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

Last updated date: 25th May 2024
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## NCERT Solutions for Class 8 Maths Chapter 3 (EX 3.3)

NCERT Solutions for Class 8 Maths Chapter 3 by Vedantu provides solutions for exercises on quadrilaterals. These solutions are written by expert teachers as per the latest CBSE guidelines. If you follow Vedantu’s NCERT Solutions for Class 8 Maths Chapter 3 PDF thoroughly, you can easily understand all the fundamental concepts related to Quadrilaterals. These solutions provide you with a detailed step by step answers for each problem. Download today NCERT Solutions for Class 8 Maths Chapter 3 to ease your exam preparation.Vedantu is a platform that provides free NCERT Solution and other study materials for students. Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

 Class: NCERT Solutions for Class 8 Subject: Class 8 Maths Chapter Name: Chapter 3 - Understanding Quadrilaterals Exercise: Exercise - 3.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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## Access NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

### Exercise 3.3

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

$\left( \text{i} \right)\text{AD=}$

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

$\left( \text{i} \right)$

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

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

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

$\left( \text{ii} \right)$

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

$\left( \text{iii} \right)$

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

$\left( \text{iv} \right)$

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

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

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

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

$\left( \text{v} \right)$

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:

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.

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).

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).

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.

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?

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 $\overline{\text{AB}}\parallel \overline{\text{DC}}$

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 $\text{m}\angle \text{P}$.)

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 $\text{m}\angle \text{P}$.

## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Our NCERT Solutions for Class 8 exercise 3.3 PDF talks about the basic theories on quadrilaterals. The NCERT solution for Class 8 Maths Chapter 3 Exercise 3.3 PDF makes it easy for the students to solidify their basic knowledge in quadrilaterals. Vedantu’s NCERT Solutions for Class 8 Maths Chapter 3 also addresses the aspects of the chapter in a step by step basis. Our NCERT solution of Maths Class 8 chapter 3 discusses Quadrilaterals in details.

### NCERT Class 8 Maths Chapter 3 Simplified - with Vedantu

In Class 8 Maths Chapter 3 Exercise 3.3 Solution discusses Quadrilaterals, explanations regarding curves, polygons, diagonals, regular and irregular polygons, convex and concave polygons etc. This introductory part of the NCERT Class 8 Maths Chapter 3 helps to form the fundamental knowledge of Quadrilaterals in a student.

Vedantu’s NCERT Maths Class 8 chapter 3 exercise 3.3 solutions PDF pertaining to Understanding Quadrilaterals is also made in such a way that students do not miss any basic knowledge about Quadrilaterals. This is the reason we explain our answers in Ex 3.3 Class 8 NCERT Solutions. You will see in our Class 8 Maths Chapter 3 exercise 3.3 solution that while answering the questions, we do not just provide the answers. Instead, we give reasons like - why the line segments AD and BC are equal. Similarly in question 7 of our NCERT Solutions for Class 8 Maths Chapter 3 exercise 3.3, we did not jump a single step to determine the values of x, y and z.

### When You Have A Thirst for A Little More - Vedantu is Your Choice

We know that this discussion about quadrilaterals is not exhaustive. After reading our Class 8 Maths Chapter 3 solutions, you may have further questions about some aspects of Quadrilaterals. Our NCERT solution Class 8 Maths Chapter 3 PDF covers all the topics. But if you want more clarity beyond the answers in Vedantu’s Maths NCERT Solutions Class 8 Chapter 3, you can do so by joining our online classes. Here, you can ask any questions regarding the chapter and get it cleared on the spot. Our NCERT Maths Solution Class 8 chapter 3 is complemented by these online classes where teachers teach with the aid of images, slides and videos. Vedantu is there for you with its online classes to answer all your queries and enhance your exam preparation.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 3: Understanding Quadrilaterals - Exercise 3.3

1. What is the meaning of Quadrilaterals according to NCERT Solutions for Class 8 Maths Chapter 3?

A quadrilateral is simply a two-dimensional figure that has four lines on all four sides, creating four vertices A, B, C, and D. The line segments, thus, are AB, BC, CD, AD. It is very important to note that the total sum of all the angles in a quadrilateral equals 360°. Some of the common quadrilaterals are rectangles, trapezoids, and squares. For more solutions, download the PDF from the link NCERT Solutions Class 8 Maths free of cost.

2. What are the main topics covered in NCERT Solutions for Class 8 Maths Chapter 3?

The main topics covered in NCERT Solutions for Class 8 Maths Chapter 3 are different types of Quadrilaterals, Rectangle & its properties, Square & its properties, Parallelogram & its properties, Rhombus & its properties, Trapezium & its properties, and all exercise questions with solutions. With NCERT Solutions for Class 8 Maths Chapter 3, you get an easy explanation of all the topics and sub-topics covered in the NCERT textbook, along with a full solution to all the exercises.

3. What are a regular polygon and irregular polygon in Class 8 Maths Chapter 3?

Polygon is a simple closed figure that has only one line segment. When polygons are equilateral (having equal sides) and equiangular (having equal angles), then the polygons can be called regular polygons. The best example of a regular polygon is a square. Irregular polygons are those polygons that are not equilateral and not equiangular, which means that they do not have equal sides and equal angles.