}\angle \text{C=9}{{\text{0}}^{\text{o}}}\](opposite angles) and \[\angle \text{D=}\angle \text{B=9}{{\text{0}}^{\text{o}}}\](opposite angles)


7. The adjacent figure \[\text{HOPE}\] is a parallelogram. Find the angle measure \[\text{x,y}\] and \[\text{z}\]. State the properties you use to find them.


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Ans: Now, observing the figure,

From alternate interior angle, we get \[\angle \text{y=4}{{\text{0}}^{\text{o}}}\],

From corresponding angles, we get 

 $\text{7}{{\text{0}}^{\text{o}}}\text{=z+4}{{\text{0}}^{\text{o}}} $ 

 $\text{z=3}{{\text{0}}^{\text{o}}}  $

And from adjacent pair of angles we get, 

 $\text{x+}\left( \text{z+4}{{\text{0}}^{\text{o}}} \right)\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\text{x+7}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\text{x=18}{{\text{0}}^{\text{o}}}\text{-7}{{\text{0}}^{\text{o}}} $ 

 $\text{x=11}{{\text{0}}^{\text{o}}}  $



8. The following figures \[\text{GUNS}\] and \[\text{RUNS}\] are parallelograms find \[\text{x}\] and \[\text{y}\] (lengths are in cm)

(i).


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Ans: As we know, the opposite sides in a parallelogram are equal to each other, from this we got,

\[\text{GU=SN}\] and \[\text{GS=UN}\], from the figure we got the length of the sides, so we get

 $\text{3y-1=26} $ 

 $\text{3y=27} $ 

 $\text{y=9}  $

 and 

 $\text{3x=18} $ 

 $\text{x=6}  $

Therefore, \[\text{x=6}\] cm and \[\text{y=9}\] cm.

(ii).


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Ans: We know that, in parallelograms, diagonals bisect each other, so we got


 $\text{y+7=20} $ 

 $\text{y=13}$ 

 and 

$\text{x+y=16} $ 

$\text{x+13=16} $ 

$\text{x=3}  $

Therefore, \[\text{x=3}\] cm and \[\text{y=13}\] cm.


9.


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In the given figure, both \[\text{RISK}\] and \[\text{CLUE}\] are parallelograms. Find the value of \[\text{x}\].

Ans: As the given figures is a parallelogram and we know that adjacent angles in parallelograms are supplementary and also opposite angles are equal.

Now, in parallelogram \[\text{RISK}\],

 $\angle \text{RKS+}\angle \text{ISK=18}{{\text{0}}^{\text{o}}} $ 

 $\text{12}{{\text{0}}^{\text{o}}}\text{+}\angle \text{ISK=18}{{\text{0}}^{\text{o}}} $ 

 $\angle \text{ISK=6}{{\text{0}}^{\text{o}}}  $

We got \[\angle \text{ISK=6}{{\text{0}}^{\text{o}}}\]

Now in parallelogram \[\text{CLUE}\],

\[\angle \text{ULC+}\angle \text{CEU=7}{{\text{0}}^{\text{o}}}\]

Applying angle sum property, we get

$\text{x+6}{{\text{0}}^{\text{o}}}\text{+7}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\text{x+13}{{\text{0}}^{\text{o}}}\text{=18}{{\text{0}}^{\text{o}}} $ 

 $\text{x=5}{{\text{0}}^{\text{o}}}  $

Therefore, the value of \[\text{x}\] is \[\text{5}{{\text{0}}^{\text{o}}}\].


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

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Ans: We know that for lines to be parallel, the transversal line should be parallel to given lines in a way in which the sum of the angles on the same side of transversal is \[\text{18}{{\text{0}}^{\text{o}}}\]

From the given figure we can see that,

\[\angle \text{NML+}\angle \text{MLK=18}{{\text{0}}^{\text{o}}}\]

Therefore, the lines \[\text{NM}\] and \[\text{LK}\] are parallel lines,

Now, we know that for a quadrilateral to be a trapezium, it should have a pair of parallel lines,

Here we have lines \[\text{NM}\] and \[\text{LK}\] are parallel

Therefore, the quadrilateral \[\text{KLMN}\] is a trapezium.


11. Find \[\text{m}\angle \text{C}\] in the following figure, if \[\overline{\text{AB}}\parallel \overline{\text{DC}}\]

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Ans: It is given that \[\overline{\text{AB}}\parallel \overline{\text{DC}}\]

We know that the angles on the same side of transversal are supplementary, so we got

 $\angle \text{B+}\angle \text{C=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{12}{{\text{0}}^{\text{o}}}\text{+}\angle \text{C=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \angle \text{C=6}{{\text{0}}^{\text{o}}}  $

Therefore, \[\text{m}\angle \text{C=6}{{\text{0}}^{\text{o}}}\]


12. Find the measure of \[\angle \text{P}\] and \[\angle \text{S}\], if \[\overline{\text{SP}}\parallel \overline{\text{RQ}}\] in the following figure. (If you find \[\text{m}\angle \text{R}\], is there more than one method to find \[\text{m}\angle \text{P}\].)

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Ans: Now, we know that sum of angles on the same side of transversal is \[\text{18}{{\text{0}}^{\text{o}}}\],

From the figure we can say that

 $\angle \text{P+}\angle \text{Q=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{13}{{\text{0}}^{\text{o}}}\text{+}\angle \text{P=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \angle \text{P=5}{{\text{0}}^{\text{o}}}  $

 and 

 $\angle \text{R+}\angle \text{S=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \text{9}{{\text{0}}^{\text{o}}}\text{+}\angle \text{S=18}{{\text{0}}^{\text{o}}} $ 

 $\Rightarrow \angle \text{S=9}{{\text{0}}^{\text{o}}}  $

Therefore, the measure of angle \[\text{m}\angle \text{P}\] and \[m\angle \text{S}\] is \[\text{5}{{\text{0}}^{\text{o}}}\] and \[\text{9}{{\text{0}}^{\text{o}}}\] respectively.

There’s one more way to find angle \[\text{m}\angle \text{P}\], as \[\text{m}\angle \text{R}\] and \[\text{m}\angle \text{Q}\] are given,

We can first find the angle \[m\angle \text{S}\] and we also know that the sum of the interior angles is \[{{360}^{\circ }}\], we can apply angle sum property to find the angle \[\text{m}\angle \text{P}\].


Conclusion

In order to help students understand the characteristics and classifications of various quadrilaterals, Vedantu's "NCERT Solutions for class 8 maths ch 3 ex 3.3 - Understanding Quadrilaterals" offers concise explanations and solutions. It emphasises the special qualities of essential geometric shapes including trapeziums, rhombuses, squares, parallelograms, and kites. Knowing the properties of each type of quadrilateral and the angle sum property are important concepts. Concentrate on working through issues involving the identification and computation of angles and sides.


Class 8 Maths Chapter 3: Exercises Breakdown

Exercise

Number of Questions

Exercise 3.1

2 Questions & Solutions

Exercise 3.2

6 Questions & Solutions (6 Short Answers)

Exercise 3.4

6 Questions & Solutions (1 Long Answer, 5 Short Answers)


CBSE Class 8 Maths Chapter 3 Other Study Materials


Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 8 Maths

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

1. What are the main properties of a parallelogram according to NCERT Solutions for Class 8 Maths Chapter 3?

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Its main properties, as per NCERT Solutions for Class 8 Maths Chapter 3, include:

  • Opposite sides are equal in length.
  • Opposite angles are equal.
  • Diagonals bisect each other.
  • The sum of the adjacent angles is 180°.

2. How is the angle sum property of a quadrilateral used in Class 8 Maths Chapter 3 NCERT Solutions?

The angle sum property of a quadrilateral states that the sum of its four interior angles is always 360°. In NCERT Solutions for Class 8 Maths Chapter 3, students apply this property to find missing angles in various quadrilaterals by setting up equations and solving for unknowns, ensuring accurate calculation in problems.

3. Why are some quadrilaterals not parallelograms even if they have equal opposite angles? (FUQ)

According to the NCERT Solutions for Class 8 Maths Chapter 3, a quadrilateral is a parallelogram only if both pairs of opposite sides are parallel and equal. Having equal opposite angles is not sufficient unless these side conditions are also met. This concept helps prevent common errors in classification.

4. What is the difference between a regular and an irregular polygon covered in Class 8 Chapter 3?

A regular polygon has all sides and angles equal, such as a square. An irregular polygon has sides and/or angles of different measures. This distinction is stressed in NCERT Solutions to build a correct understanding of geometric shape properties in Chapter 3.

5. How do NCERT Solutions for Class 8 Maths Chapter 3 help in solving questions about trapeziums and their parallel sides?

The solutions guide students to identify that a trapezium is a quadrilateral with exactly one pair of parallel sides. Students are taught to use properties and angle relationships to distinguish trapeziums from other quadrilaterals and to pinpoint the parallel sides when analyzing figures, as outlined in Chapter 3.

6. In what situations can two adjacent angles of a parallelogram be equal, based on NCERT Solutions for Class 8 Maths?

When two adjacent angles of a parallelogram are equal, each must measure 90°. This means the shape is a rectangle (a special parallelogram). NCERT Solutions explain that this only happens when all the angles of the parallelogram are right angles, not in general parallelograms.

7. How does Exercise 3.3 of Class 8 Maths reinforce understanding of parallelogram properties? (FUQ)

Exercise 3.3 provides students with questions requiring them to apply definitions and properties of parallelograms to compute sides, angles, and diagonals. It deepens conceptual clarity by linking theory to practical problem-solving and highlights the relationships among angles and sides.

8. What steps are recommended for solving angle-based questions in quadrilaterals as per NCERT Solutions for Class 8 Maths Chapter 3? (FUQ)

Students should:

  • Identify the type of quadrilateral (parallelogram, trapezium, etc.).
  • Apply specific properties (like sum of adjacent angles or diagonals bisecting).
  • Use the angle sum property to form and solve equations.
  • Clearly write out each step as per CBSE pattern to ensure full marks.

9. Why is it important to distinguish between different types of quadrilaterals in exams? (FUQ)

Accurately identifying quadrilateral types (parallelogram, rhombus, rectangle, trapezium, kite) allows students to apply correct formulas and properties, preventing calculation errors and helping secure maximum marks in both objective and application-based questions as per the CBSE 2025-26 pattern.

10. What are common mistakes students should avoid while using NCERT Solutions for Class 8 Maths Chapter 3?

Common errors include:

  • Assuming all quadrilaterals with equal angles are parallelograms
  • Ignoring the requirement that opposite sides must also be equal and parallel
  • Misapplying properties between different types of quadrilaterals
  • Not verifying the sum of angles equals 360°
Careful reading and step-by-step verification help avoid these mistakes.

11. How do the properties of diagonals differ among quadrilaterals in Class 8 Maths Chapter 3?

In a parallelogram, diagonals bisect each other but are not generally equal. In a rectangle, diagonals are equal and bisect each other. In a rhombus, diagonals bisect at right angles. Recognizing these distinctions is crucial in solving related NCERT Solutions questions.

12. What is the process to classify a given figure as a quadrilateral or not as per NCERT Solutions for Class 8 Maths Chapter 3?

Check if the figure has four sides, four vertices, and is a closed shape. Then confirm by summing angles to see if they total 360°. Only if both conditions are met, it is a quadrilateral, according to the rules in the chapter.

13. Can a quadrilateral with only one pair of equal sides be a parallelogram according to NCERT Class 8 Chapter 3? (FUQ)

No. A parallelogram must have both pairs of opposite sides equal and parallel. If only one pair is equal, it cannot be classified as a parallelogram as per NCERT Solutions for Class 8 Maths Chapter 3 guidelines.

14. How do NCERT Solutions for Class 8 Maths Chapter 3 promote stepwise problem-solving skills?

The solutions demonstrate how to break down a problem into smaller, logical steps aligned with CBSE's marking scheme. This includes identifying the figure, applying relevant properties, forming equations, and solving for missing values, making solutions easy to follow and replicate in exams.