NCERT Archives &#8211; Mathemafia

Click here to watch proof of √3

Chapter 1 - Real Numbers

Square root of any prime number is irrational.

E.g. – $ \begin{aligned} &\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7} \end{aligned}$

$ \begin{aligned} LCM (a, b) \times HCF (a, b) = a \times b \end{aligned}$

Fundamental Theorem of Arithmetic

Every composite number can be factorised as a product of primes, and this factorisation is unique.

Theorem

If p is a prime and p divides a², then p divides a, where a is a positive integer.

Chapter 2 - Polynomials

Linear Polynomial

Degree = 1

No. of real zeros: At most 1

E.g: x

Quadratic Polynomial

Degree = 2

No. of real zeros: At most 2

E.g: x²

Cubic Polynomial

Degree = 3

No. of real zeros: At most 3

E.g: x³

Monomial

No: of terms = 1

E.g: 2x

Binomial

No: of terms = 2

E.g: 2x + 3

Trinomial

No: of terms = 3

E.g: 2x² + 3x + 2

Quadratic Polynomials

Standard form of Quadratic Polynomial

$ \begin{aligned} &ax^2+bx+c, \ \ a\ne0\\ \end{aligned}$

Zeros of Quadratic Polynomial

If α is a zero of polynomial

$ \begin{aligned} &p(x) = ax^2+bx+c, \ \ a\ne0\\ &then \ \ p(α) = 0 \end{aligned}$

Sum of Zeros

$ \begin{aligned} &α + β = -\frac{b}{a} \end{aligned}$

Product of Zeros

$ \begin{aligned} &α \times β = \frac{c}{a} \end{aligned}$

Add Graphs – Later XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Chapter 3 - Pair of Linear Equations in Two Variables

Standard form of Linear Equation in Two Variables

$ \begin{aligned} &ax+by+c =0,\\ &\text{where a and b are not both 0.} \end{aligned}$

$ \begin{aligned} &ax+by+c =0, \ \ where \ \ a^2+b^2\ne 0 \end{aligned}$

Standard form of Pair of Linear Equation in Two Variables

$ \begin{aligned} &a_{1}x+b_{1}y+c_{1} =0, \ \ where \ \ a_{1}^2+b_{1}^2\ne 0 \\&a_{2}x+b_{2}y+c_{2} =0, \ \ where \ \ a_{2}^2+b_{2}^2\ne 0 \end{aligned}$

Solution of Pair of LE in 2 Variables

Intersecting Lines
➤ Consistent Pair of LE

➺ Lines Intersect at 1 point
🎯 No. of Solutions = 1

$ \begin{aligned} &\frac{a_{1}}{a_{2}} \ne \frac{b_{1}}{b_{2}} \end{aligned}$

Coincident Lines

➤ Consistent Pair of LE
➺ Lines overlap each other

🎯 No. of Solutions = Infinite

$ \begin{aligned} &\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{aligned}$

Parallel Lines
➤ inconsistent Pair of LE

➺ No common point

🎯 No. of Solutions = 0

$ \begin{aligned} &\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \ne \frac{c_{1}}{c_{2}} \end{aligned}$

Click here to watch Introduction to pair of Linear equations

Chapter 4 - Quadratic Equations

Standard form of Quadratic Equation

$ \begin{aligned} &ax^2+bx+c=0, \ \ a\ne0\\ \end{aligned}$

Roots of Quadratic Equation

Zeros of quadratic polynomial $ \begin{aligned} &ax^2+bx+c \end{aligned}$

➤ Roots of the quadratic equation $ \begin{aligned} &ax^2+bx+c=0 \end{aligned}$

are the same.

Sum of Roots

$ \begin{aligned} &α + β = -\frac{b}{a} \end{aligned}$

Product of Roots

$ \begin{aligned} &α \times β = \frac{c}{a} \end{aligned}$

Nature of Roots

Discriminant

$ \begin{aligned} &D = b^2-4ac \end{aligned}$

D > 0 2 Distinct, Real Roots

D = 0

2 Equal, Real Roots

(coincident roots)

D < 0

No Real Roots

➤ Roots of the QE are

$ \begin{aligned} &x=\frac{-b ± \sqrt{D}}{2a} \end{aligned}$

Click here to watch the Derivation of Quadratic Formula

Chapter 5 - Arithmetic Progression

General Form of AP

a, a + d, a + 2d, a + 3d, . . .

First Term: a

Common Difference: d

➺ d can be +ve, -ve or 0

$ \begin{aligned} &a_{2} -a_{1}=a_{3} -a_{2}=a_{4} -a_{3}=…=a_{n} -a_{n-1}=d \end{aligned}$

nth Term of AP

$ \begin{aligned} &a_{n} = a +(n-1)d \end{aligned}$

Sum of first n Terms

$ \begin{aligned} &S_{n} =\frac{n}{2}[2a+(n-1)d] \end{aligned}$

Sum of first n Terms

$ \begin{aligned} &S_{n} =\frac{n}{2}(a+l) \\&\text{where }l \text{ is the last term}\end{aligned}$

📌 Super important 🔥

Total terms in AP: 25

11th term from the end = 25 – 11 + 1 = 15th Term

If a, b, c are in AP, then

$ \begin{aligned} &b=\frac{a+c}{2} \end{aligned}$

Chapter 6 - Triangles

Similar Figures

➺ Same shape

➺ Not necessarily the same size

🦆 🦆 🔺🔺

➺ All Circles are similar 🔵

➺ All Squares are similar 🟩

➺ All Equilateral Triangles are similar 🔺

➺ All Congruent figures are similar ✅

❌ All similar figures are not congruent

🔺 🔺 Similar Triangles 🔺🔺

Two Triangles ABC and DEF are similar if

➺ their corresponding angles are equal

$ \begin{aligned} &\angle A = \angle D, \ \ \angle B = \angle E, \ \ \angle C = \angle F\end{aligned}$

➺ & their corresponding sides are in the same ratio (proportion).

$ \begin{aligned} &\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\end{aligned}$

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Converse of Basic Proportionality Theorem

If a line divides any two sides of a triangle in the same ratio then the lines is parallel to the third side.

Triangles Similarity Criteria

SSS

If Corresponding Sides are in the same ratio, then the ▲s are similar.

$ \begin{aligned} &\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\end{aligned}$

AAA

If Corresponding Angles are equal, then the ▲s are similar.

$ \begin{aligned} \angle A &= \angle D,\\ \angle B &= \angle E,\\ \ \ \angle C &= \angle F\end{aligned}$

SAS

If one angle of a triangle is equal to one angle of another triangle and the sides including this angle are in the same ratio, then the ▲s are similar. For e.g.

$ \begin{aligned} \frac{AB}{DE}&=\frac{BC}{EF} \ \ \\ \angle B &= \angle E \end{aligned}$

RHS

If in two right angled ▲s, hypotenuse and one side of a ▲ are proportional to hypotenuse and one side of the other ▲, then the ▲s are similar.

If two angles of one ▲ are respectively equal to the two angles of the other ▲, then the two ▲s are similar.

$ \begin{aligned} \angle A &= \angle D,\\ \angle B &= \angle E\end{aligned}$

This case is same as AAA because 3rd angle will be equal as well.

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Chapter 7 - Coordinate Geometry

Abscissa / x-coordinate

The distance of a point from the y-axis

Ordinate / y-coordinate

The distance of a point from the x-axis

Distance Formula

Distance between points A (x₁, y₁) and B (x₂, y₂)

Distance from origin

Distance of point A (x, y) from origin O (0, 0)

Section Formula

Mid-Point Formula

Chapter 8 - Introduction to Trigonometry

For Angle A

BC – Perpendicular

AB – Base

AC – Hypotenuse

For Angle C

AB – Perpendicular

BC – Base

AC – Hypotenuse

Trigonometric Ratios

$\begin{align*}&\color{red}{sin} \ \ \color{green}{cos} \ \ \color{blue}{tan}\\ &\boxed{\color{red}{\frac{P}{H}}}\boxed{\color{green}{\frac{B}{H}}}\boxed{\color{blue}{\frac{P}{B}}}\end{align*}$

P B P – Pandit Badri Prasad

H H B – Har Har Bole

$ \begin{aligned} sinθ = \frac{P}{H}\end{aligned}$

$ \begin{aligned} &cosθ = \frac{B}{H}\end{aligned}$

$ \begin{aligned} &tanθ = \frac{P}{B}\end{aligned}$

$ \begin{aligned} &cosecθ = \frac{H}{P}\end{aligned}$

$ \begin{aligned} &secθ = \frac{H}{B}\end{aligned}$

$ \begin{aligned} &cotθ = \frac{B}{P}\end{aligned}$

Imp Relationships b/w ratios

$ \begin{aligned} &sin θ= \frac{1}{cosecθ}\end{aligned}$

$ \begin{aligned} &cos θ= \frac{1}{secθ}\end{aligned}$

$ \begin{aligned} &tan θ= \frac{1}{cotθ}\end{aligned}$

Trigonometric ratios of standard angles

∠A	0°	30°	45°	60°	90°
sin A	0	$\begin{aligned} \frac {1}{2}\end{aligned} $	$\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} $	$\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} $	1
cos A	1	$\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} $	$\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} $	$\begin{aligned} \frac {1}{2}\end{aligned} $	0
tan A	0	$\begin{aligned} \frac {1}{\sqrt {3}}\end{aligned} $	1	$\begin{aligned} \sqrt {3} \end{aligned} $	N.D.
cosec A	N.D.	2	$\begin{aligned} \sqrt {2} \end{aligned} $	$\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} $	1
sec A	1	$\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} $	$\begin{aligned} \sqrt{2}\end{aligned} $	2	N.D.
cot A	N.D.	$\begin{aligned} \sqrt{3}\end{aligned} $	1	$\begin{aligned} \frac {1}{\sqrt{3}}\end{aligned} $	1

N.D. – Not Defined

Trigonometric Identities

$ \begin{aligned} &sin^2θ+cos^2θ =1\end{aligned}$

$ \begin{aligned} &1+tan^2θ =sec^2θ\end{aligned}$

$ \begin{aligned} &1+cot^2θ =cosec^2θ\end{aligned}$

Chapter 9 - Some Applications of Trigonometry

This chapter is about the applications of the formulas that we learnt in the previous chapter. You Students are recommended to practice different types of questions to master this chapter.

Chapter 10 - Circles

Tangent

Line that intersects the circle at only one point

Important

➺ Diameter is the longest chord

➺ Tangent is perpendicular to the radius through the point of contact

➺ A circle has infinitely many tangents

Theorem

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Theorem

The lengths of tangents drawn from an external point to a circle are equal.

Chapter 11 - Areas Related to Circles

Area of Circle = πr²

Circumference of Circle = 2πr

Area of a Sector

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{θ}{360}} \color{black}\times \color{blue}{\pi r^2}}\end{aligned}$

Length of Arc

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{θ}{360}} \color{black}\times \color{blue}{2\pi r}}\end{aligned}$

Area of a Segment

Area of the corresponding sector

– Area of the corresponding triangle

Area of Major Sector = Area of Circle – Area of Minor Sector

Area of Major Segment = Area of Circle – Area of Minor Segment

Chapter 12 - Surface Areas and Volumes

HS – Hemisphere
SA – Surface Area
CSA – Curved Surface Area
TSA – Total Surface Area
LSA – Lateral Surface Area

Surface Areas

TSA of a Cube

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{6} \color{blue}{a^2}}\end{aligned}$

LSA of a Cube

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{4} \color{blue}{a^2}}\end{aligned}$

TSA of a Cuboid

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{(lb + bh + hl)}}\end{aligned}$

LSA of a Cuboid

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2h} \color{blue}{(l + b)}}\end{aligned}$

SA of a Sphere

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{4} \color{blue}{\pi r^2}}\end{aligned}$

CSA of a Solid HS

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{\pi r^2}}\end{aligned}$

TSA of a Solid HS

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{\pi r^2} + \color{magenta}{\pi r^2}} \end{aligned}$

TSA of a Hollow HS

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{\pi r^2} + \color{red}{2} \color{blue}{\pi r^2} } \end{aligned}$

CSA of a Solid Cylinder

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{ \color{blue}{2 \pi r} \color{red}{h} } \end{aligned}$

TSA of a Solid Cylinder

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{ \color{blue}{2 \pi r} \color{red}{h} + \color{red}{2}\color{blue}{\pi r^2} } \end{aligned}$

CSA of a Hollow Cylinder

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{ \pi r} \color{red}{h} + \color{red}{2} \color{blue}{ \pi r} \color{red}{h}} \end{aligned}$

TSA of a Hollow Cylinder

with Base

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{ \color{red}{4} \color{blue}{\pi r} \color{red}{h} +\color{red}{2} \color{blue}{\pi r^2} } \end{aligned}$

CSA of a Solid Cone

$\begin{aligned}&\bbox[15px, border: 3px solid green;] {\color{blue}{\pi rl}} \end{aligned}$

TSA of a Solid Cone

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}{\pi rl} + \color{red}{\pi r^2}} \end{aligned}$

CSA of a Hollow Cone

$\begin{aligned}&\bbox[15px, border: 3px solid green;] {\color{red}{2} \color{blue}{\pi rl}} \end{aligned}$

TSA of a Hollow Cone

$\begin{aligned}&\bbox[15px, border: 3px solid green;] {\color{red}{2} \color{blue}{\pi rl}} \end{aligned}$

Volumes

Volume of a Cube

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}{a^3} } \end{aligned}$

Volume of a Cuboid

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}l \color{black} \times \color{green}b \color{black}\times \color{red}h} \end{aligned}$

Volume of Sphere

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{4}{3}}\color{blue}{\pi r^3}} \end{aligned}$

Volume of HS

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{2}{3}}\color{blue}{\pi r^3}} \end{aligned}$

Volume of Cylinder

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}{\pi r^2} \color{red}{h} } \end{aligned}$

Volume of Cone

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{1}{3}}\color{blue}{\pi r^2} \color{red}{h} } \end{aligned}$

Chapter 13 - Statistics

Direct Method

$\begin{aligned} &\bbox[15px, border: 3px solid green;]{\overline{x} = \frac{\sum {f_ix_i}}{\sum{f_i}}} \end{aligned}$

Assumed Mean Method

$\begin{aligned} &\bbox[15px, border: 3px solid green;]{\overline{x} = a +\frac{\sum {f_id_i}}{\sum{f_i}}} \end{aligned}$

Step Deviation Method

$\begin{aligned} &\bbox[15px, border: 3px solid green;]{\overline{x} = a +\frac{\sum {f_iu_i}}{\sum{f_i}}\times h }\end{aligned}$

Mode of Grouped Data

$\begin{aligned} &\bbox[15px, border: 3px solid green;]{l +\frac{f_1-f_0}{2f_1-f_0-f_2}\times h }\end{aligned}$

Median of Grouped Data

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{l +\frac{\frac{n}{2}-cf}{f}\times h }\end{aligned}$

Class Marks

$\begin{aligned}&\bbox[15px, border: 3px solid green;]{\frac{\text{Upper Class Limit}- \text{Lower Class Limit}}{2}}\end{aligned}$

Chapter 14 - Probability

Experiment: Throwing a dice 🎲

Outcomes: { 1, 2, 3, 4, 5, 6 }

Total no. of outcomes = 6

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

➺ Outcomes are Equally Likely

Examples of Events

E1: Getting 3

E2: Getting a number > 4

E3: Getting a number < 7

E4: Getting a prime number

E5: Getting a number < 1

E – An Event 0 ≤ P(E) ≤ 1

$ \begin{aligned} &P(E) = \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}} \end{aligned}$

Probability of Events (E1 – E5)

P(E1)

E1: Getting 3

Favourable Outcome: { 3 }

No. of favourable outcomes = 1

$ \begin{aligned} P(E1)=\frac{1}{6} \end{aligned}$

P(E2)

E2: Getting a number > 4

Numbers > 4: { 5, 6 }

No. of favourable outcomes = 2

$ \begin{aligned} P(E2)=\frac{2}{6} =\frac{1}{3} \end{aligned}$

P(E3)

E3: Getting a number < 7

Numbers < 7: { 1, 2, 3, 4, 5, 6 }

No. of favourable outcomes = 6

$ \begin{aligned} P(E3)=\frac{6}{6} =1 \end{aligned}$

Sure Event

P(E4)

E4: Getting a prime number

Prime numbers: { 2, 3, 5 }

No. of favourable outcomes = 3

$ \begin{aligned} P(E4)=\frac{3}{6} =\frac{1}{2} \end{aligned}$

P(E5)

E5: Getting a number < 1

Number < 1: None

No. of favourable outcomes = 0

$ \begin{aligned} P(E5)=\frac{0}{6} =0 \end{aligned}$

Impossible Event

Sure Event 👍

E = It will rain OR it will not rain today

P(E) = 1

Impossible Event 👎

E = It will rain AND it will not rain today

P(E) = 0

Complementary Events 🎭

E = It will rain today

not E = It will not rain today

P(E) + P(not E) = 1

Equally Likely & Not Equally Likely Outcomes

Randomly take a ball out of a bag of same sized 4 red and 2 blue balls 🔴 🔴 🔴 🔴 🔵 🔵

Experiment 1

Outcomes: Red Ball 🔴, Blue Ball 🔵

Total possible outcomes = 2

P(Red Ball) = 4/6 = 2/3

P(Blue Ball) = 2/6 = 1/3

➺ Outcomes are Not Equally Likely

Experiment 2

Outcomes: A Ball of any color

(Color doesn’t matter)

Total possible outcomes = 6

P(Ball 1) = P(Ball 2) = P(Ball 3) = P(Ball 4) = P(Ball 5) = P(Ball 6) = 1/6

➺ Outcomes are Equally Likely

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The post Class 10 – All Maths Formulas – Complete List appeared first on Mathemafia.

Exercise 2.1 (NCERT) – Polynomials, Class 10 – All Solutions

Sunil Kumar — Thu, 12 Jun 2025 12:20:08 +0000

Exercise 2.1 (NCERT) - Polynomials, Class 10

Chapter 2	Exercise 2.1	Class 10

Question 1

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

The number of zeros of a graph is the number of times the graph intersects the x-axis.

The answers are written on the graphs below in red circle.

The post Exercise 2.1 (NCERT) – Polynomials, Class 10 – All Solutions appeared first on Mathemafia.

Exercise 11.1 (NCERT) – Areas Related To Circles, Class 10 – All Solutions

Sunil Kumar — Thu, 12 Jun 2025 11:48:35 +0000

Exercise 11.1 (NCERT) - Areas Related To Circles, Class 10

Chapter 11	Exercise 11.1	Class 10

Below are the Quick links for all questions of Exercise 11.1, Areas Related To Circles, Class 10.

Click a link to view the solution of corresponding question.

Ex: 11.1, Question 1
Ex: 11.1, Question 2
Ex: 11.1, Question 3
Ex: 11.1, Question 4
Ex: 11.1, Question 5
Ex: 11.1, Question 6
Ex: 11.1, Question 7
Ex: 11.1, Question 8
Ex: 11.1, Question 9
Ex: 11.1, Question 10
Ex: 11.1, Question 11
Ex: 11.1, Question 12
Ex: 11.1, Question 13
Ex: 11.1, Question 14

Below are the solutions of every question of exercise 11.1 of chapter 11, Circles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Unless stated otherwise, use 𝛑 = 22/7.

Question 1

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Question 2

Find the area of a quadrant of a circle whose circumference is 22 cm.

Given

Circumference of circle, C = 22cm

We know that

Question 3

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Given

Length of minute hand (radius) = r = 14cm

Angle swept in 5 minutes

Question 4

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment (ii) major sector. (Use 𝝅 = 3.14)

Triangle AOB is a right angled triangle

=> Base = 10cm, Height = 10cm

Area = 1/2 (Base * Height)

Question 5

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc (ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

Given

(i) Length of the Arc

(ii) Area of the Sector

(iii) Area of the Segment

You can finish the question here but in case you are asked to use the value of √3 = 1.732 then solve it further.

Question 6

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use pi = 3.14 and √3 = 1.73)

Given

Question 7

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

(Use pi = 3.14 and √3 = 1.73)

Given

To find the area of triangle AOB, split the triangle into 2 equal right angled triangles by drawing a perpendicular OD on AB.

∠AOD = ∠BOD = 60°

See the video.

Question 8

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m.

(Use pi = 3.14)

Given

Horse is tied at one corner, so grazing area is a quarter circle.

Question 9

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Given

5 diameters divide the circle into 10 equal sectors.

(i) Total Length of Silver Wire

(ii) Area of Each Sector

Question 10

An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Given

Umbrella has 8 equally spaced ribs. This means the circle is divided into 8 equal sectors.

Since there are equally spaced 8 sectors, hence angle formed by every sector is

Question 11

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Given

Area swept by one wiper = Area of sector formed at angle 115 degrees

Question 12

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use pi = 3.14)

Given

Area of sea over which ships are warned

Question 13

A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm².

(√3 = 1.7)

Given

Step 2: Area of triangle formed by 2 radii and the chord

This triangle is an equilateral triangle.

Question 14

Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is

Given

p = Angle of the sector in degrees
R = Radius of the circle

Answer: D

$\begin{aligned} \end{aligned}$

The post Exercise 11.1 (NCERT) – Areas Related To Circles, Class 10 – All Solutions appeared first on Mathemafia.

Exercise 10.2 (NCERT) – Circles, Class 10 – All Solutions

Sunil Kumar — Wed, 11 Jun 2025 08:57:39 +0000

Exercise 10.2 (NCERT) - Circles, Class 10

Chapter 10	Exercise 10.2	Class 10

Below are the Quick links for all questions of Exercise 10.1, Circles, Class 10.

Click a link to view the solution of corresponding question.

Ex: 10.2, Question 1
Ex: 10.2, Question 2
Ex: 10.2, Question 3
Ex: 10.2, Question 4
Ex: 10.2, Question 5
Ex: 10.2, Question 6
Ex: 10.2, Question 7
Ex: 10.2, Question 8
Ex: 10.2, Question 9
Ex: 10.2, Question 10
Ex: 10.2, Question 11
Ex: 10.2, Question 12
Ex: 10.2, Question 13

Below are the solutions of every question of exercise 10.2 of chapter 10, Circles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm

Given

Length of tangent, PQ = 24 cm

OQ = 25 cm

Let the radius = r = OP

Since PQ is a tangent to the circle at P

Hence ∠OPQ = 90

In right angled △OPQ

Using Pythagoras Theorem

Question 2

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

(A) 60° (B) 70° (C) 80° (D) 90°

Given

∠POQ = 110°

TP and TQ are tangents from point T to the circle.

So, ∠OPT = ∠OQT = 90° (radius ⟂ tangent)

Now, in quadrilateral POQT:

∠OPT = 90°
∠OQT = 90°
∠POQ = 110°
∠PTQ = ?

Using the angle sum property of a quadrilateral:

∠PTQ = 360° – (90° + 90° + 110°) = 70°

Answer: B

Question 3

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to

(A) 50° (B) 60° (C) 70° (D) 80°

Given

∠APB = 80°

PA and PB are tangents from point P to the circle.

OA = radius

Hence ∠OAP = 90°

Join OP

∠APB = 80°

Line OP bisects ∠APB

Therefore, ∠OPA = ∠OPB = 40°

In right angled triangle △OPA

sum of angles = 180°

So, ∠POA = 180° − 90° − 40° = 50°

Answer: A

Question 4

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Given

A circle with diameter AB. Tangents are drawn at points A and B.

To prove: The tangents at A and B are parallel.

Construction: Let O be the centre of the circle. Join OA and OB.

Since AB is the diameter => AOB is a straight line
Therefore, both tangents are perpendicular to the same line AOB.

Question 5

Prove that the perpendicular at the point of contact to the tangent to a circle passes

through the centre.

Assume that ⊥ on point P, does not pass through the centre of the circle, O.

Consider it passes through point O’.

=> ∠APO’ = 90°

Also ∠BPO = 90° (∵ tangent is ⊥ to the radius through point of contact)

Also ∠APB = 180° (Straight angle)

This means

This is possible only when OP and O’P coincide.

Hence our assumption is wrong.

This implies perpendicular at the point of contact to tangent passes through the centre of the circle.

Question 6

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Given
OA = 5 cm
AP = 4 cm

To find

Radius (r) = OP

AP is tangent and OP is radius, hence

∠OPA = 90°

In right angled △OPA

Using Pythagoras Theorem

Question 7

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Let the chord AB touches the inner circle at point P. This means:

OP = 3 cm (Radius of inner circle)

OA = 5 cm (Radius of outer circle)

APB is tangent and OP is radius of inner circle, hence

∠OPA = 90°

In right angled △OPA

Using Pythagoras Theorem

Question 8

A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Let the circle touches sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

Let the lengths of tangents from each vertex to the points of contact be:

Using the above information

Question 9

In Fig. 10.13, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Watch this video to learn the easiest method to solve this question.

XY and X’Y’ are two || tangents.

=> PAQ is a diameter of the circle.

∠1 = ∠2 = 90°

Because AP ⊥ OP and BQ ⊥ OQ

In quadrilateral APBQ

Now, AP and AC are two tangents from the same point, A.

This implies, OA bisects ∠PAC

Similarly, OB bisects ∠QBC

In triangle AOB

Question 10

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

To Prove: ∠1 + ∠3 = 180°

Let the circle have centre O &

A be a point outside the circle.

Draw tangents AP and AQ.

Join OP and OQ.

PA is tangent and OP is radius, hence

∠OPA = ∠4 = 90°

Similarly AQ is tangent and OQ is radius, hence

∠OQA = ∠2 = 90°

Now, OPAQ is a quadrilateral.

Applying Angle sum property of quadrilateral

∠1 + ∠2 + ∠3 + ∠4 = 360°

∠1 + 90° + ∠3 + 90° = 360°

∠1 + ∠3 + 180° = 360°

∠1 + ∠3 = 180°

Hence proved.

Question 11

Prove that the parallelogram circumscribing a circle is a rhombus.

Let ABCD be a parallelogram circumscribing a circle.

Since tangents from a point to a circle are equal:

Using above information

This means

Now AB = CD and BC = AD because ABCD is a parallelogram. Therefore

Hence ABCD is a rhombus.

Question 12

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Method 1

Given

BD = 8cm, DC = 6cm

=> BC = 14cm

Area of triangle using Heron’s Formula & inradius

Method 2

Given

BD = 8cm, DC = 6cm

=> BC = 14cm

Area of triangle ABC = Finding areas of 3 small triangles

Area of triangle ABC by Heron’s Formula

Question 13

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

To prove

Given

ABCD is a quadrilateral circumscribing a circle.

Let centre of circle is O.

Let the circles touches the quadrilateral at points P, Q, R and S as shown in the figure.

This means APB, BQC, CRD and DSA are tangents to the circle.

Join OA, OB, OC, OD, OP, OQ, OR and OS.

Use numbers (1 to 8) to name the angles as shown in the figure.

Now we know that

Also

Similarly

Hence proved.

$\begin{aligned} \end{aligned}$

The post Exercise 10.2 (NCERT) – Circles, Class 10 – All Solutions appeared first on Mathemafia.

Exercise 10.1 (NCERT) – Circles, Class 10 – All Solutions

Sunil Kumar — Mon, 09 Jun 2025 07:46:09 +0000

Exercise 10.1 (NCERT) - Circles, Class 10

Chapter 10	Exercise 10.1	Class 10

Below are the Quick links for all questions of Exercise 10.1, Circles, Class 10.

Click a link to view the solution of corresponding question.

Ex: 10.1, Question 1
Ex: 10.1, Question 2
Ex: 10.1, Question 3
Ex: 10.1, Question 4

Below are the solutions of every question of exercise 10.1 of chapter 10, Circles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

How many tangents can a circle have?

Infinitely many.

Reason: There are infinitely many points on the circumference of a circle. Hence infinitely many tangents.

Still confused? Watch the video.

Question 2

Fill in the blanks :

(i) A tangent to a circle intersects it in 1 point (s).

(iii) A line intersecting a circle in two points is called a secant.

(ii) A circle can have 2 parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called point of contact.

Question 3

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm.

OP = 5 cm (radius)

OQ = 12 cm

Since PQ is a tangent to the circle at P

∠OPQ = 90

In right triangle OPQ, apply the Pythagoras Theorem

Question 4

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

You need to watch the video 😉

$\begin{aligned} \end{aligned}$

The post Exercise 10.1 (NCERT) – Circles, Class 10 – All Solutions appeared first on Mathemafia.

Exercise 7.2 (NCERT) – Coordinate Geometry, Class 10 – All Solutions

Sunil Kumar — Thu, 05 Jun 2025 11:43:15 +0000

Exercise 7.2 (NCERT) – Coordinate Geometry, Class 10

Chapter 7	Exercise 7.2	Class 10

Below are the Quick links for all questions of Exercise 7.2, Coordinate Geometry, Class 10. Click a link to view the solution of corresponding question.

Ex: 7,2 Q1
Ex: 7.2, Q2
Ex: 7.2, Q3
Ex: 7.2, Q4
Ex: 7.2, Q5
Ex: 7.2, Q6
Ex: 7.2, Q7
Ex: 7.2, Q8
Ex: 7.2, Q9
Ex: 6.1, Q10

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Below are the solutions of every question of exercise 7.2 of chapter 7, Coordinate Geometry, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Complete Chapter – Video Tutorial

Question 2

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Let the given points be A(4, -1) and B(-2, -3).

We want to find the points of trisection of AB.

Let P and Q be the points dividing AB in the ratio 1:2 and 2:1 respectively.

Using section formula
If P divides the line joining A(x₁, y₁) and B(x₂, y₂) in ratio m:n

Question 3

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown 1 in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Given

AD = 100m (since 100 pots placed 1 m apart)

Distance between flags G and R

Midpoint (M) coordinates where Rashmi posts the blue flag

Question 4

Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Question 5

Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Question 6

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Question 7

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Question 8

If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

Question 9

Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Question 10

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

[Hint : Area of a rhombus = (product of its diagonals)/2]

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Exercise 7.1 (NCERT) – Coordinate Geometry, Class 10 – All Solutions

Sunil Kumar — Sat, 10 May 2025 12:35:12 +0000

Exercise 7.1 (NCERT) – Coordinate Geometry, Class 10

Chapter 7	Exercise 7.1	Class 10

Below are the Quick links for all questions of Exercise 7.1, Coordinate Geometry, Class 10. Click a link to view the solution of corresponding question.

Ex: 7.1, Q1
Ex: 7.1, Q2
Ex: 7.1, Q3
Ex: 7.1, Q4
Ex: 7.1, Q5
Ex: 7.1, Q6
Ex: 7.1, Q7
Ex: 7.1, Q8
Ex: 7.1, Q9
Ex: 7.1, Q10

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Below are the solutions of every question of exercise 7.1 of chapter 7, Coordinate Geometry, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

Find the distance between the following pairs of points :

(i) (2, 3), (4, 1)

(ii) (– 5, 7), (– 1, 3)

(iii) (a, b), (– a, – b)

Complete Chapter – Video Tutorial

Question 2

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

The problem given in section 7.2 has the same solution. Watch the video to know more about it.

Question 3

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Question 4

Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Since two sides are equal, the triangle is isosceles.

Question 5

In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Question 6

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

All sides are equal.

Now, checking the diagonals:

Diagonals are also equal.

Hence it is square.

(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Now, checking the diagonals:

Since opposite sides are equal, and the diagonals are not equal, the quadrilateral is a parallelogram.

Question 7

Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Question 8

Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Question 9

If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Question 10

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

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Exercise 6.1 (NCERT) – Triangles, Class 10 – All Solutions

Sunil Kumar — Fri, 09 May 2025 11:26:35 +0000

Exercise 6.1 (NCERT) – Triangles, Class 10

Chapter 6	Exercise 6.1	Class 10

Below are the Quick links for all questions of Exercise 6.1, Triangles, Class 10. Click a link to view the solution of corresponding question.

Ex: 6.1, Q1
Ex: 6.1, Q2
Ex: 6.1, Q3

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Below are the solutions of every question of exercise 6.1 of chapter 6, Triangles, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

Fill in the blanks using the correct word given in brackets :

(i) All circles are similar.

(ii) All squares are similar.

(iii) All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar, if

(a) their corresponding angles are equal and

(b) their corresponding sides are proportional.

Complete Exercise – Video Tutorial

Question 2

Give two different examples of pair of

(i) similar figures

All circles are similar
All squares are similar
All equilateral triangles are similar

(ii) non-similar figures

A circle and a square
A square and a triangle
A right angled triangle and an equilateral triangle

Question 3

State whether the following quadrilaterals are similar or not:

No, because all the corresponding angles in the figures are not equal.

Detail (Optional): All angles of a square are equal to 90 degrees where all of the angles in the rhombus are not equal to 90 degree. Hence the figures are not similar.

The post Exercise 6.1 (NCERT) – Triangles, Class 10 – All Solutions appeared first on Mathemafia.

Exercise 5.2 (NCERT) – Arithmetic Progressions, Class 10 – All Solutions

Sunil Kumar — Fri, 09 May 2025 10:28:18 +0000

Exercise 5.2 (NCERT) – Arithmetic Progressions, Class 10

Below are the Quick links for all questions of Exercise 5.2, Arithmetic Progressions, Class 10. Click a link to view the solution of corresponding question.

Chapter 5	Exercise 5.2	Class 10

Ex: 5.2, Q1
- Part (i)
- Part (ii)
- Part (iii)
- Part (iv)
- Part (v)
Ex: 5.2, Q2
- Part (i)
- Part (ii)
Ex: 5.2, Q3
- Part (i)
- Part (ii)
- Part (iii)
- Part (iv)
- Part (v)
Ex: 5.2, Q4
Ex: 5.2, Q5
Ex: 5.2, Q6
Ex: 5.2, Q7
Ex: 5.2, Q8
Ex: 5.2, Q9
Ex: 5.2, Q10
Ex: 5.2, Q11
Ex: 5.2, Q12
Ex: 5.2, Q13
Ex: 5.2, Q14
Ex: 5.2, Q15
Ex: 5.2, Q16
Ex: 5.2, Q17
Ex: 5.2, Q18
Ex: 5.2, Q19
Ex: 5.2, Q20

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Below are the solutions of every question of exercise 5.2 of chapter 5, Arithmetic Progressions, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

Fill in the blanks in the following table, given that a is the first term, d the common difference and a the nth term of the AP:

(i) a=7, d=3, n=8, a_n=?

nth term of an AP

(ii) a=-18, d=?, n=10, a_n=0

(iii) a=?, d=-3, n=18, a_n=-5

(iv) a=-18.9, d=2.5, n=?, a_n=3.6

(v) a=3.5, d=0, n=105, a_n=?

Question 2

Choose the correct choice in the following and justify :

(i) 30th term of the AP: 10, 7, 4, . . . , is

(A) 97 (B) 77 (C) –77 (D) – 87

First Term, a = 10

Common Difference, d = 7 – 10 = -3

n = 30

nth terms of an AP

Answer: C

(ii) 11th term of the AP: – 3, -1/2, 2, … is

(A) 28 (B) 22 (C) -38 (D) -48 + 1/2

Answer: B

Question 3

In the following APs, find the missing terms in the boxes :

(i) 2, ? , 26

Given AP: 2, ? , 26
Middle term is the mean of adjacent terms in an AP series.

Answer: 14

(ii) ? , 13, ?, 3

Let us write the viven AP as : x, 13, y, 3

Middle term is the mean of adjacent terms in an AP series.

Now the series is : x, 13, 8, 3

Now 13 is mean of x and 8.

Answer: 18, 8

(iii) 5, ?, ?, 19/2

First term, a = 5

Let the common difference = d

nth terms of an AP

Let us now find the missing terms

Answer: 13/2, 8

(iv) -4, ?, ?, ?, ?, 6

First term, a = -4

Let the common difference = d

nth terms of an AP

Let us now find the missing terms

Answer: -2, 0, 2, 4

(v) ?, 38, ?, ?, ?, -22

Let the First term = a

Let the common difference = d

nth terms of an AP

Subtract (2) from (1)

Substitute value of d in (2)

Let us now find the missing terms

Answer: 53, 23, 8, -7

Question 4

Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

First term (a) = 3
Common difference (d) = 8 – 3 = 5

nth terms of an AP

Answer: 16th term

Question 5

Find the number of terms in each of the following APs :

(i) 7, 13, 19, . . . , 205

First term (a) = 7
Common difference (d) = a₂ – a₁ = 13 – 7 = 6

nth terms of an AP

This means

Answer: n = 34

(ii) 18, 15+1/2, 13, . . . , – 47

First term (a) = 18

Common difference

nth terms of an AP

This means

Answer: n = 27

Question 6

Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .

First term (a) = 11
Common difference (d) = a₂ – a₁ = 8 – 11 = -3

nth terms of an AP

This means

Since n is not an integer this implies -150 is not a term of the AP.

Answer: -150 is not a term of given AP

Question 7

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

nth terms of an AP

This means

Subtract (1) from (2)

Substitute value of d in (1)

Let’s find 31st term.

Answer: a₃₁ = 178

Question 8

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

nth terms of an AP

This means

Subtract (1) from (2)

Substitute value of d in (1)

Let’s find 29th term.

Answer: a₂₉ = 64

Question 9

If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

nth terms of an AP

This means

Subtract (1) from (2)

Substitute value of d in (1)

Let nth term is 0, this implies

Answer: 5th term

Question 10

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

nth terms of an AP

This means

Answer: d = 1

Question 11

Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

nth terms of an AP

Also given

Let kth term is 132 more than 54th term, this implies

Answer: 65th term

Question 12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Let the two APs be

First AP: a₁, a₂, a₃, …

Second AP: b₁, b₂, b₃, …

Let Common difference for both APs = d

nth terms of the APs will be

Difference between 100th terms = 100

Let’s find difference between 1000th terms

Answer: 100

Question 13

How many three-digit numbers are divisible by 7?

Three-digit numbers range from 100 to 999

The smallest 3-digit number divisible by 7 = 105

The largest 3-digit number divisible by 7 = 994

All 3-digit numbers divisible by 7 are:

105, 112, 119, ….. , 994

This is an AP series.

First term, a = 105

Common Difference, d = 7

Let the number of terms in this AP = n

We know that, nth terms of an AP

This implies

Answer: 128

Question 14

How many multiples of 4 lie between 10 and 250?

Smallest multiple of 4 > 10 = 12

Largest multiple of 4 < 250 = 248

All multiples of 4 between numbers between 10 and 250 are:

12, 16, 20, …., 248

This is an AP series.

First term, a = 12

Common Difference, d = 4

Let the number of terms in this AP = n

We know that, nth terms of an AP

This implies

Answer: 60

Question 15

For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

First AP: 63, 65, 67, …

Second AP: 3, 10, 17, …

nth Terms of both APs are equal, this implies

Answer: 13

Question 16

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Let the AP be

a, a + d, a + 2d, a + 3d,…

3rd term = 16

7th term – 5th term = 12

Substitute the value of d in (1)

AP series:

a, a + d, a + 2d, a + 3d,…

4, 10, 16, 22, …

Answer: 4, 10, 16, 22, …

Question 17

Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Given AP

3, 8, 13,…, 253
a = 3, d = 5, l = 253

nth terms of an AP

Let us find n for for last term

20th term from the last = (51 – 20 + 1)th term from start = 32nd term from start

Answer: 158

Question 18

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Let the AP be:

a, a + d, a + 2d,…

Sum of 4th and 8th terms = 24

Sum of 6th and 10th terms = 44

Subtract (1) from (2)

Substitute value of d in (1)

AP series:

a, a + d, a + 2d, …

-13, -8, -3, …

Answer: -13, -8, -3, …

Question 19

Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?

Salary in 1995 = 5000

Salary in 1996 = 5000 + 200 = 5200

Salary in 1997 = 5200 + 200 = 5400

and so on …

5000, 5200, 5400, … is an AP series

First term, a = 5200

Common difference, d = 200

Consider nth term will be 7000

n = 11 means in 11th year the salary will be 7000.

=> 1995 + 11 – 1 = 2005

In year 2005, the salary will be Rs. 7000.

Answer: 11th years, 2005

Why 2005 and not 2006? Watch the video.

Question 20

Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.

Savings in the 1st week = Rs. 5
Savings in the 2nd week = 5 + 1.75 = Rs. 6.75
Savings in the 3rd week = 6.75 + 1.75 = Rs. 8.5
Savings in the 4th week = 8.5 + 1.75 = Rs. 10.25
and so on…

5, 6.75, 8.5, 10.25, … is an AP series

First term, a = 5

Common difference, d = 1.75

Savings in the nth week are Rs. 20.75.

Answer: 10

The post Exercise 5.2 (NCERT) – Arithmetic Progressions, Class 10 – All Solutions appeared first on Mathemafia.

Exercise 5.1 (NCERT) – Arithmetic Progressions, Class 10 – All Solutions

Sunil Kumar — Wed, 07 May 2025 07:54:05 +0000

Exercise 5.1 (NCERT) – Arithmetic Progressions, Class 10

Below are the Quick links for all questions of Exercise 5.1, Arithmetic Progressions, Class 10. Click a link to view the solution of corresponding question.

Chapter 5	Exercise 5.1	Class 10

Ex: 5.1, Q1
Ex: 5.1, Q2
Ex: 5.1, Q3
Ex: 5.1, Q4

Mathemafia

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Below are the solutions of every question of exercise 5.1 of chapter 5, Arithmetic Progressions, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.

Question 1

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

Fare for

1 km = ₹15

2 km = 15 + 8 = ₹23

3 km = 15 + 8 + 8 = ₹31

4 km = 15 + 8 + 8 + 8 = ₹39

and so on…

We get the list:

15, 23, 31, 39, …

This is an AP series because the difference between successive terms is constant = 8.

Answer: Yes, It is an AP series

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Considering the original amount of air (a₁)= 1 Unit

After the first removal, the remaining amount of air is:

After the second removal, the remaining amount of air is:

Let us find the common difference

Since a₂ − a₁ ≠ a₃ − a₂, the sequence is not an Arithmetic Progression (AP).

Answer: No, It is not an AP series

(iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.

Digging cost

For 1 metre: ₹150

For 2 metres: 150 + 50 = ₹200

For 3 metres: 150 + 50 + 50 = ₹250

For 4 metres: 150 + 50 + 50 + 50 = ₹300

and so on…

We get the list:

150, 200, 250, 300, …

This is an AP series because the difference between successive terms is constant = 50.

Answer: Yes, It is an AP series

(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.

Amount in the account

For the 1st year:

For the 2nd year:

For the 3rd year:

We get the list:

10000, 10800, 11664, 12597, …

We can see that

A₂– A₁ = 11664 – 10800 = 864

A₃– A₂ = 12597 – 11664 = 933

Since a₂ − a₁ ≠ a₃ − a₂, the sequence is not an Arithmetic Progression (AP).

Answer: No, It is not an AP series

Question 2

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

First term $a = 10$
Common difference $d = 10$

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = 10
a₂ = 10 + 10 = 20
a₃ = 10 + 20 = 30
a₄ = 10 + 30 = 40

Answer: 10, 20, 30, 40

(ii) a = –2, d = 0

First term $a = -2$
Common difference $d = 0$

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = –2
a₂ = –2 + 0 = –2
a₃ = –2 + 0 = –2
a₄ = –2 + 0 = –2

Answer: -2, -2, -2, -2

(iii) a = 4, d = – 3

First term $a = 4$
Common difference $d = -3$

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = 4
a₂ = 4 + (–3) = 1
a₃ = 1 + (–3) = –2
a₄ = –2 + (–3) = –5

Answer: 4, 1, –2, –5

(iv) a = – 1, d = 1/2

First four terms are: a, a + d, a + 2d, a + 3d

Answer: –1, –1/2, 0, 1/2

(v) a = – 1.25, d = – 0.25

First term $a = -1.25$
Common difference $d = -0.25$

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = –1.25
a₂ = –1.25 + (–0.25) = –1.50
a₃ = –1.50 + (–0.25) = –1.75
a₄ = –1.75 + (–0.25) = –2.00

Answer: –1.25, –1.50, –1.75, –2.00

Question 3

For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .

First term (a) = 3
Common difference (d) = a₂ – a₁ = 1 – 3 = –2

(ii) – 5, – 1, 3, 7, . . .

First term (a) = –5
Common difference (d) = a₂ – a₁ = (–1) – (–5) = –1 + 5 = 4

(iii) 1/3, 5/3, 9/3, 13/3

(iv) 0.6, 1.7, 2.8, 3.9, . . .

First term (a) = 0.6
Common difference (d) = a₂ – a₁ = 1.7 – 0.6 = 1.1

Question 4

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, . . .