The NCERT solutions for class 8 maths chapter 3 are provided here to help students clear their doubts easily, understand the chapter and the topics clearly and prepare well for exams. The class 8 maths chapter 3 deals with the topic “understanding quadrilaterals”. Students also get introduced to important topics like polygons, trapezium, kite, parallelogram, rhombus, rectangle, square etc. While students have to cover a lot of topics, the NCERT solutions given here covers all the questions and concepts related to this chapter and helps students understand the topics in an easy manner. Besides, these solutions consist of important solved exercises which students can refer and prepare well for exams or they can use it for doing their assignments.

The understanding quadrilaterals solutions provided here are one of the perfect study materials as it provides a detailed solution for all the questions provided in this chapter. These solutions have been solved and reviewed by a specialized team of our subject experts. These solutions help students to learn all the formulas used for finding the measurement of the particularly given angles. By making use of these solutions, they will give students an idea about the exam pattern and marking scheme of the exam paper. The solutions are available in a PDF format and students can either study online or download it from our website for free. NCERT class 8 maths solutions for Understanding Quadrilaterals chapter 3 PDF is also available here which will help students understand the concept clearly, study effectively and even help them develop better math problem-solving skills.

### NCERT Maths Solutions for Class 8 Chapter 3 Exercises

- NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1
- NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2
- NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3
- NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4

**Exercise 3.1**

**Q1) Given below are some shapes.**

**Identify the below diagrams based on the following category**

**(a) Concave polygon (b) Simple curve (c)Convex polygon**

**(d)Simple closed curve (e) Polygon**

**Ans.)**

(a) Concave polygon: 1

(b) Simple curve: 1, 2, 5, 6 and 7

(c) Convex polygon: 2

(d) Simple closed curve: 1, 2, 5, 6 and 7

(e) Polygon: 1 and 2

**Q2) Identify the number of diagonals does each figure have?**

**(a)Regular hexagon (b)Triangle (c)Convex quadrilateral**

**Ans.)**

(a)Regular hexagon

It has 9 diagonals

(b) Triangle

It has no diagonals.

(c) Convex quadrilateral

It has 2 diagonals.

**Q3) In a convex quadrilateral, determine the sum of measures of angles? If the quadrilateral is not convex, will the property be the same?**

**Ans.)**

Let the sides of the quadrilateral be ABCD. We can see that the quadrilateral is formed by the combination of 2 triangles that is \(\triangle ADC\) and \(\triangle ABC\)** .**

Since ,we are aware that the total sum of the interior angles of triangle will be \(180^{\circ}\)

Thus, the sum of the interior angles of both the triangles are ** **\(180^{\circ}\)

+\(180^{\circ}\) = \(360^{\circ}\)

Let us consider another quadrilateral ABCD which is not convex and join BC which divides it into two triangles \(\triangle BCD\) and \(\triangle ABC\)** .**

In ΔABC,

∠1 + ∠2 + ∠3 = 180° (angle sum property of triangle)

In ΔBCD,

∠4 + ∠5 + ∠6 = 180° (angle sum property of triangle)

Therefore, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180°

⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Thus, this property is valid only if the quadrilateral is not convex.

**Q4) Check the table (Each diagram is separated into triangles and the sum of the angles is taken out from that)**

**Find out the angle sum of convex polygon with following number of sides?**

**(a)n (b)11 (c)9 (d)7**

**Ans. **

(a)n

Given n = n

Therefore, angle sum = \((n-2)\times 180^{\circ}\)

(b)11

Given, n=11

Therefore, angle sum = \((11-2)\times 180^{\circ}=1620^{\circ}\)

(c)9

Given, n=9

Therefore, angle sum = \((9-2)\times 180^{\circ}=1260^{\circ}\)

(d)7

Given, n=7

Therefore, angle sum = \((7-2)\times 180^{\circ}=900^{\circ}\)

**Q5) What do mean by a regular polygon?**

**Tell the name of the regular polygon which has**

**(i) 6 sides (ii) 3 sides (iii) 4 sides**

**Ans.) **A polygon which has sides of equal length and the angles whose measures are equal is called a regular polygon.

(i) Regular polygon which has 6 sides is called regular hexagon.

(ii) Regular polygon which has 3 sides is called equilateral triangle.

(iii) Regular polygon which has 4 sides is square.

**Q6) Calculate the measure of the angle x in the figures given below:**

**Ans.)**

(a) The diagram has 4 sides. Hence, Quadrilateral.

Total sum of the angles of quadrilateral = \(360^{\circ}\) \(50^{\circ}+130^{\circ}+120^{\circ}+x=360^{\circ}\) \(300^{\circ}+x=360^{\circ}\) \(x=360^{\circ}-300^{\circ}=60^{\circ}\)

(b) The diagram is having four sides. It’s a quadrilateral.

And , one side is \(90^{\circ}\)

Sum of the interior angles of the quadrilateral = \(360^{\circ}\) \(90^{\circ}+70^{\circ}+60^{\circ}+x=360^{\circ}\) \(220^{\circ}+x=360^{\circ}\) \(x=360^{\circ}-220^{\circ}=140^{\circ}\)

(c) There are 5 sides in the figure. It’s a pentagon.

Sum of the interior angles of pentagon = \(540^{\circ}\)

Angles which are at the bottom are a linear pair

Hence, \(180^{\circ}-70^{\circ}=110^{\circ}\) \(180^{\circ}-60^{\circ}=120^{\circ}\) \(30^{\circ}+110^{\circ}+120^{\circ}+x+x=540^{\circ}\) \(260^{\circ}+2x=540^{\circ}\) \(2x=540^{\circ}-260^{\circ}=280^{\circ}\) \(x=280^{\circ}/2=140^{\circ}\)

(d) The diagram given has five equal sides. Therefore a regular pentagon .Thus, all interior angles are equal.

\(5x=540^{\circ}\) \(x=540^{\circ}/5\) \(x=108^{\circ}\)** **

**Q7) ****(a) Find x + y + z + w**

**(b)Find x + y + z**

**Ans.**

**(a)** Sum of all interior angles of quadrilateral = \(360^{\circ}\)

Single side of quadrilateral = \(360^{\circ}-(60^{\circ}+80^{\circ}+120^{\circ})=360^{\circ}-260^{\circ}\) \(x+120^{\circ}=180^{\circ}\Rightarrow 180^{\circ}-120^{\circ}=60^{\circ}\) \(y+80^{\circ}=180^{\circ}\Rightarrow y=180^{\circ}-80^{\circ}=100^{\circ}\) \(z+60^{\circ}=180^{\circ}\Rightarrow z=180^{\circ}-60^{\circ}=120^{\circ}\) \(w+100^{\circ}=180^{\circ}\Rightarrow w=180^{\circ}-100^{\circ}=80^{\circ}\) \(x+y+z+w=60^{\circ}+100^{\circ}+120^{\circ}+80^{\circ}=360^{\circ}\)

** **

**(b)** Sum of interior angles of triangle = \(180^{\circ}\)

Single side of triangle = \(180^{\circ}-(90^{\circ}+30^{\circ})=60^{\circ}\) \(x+90^{\circ}=180^{\circ}\Rightarrow x=180^{\circ}-90^{\circ}=90^{\circ}\) \(y+60^{\circ}=180^{\circ}\Rightarrow y=180^{\circ}-60^{\circ}=120^{\circ}\) \(z+30^{\circ}=180^{\circ}\Rightarrow z=180^{\circ}-30^{\circ}=150^{\circ}\) \(x+y+z=90^{\circ}+120^{\circ}+150^{\circ}=360^{\circ}\)

**EXERCISE – 3.2**

** ****QUESTIONS-:**

*1Find the value of ‘p’ from the following figures*

*Ans.-(a)*

Here, 125° + a = 180° => 180° – 125° = 55° (Linear pair)

125° + b = 180° =>180° – 125° = 55° (Linear pair)

p = a + b (exterior angle of a triangle is equal to the sum of two opposite interior two angles)

=> p = 55° + 55° = 110°

(b)

Two interior angles are right angles = 90°

70° + a = 180° => a = 180° – 70° = 110° (Linear pair)

60° + b = 180° =>b = 180° – 60° = 120° (Linear pair)

The given figure consisting of five sides and it is a pentagon.

Hence, the sum of the angles of a pentagon = 540°

90° + 90° + 110° + 120° + q = 540°

=>410° + y = 540° => q = 540° – 410° = 130°

p + q = 180° (Linear pair)

=> p + 130° = 180°

=> p = 180° – 130° = 50°

*2. Obtain the value for each of the exterior angle of a regular polygon with*

* (i) 10 sides (ii) 25 sides*

* *

** Ans.- **Sum of angles a regular polygon having side a = (a-2)×180°

(i) Sum of angles a regular polygon having side 10 = (10-2)×180°

= 8×180° = 1440°

Each interior angle = 1440°/10 = 144°

Each exterior angle = 180° – 144° = 36°

Or,

Each exterior angle = Sum of exterior angles/Number of sides = 360°/10 = 36°

(i) Sum of angles a regular polygon having side 25 = (25-2)×180°

= 23×180° = 4140°

Each interior angle = 4140°/25 = 165.6°

Each exterior angle = 180° – 165.6° = 14.4°

Or,

Each exterior angle = Sum of exterior angles/Number of sides = 360°/25 = 14.4°

*3. Calculate the number of sides a regular polygon will have if the value of an exterior angle is 36°?*

** Ans.-**As we know that, Each exterior angle = Sum of exterior angles/Number of sides

So, 36° = 360°/Number of sides

=>Number of sides = 360°/36° = 10

Hence therefore, the regular polygon have 15 sides.

*4. A regular polygon having each of its interior angles 144°. Calculate the number of sides it will have.*

** Ans.-**Given, Interior angle = 144°

Exterior angle = 180° – 144° = 36°

Number of sides = Sum of exterior angles/exterior angle

=>Number of sides = 360°/36° = 10

Thus, the regular polygon have 10 sides.

* *

*5. (a) Can a regular polygon have each exterior angle with a measure of 37°?*

*(b) Can it be an interior angle of a regular polygon? Why?*

* *

** Ans.-**(a) Exterior angle = 22°

Number of sides = Sum of exterior angles/exterior angle

=>Number of sides = 360°/37° = 9.72

No, we cannot have a regular polygon with each exterior angle as 37° as it is not divisor of 360.

(b) Interior angle = 37°

Exterior angle = 180° – 37°= 143°

No, we can’t have a regular polygon with each exterior angle as 143° as it is not divisor of 360.

* *

*6. (a) Find the minimum possible interior angle for a regular polygon? Why?
(b) Find the maximum possible exterior angle for a regular polygon?*

* *

** Ans.-**(a)An equilateral triangle is a regular polygon having 3 sides with the least possible minimum interior angle as the regular polygon with minimum sides can be constructed with 3 sides at least..

Since, sum of the interior angles of a triangle = 180°

Each interior angle = 180°/3 = 60°

(b) Equilateral triangle is regular polygon with 3 sides has the maximum exterior angle because the regular polygon consisting least number of sides have the maximum exterior angle possible.

Maximum exterior possible = 180 – 60° = 120°

**Exercise 3.3**

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

(i) AD = …… (ii) \(\angle DCB \) = ……

(iii) OC = …… (iv) \(\angle DAB \) + \(\angle CDA \) = ……

**Answer**

(i) AD = BC (Opposite sides of a parallelogram are equal)

(ii) \(\angle DCB \) = \(\angle DAB \) (Opposite angles of a parallelogram are equal)

(iii) OC = OA (Diagonals of a parallelogram are equal)

(iv) m \(\angle DAB \) + m \(\angle CDA \) = \(180^{\circ}\)

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

**Answer**

(a)

y = \(100^{\circ}\) (opposite angles of a parallelogram)

x + \(100^{\circ}\)= \(180^{\circ}\) (Adjacent angles of a parallelogram)

⇒ x = \(180^{\circ}\)-\(100^{\circ}\)= \(80^{\circ}\)

x = z = \(80^{\circ}\) (opposite angles of a parallelogram)

Thus, x = \(80^{\circ}\), y = \(100^{\circ}\)and y = \(100^{\circ}\)

(b)

\(50^{\circ}\)+ x = \(180^{\circ}\)⇒ x = \(180^{\circ}\) – \(50^{\circ}\) = \(130^{\circ}\) (Adjacent angles of a parallelogram)x = y = \(130^{\circ}\) (opposite angles of a parallelogram)

x = z = \(130^{\circ}\) (corresponding angle)

(c)

x = \(90^{\circ}\) (vertical opposite angles)

x + y + \(30^{\circ}\) = \(180^{\circ}\) (angle sum property of a triangle)

⇒\(90^{\circ}\) + y + \(30^{\circ}\) = \(180^{\circ}\) ⇒ y = \(180^{\circ}\) – \(120^{\circ}\) = \(60^{\circ}\) also, y = z = \(60^{\circ}\) (alternate angles)

(d)

z = \(80^{\circ}\) (corresponding angle)

z = y = \(80^{\circ}\) (alternate angles)

x + y = \(180^{\circ}\) (adjacent angles)

⇒ x + \(80^{\circ}\) = \(180^{\circ}\) ⇒ x \(180^{\circ}\) – \(80^{\circ}\) = \(100^{\circ}\)

(e)

y = \(112^{\circ}\) [Opposite angles are equal in a parallelogram

\(40^{\circ}+y+x=180^{\circ}\) \(40^{\circ}+112^{\circ}+x=180^{\circ}\) \(152^{\circ}+x=180^{\circ}\) \(x=180^{\circ}-152^{\circ}=28^{\circ}\)and z = x = \(28^{\circ}\) [Alternate angles]

**3. Can a quadrilateral ABCD be a parallelogram if**

**(i) \(\angle D\) + \(\angle B \) = \(180^{\circ}\)? (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) \(\angle A \) = \(70^{\circ}\)and \(\angle C \) = \(65^{\circ}\)?**

** ****Answer**

(i)Yes, a quadrilateral ABCD be a parallelogram if \(\angle D\) + \(\angle B \) = \(180^{\circ}\)but it should also fulfilled some conditions which are:

- The sum of the adjacent angles should be \(180^{\circ}\)
- Opposite angles must be equal.

(ii) No, opposite sides should be of same length. Here, AD ≠ BC

(iii) No, opposite angles should be of same measures. \(\angle A \) ≠ \(\angle C \)

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

**Answer**

ABCD is a figure of quadrilateral that is not a parallelogram but has exactly two opposite angles that is \(\angle B \) = \(\angle D \) of equal measure. It is not a parallelogram because \(\angle A \) ≠ \(\angle C \)

**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.**

** ****Answer**

Let the measures of two adjacent angles \(\angle A \) and \(\angle B \) be 3x and 2x respectively in parallelogram ABCD.

\(\angle A \) + \(\angle B\) = \(180^{\circ}\)⇒ 3x + 2x = \(180^{\circ}\)

⇒ 5x = \(180^{\circ}\)

⇒ x = \(36^{\circ}\)

We know that opposite sides of a parallelogram are equal.

\(\angle A \) = \(\angle C \) = 3x = 3 × \(36^{\circ}\) = \(108^{\circ}\) \(\angle B \) = \(\angle D \) = 2x = 2 × \(36^{\circ}\) = \(72^{\circ}\)

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

** ****Answer**

Let ABCD be a parallelogram.

Sum of adjacent angles of a parallelogram = \(180^{\circ}\) \(\angle A \) + \(\angle B \) = \(180^{\circ}\)

⇒ 2\(\angle A \) = \(180^{\circ}\)

⇒\(\angle A \) = \(90^{\circ}\)

also, \(90^{\circ}\)+ \(\angle B \) = \(180^{\circ}\)

⇒\(\angle B \) = \(180^{\circ}\)- \(90^{\circ}\)= \(90^{\circ}\) \(\angle A \) = \(\angle C \) = \(90^{\circ}\) \(\angle B\) = \(\angle D \) = \(90^{\circ}\)

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

**Answer**

y = \(40^{\circ}\) (alternate interior angle)

\(\angle P \) = \(70^{\circ}\) (alternate interior angle) \(\angle P \) = \(\angle H \) = \(70^{\circ}\) (opposite angles of a parallelogram)z = \(\angle H \) – \(40^{\circ}\) = \(70^{\circ}\) – \(40^{\circ}\) = \(30^{\circ}\) \(\angle H \) + x = \(180^{\circ}\)

⇒\(70^{\circ}\) + x = \(180^{\circ}\)

⇒ x = \(180^{\circ}\)- \(70^{\circ}\) = \(110^{\circ}\)

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

** **

**Answer**

(i) SG = NU and SN = GU (opposite sides of a parallelogram are equal)

3x = 18

⇒ x = 18/3 = 6

3y – 1 = 26 and,

⇒ 3y = 26 + 1

⇒ y = 27/3 = 9

x = 6 and y = 9

(ii) 20 = y + 7 and 16 = x + y (diagonals of a parallelogram bisect each other)

y + 7 = 20

⇒ y = 20 – 7 = 13 and,

x + y = 16

⇒ x + 13 = 16

⇒ x = 16 – 13 = 3

x = 3 and y = 13

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

**Answer**

⇒ \(120^{\circ}\) + \(\angle R \) = \(180^{\circ}\)

⇒\(\angle R \) = \(180^{\circ}\)- \(120^{\circ}\) = 60°

also, \(\angle R \) = \(\angle SIL \) (corresponding angles)

⇒\(\angle SIL \) = \(60^{\circ}\)

also, \(\angle ECR \) = \(\angle L \) = \(70^{\circ}\)(corresponding angles)

x + \(60^{\circ}\)+ \(70^{\circ}\)= \(180^{\circ}\) (angle sum of a triangle)

⇒ x + \(130^{\circ}\) = \(180^{\circ}\)

⇒ x = \(180^{\circ}\)- \(130^{\circ}\) = \(50^{\circ}\)

**10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32)**

** **

**Answer**

When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is \(180^{\circ}\)

then the lines are parallel to each other.

Here, \(\angle M \) + \(\angle L \) =\(100^{\circ}\)

+ \( 80^{\circ}\)

= \(180^{\circ}\)

Thus, MN || LK

As the quadrilateral KLMN has one pair of parallel line therefore it is a trapezium.

MN and LK are parallel lines.

**11. Find m\(\angle C \) in Fig 3.33 if AB || DC ?**

** **

**Answer**

m\(\angle C \) + m\(\angle B \) = \(180^{\circ}\)

(angles on the same side of transversal)

⇒m\(\angle C \) + \(60^{\circ}\)

= \(180^{\circ}\)

⇒m\(\angle C \) \(180^{\circ}\)

– \(60^{\circ}\)

= \(120^{\circ}\)

**12. Find the measure of \(\angle p \) and\(\angle S \) if SP || RQ ? in Fig 3.34. (If you find m\(\angle R \) , is there more than one method to find m\(\angle p \) ?)**

** **

**Answer**

\(\angle p \) +\(\angle Q \) = \(180^{\circ}\)

(angles on the same side of transversal)

⇒\(\angle p \) +\(130^{\circ}\)= \(180^{\circ}\)

⇒\(\angle p \) =\(180^{\circ}\) \(-130^{\circ}\)= 50°

also, \(\angle R \) + \(\angle S \) = \(180^{\circ}\)

(angles on the same side of transversal)

⇒\(90^{\circ}\)+ \(\angle S \) = \(180^{\circ}\)

⇒\(\angle S \) =\(180^{\circ}\)

– \(90^{\circ}\)= \(90^{\circ}\)

Thus, \(\angle p \) =\(50^{\circ}\)and \(\angle S \) = \(90^{\circ}\)

Yes, there are more than one method to find m\(\angle p \) .

PQRS is a quadrilateral. Sum of measures of all angles is \(360^{\circ}\)

Since, we know the measurement of \(\angle Q \) ,\(\angle R \) and \(\angle S \) .

\(\angle Q \) = \(130^{\circ}\), \(\angle R \) = \(90^{\circ}\)and \(\angle S \) = \(90^{\circ}\) \(\angle p \) + \(130^{\circ}\)+ \(90^{\circ}\)+ \(90^{\circ}\)= \(360^{\circ}\)⇒\(\angle p \) +\(310^{\circ}\)= \(360^{\circ}\)

⇒\(\angle p \) =\(360^{\circ}\) \(-310^{\circ}\) = \(50^{\circ}\)

**Exercise 3.4**

**1. State whether the following questions given below are True or False.**

(a) All rectangles are squares.

(b) All rhombuses are parallelograms.

(c) All squares are rhombuses and also rectangles.

(d) All squares are not parallelograms.

(e) All kites are rhombuses.

(f) All rhombuses are kites.

(g) All parallelograms are trapeziums.

(h) All squares are trapeziums.

**Answer**

(a) False. All square are rectangles but all rectangles are not square.

(b) True

(c) True

(d) False. All squares are parallelograms as opposite sides are parallel and opposite angles are equal.

(e) False. A kite doesn’t have all sides of same length.

(f) True

(g) True

(h) True

**2. Identify all the quadrilaterals that have.**

(a) four sides of equal length

(b) four right angles

**Answer**

(a) Rhombus and square have four sides of equal length.

(b) Square and rectangle have four right angles.

**3. Explain how a square is.**

(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle

**Answer**

(i) Square is a quadrilateral because it has four sides.

(ii) Square is a parallelogram because its opposite sides are parallel and opposite angles are equal.

(iii) Square is a rhombus because all four sides are of equal length and diagonals bisect at right angles.

(iv) Square is a rectangle because its interior angle is 90°

**4. Name the quadrilaterals whose diagonals are equal to each other.
**(i) bisect each other

(ii) are perpendicular bisectors of each other

(iii) are equal

**Answer**

(i) Parallelogram, Rhombus, Square and Rectangle

(ii) Rhombus and Square

(iii) Rectangle and Square

**5. Explain why a rectangle is a convex quadrilateral.**

**Answer**

Rectangle is a convex quadrilateral because its both diagonals lie inside the rectangle.

**6. PQR is a right-angled triangle and W is the midpoint of the side opposite to the right angle. Explain why W is equidistant from P, Q and R. (The dotted lines are drawn additionally to help you).**

** **

**Answer**

PS and SR are drawn so that PS || QR and PQ || SR

PS = QR and PQ = SR

ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90°.

In a rectangle, diagonals are of equal length and also bisects each other.

Hence, PW = WR = QW = WS

Thus, W is equidistant from P, Q and R.

**About BYJU’s Solutions**

BYJU’s has only one main purpose, to create the best learning content for students and help them succeed in exams be it normal school exams or board or competitive exams. The solutions that we offer have been designed carefully to boosts student’s learning capabilities. As such, students not only get access solved exercises but they can access educational videos, interactive games and more which makes learning more fun. However, the NCERT solutions which is a part of the many solutions that we offer has been created thoughtfully to help students make the most out of it.

BYJU’s host some of the top and experienced subject experts that can help students in their learning journey. The teacher can guide the students to learn the subject and the different chapters as well as topics in a more engaging, simple and clear manner. Students who sign up with BYJU’s have more benefits as they are given regular feedback regarding their performance and their progress. Apart from these we also have a responsive support team that is always ready and prompt to clear all the doubts of the students. Meanwhile, students can download and try BYJU’s – The Learning app for a more interactive and seamless learning experience.