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Alen Mathew 5 years, 3 months ago
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Student ✍️✍️✍️ 5 years, 3 months ago
Posted by Mahend Mishra 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
CBSE Class 10 Mathematics (041) - Deleted portion:
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UNIT I-NUMBER SYSTEMS |
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Chapter |
Topics |
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REAL NUMBERS |
Euclid’s division lemma |
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UNIT II-ALGEBRA |
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Chapter |
Topics |
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POLYNOMIALS |
Statement and simple problems on division algorithm for polynomials with real coefficients. |
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES |
cross multiplication method |
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QUADRATIC EQUATIONS |
Situational problems based on equations reducible to quadratic equations |
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ARITHMETIC PROGRESSIONS |
Application in solving daily life problems based on sum to n terms |
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UNIT III-COORDINATE GEOMETRY |
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Chapter |
Topics |
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COORDINATE GEOMETRY |
Area of a triangle |
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UNIT IV-GEOMETRY |
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Chapter |
Topics |
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TRIANGLES |
Proof of the following theorems are deleted · The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. · In a triangle, if the square on one side is equal to sumof the squares on the other two sides, the angle opposite to the first side is a right angle. |
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CIRCLES |
No Deletion |
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CONSTRUCTIONS |
Construction of a triangle similar to a given triangle. |
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UNIT V- TRIGONOMETRY |
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Chapter |
Topics |
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INTRODUCTION TO TRIGONOMETRY |
Motivate the ratios whichever are defined at 0o and 90o |
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TRIGONOMETRIC IDENTITIES |
Trigonometric ratios of complementary angles |
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HEIGHTS AND DISTANCES |
No deletion |
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UNIT VI-MENSURATION |
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AREAS RELATED TO CIRCLES |
Problems on central angle of 120° |
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SURFACE AREAS AND VOLUMES |
Frustum of a cone |
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UNIT VI-STATISTICS & PROBABILITY |
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Chapter |
Topics |
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STATISTICS |
· Step deviation Method for finding the mean · Cumulative Frequency graph |
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PROBABILITY |
No deletion |
Posted by Bharat Kol 5 years, 3 months ago
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Niharika Raj 5 years, 3 months ago
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Lakhan Jee Madhav 5 years, 3 months ago
Posted by Prakriti Dubey 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
Given, in two Δ ABC and Δ PQR, AB/QR = BC/PR = CA/PQ
which shows that sides of one triangle are proportional to the side of the other triangle, then their corresponding angles are also equal, so by SSS similarity, triangles are similar.
i.e., Δ CAB ∼ Δ PQR
Posted by Shubham Rai 5 years, 4 months ago
- 2 answers
Gaurav Seth 5 years, 4 months ago
let us assume that √7 be rational.
then it must in the form of p / q [q ≠ 0] [p and q are co-prime]
√7 = p / q
=> √7 x q = p
squaring on both sides
=> 7q2= p2 ------ (1)
p2 is divisible by 7
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p2 = 49 c2 --------- (2)
Subsitute p2 in equ (1) we get
7q2 = 49 c2
q2 = 7c2
=> q is divisible by 7
thus q and p have a common factor 7.
there is a contradiction
as our assumsion p & q are co prime but it has a common factor.
So that √7 is an irrational.
Posted by Aayush Ranjan 5 years, 4 months ago
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Sakshi Sehrawat 5 years, 4 months ago
Gaurav Seth 5 years, 4 months ago
Let us consider two right angle triangles right angled at Q and S respectively.
Now,


But cos A = cos B (given)
∴ 

Let 

In ΔRAS,
Using Pythagoras theorem, we have

In APBQ,
Using Pythagoras theorem, we have

So, 
Comparing (i) and (ii), we get

So, by using SSS similar condition

∴ 
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Prakriti Dubey 5 years, 4 months ago
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Prakriti Dubey 5 years, 4 months ago
Posted by Sharon Elsa 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
Solution :
Now, We have to make up a pair of linear equations with solution: x=-1 and y =3.
Let, and
1st Equation form is
Let, and
2nd Equation form is
Therefore, The required equations are and
Posted by Md Nadim Sarbar 5 years, 4 months ago
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Posted by Veshu Pathak 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
The given system of equations:
kx + 2y = 5
⇒ kx + 2y - 5 = 0 ….(i)
3x - 4y = 10
⇒3x - 4y - 10 = 0 …(ii)
These equations are of the forms:
Thus for all real values of k other than −3/2 , the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
Posted by Ritika Talwar 5 years, 4 months ago
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Santosh Suhana 5 years, 4 months ago
Meghna Thapar 5 years, 4 months ago
If the systems of equations are dependent, it means that there are an infinite number of solutions. So in order to determine a single solution (out of the infinite possibilities), the value of x will depend on what you choose as the value of y. That is, x varies with y (and y varies with x). In mathematics and particularly in algebra, a linear or nonlinear system of equations is called consistent if there is at least one set of values for the unknowns that satisfies each equation in the system—that is, when substituted into each of the equations, they make each equation hold true as an identity.
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Harshini V.M 5 years, 4 months ago
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Yogita Ingle 5 years, 3 months ago
Let a , d are first term and common
difference of an A.P
nth term = Last term = a + ( n - 1 )d
an = a + ( n - 1 )d
Now ,
It is given that ,
Third term = 12
a + 2d = 12 ------( 1 )
Last term = 106
a + 49d = 106 ---( 2 )
Subtract ( 1 ) from ( 2 ) , we get
47d = 94
d = 2
Substitute d value in equation ( 1 ) ,
We get
a + 2 × 2 = 12
a = 12 - 4
a = 8
29th term = a + 28d
a29 = 8 + 28 × 2
= 8 + 56
= 64
Posted by Manishra, M 5 years, 4 months ago
- 2 answers
Yogita Ingle 5 years, 4 months ago
(√3-2) (√3+2)
=(√3)2 -(2)2 {(a - b)(a+b) = a2 - b2}
= 3 - 4
= -1

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Vikram Singh 5 years, 3 months ago
2Thank You