Is maths difficult?

 Is maths difficulta

This is said that maths is very difficult, I think iit is not right as maths is the subject which becomes  very easy after understanding once. 

But every subject becomes easy after understanding but there is difference between Maths and Other subject. 

In maths after applying the appropriate methods we get the right answer os it is objective but on the other hand other subject may be subjective. 

Maths is logical but humanities are  also logical but it affects by personal thinking also. 

First you want to become expert in  maths then first learn. 

Learn Basic Mathematical operation like plus, minus, multiply, divide

Learn table

Understand basic maths topic such LCM, HCF

Learn Plus, minus of fraction

Understand some maths tricks  to solve the  question easily. 

Students must read  the question atleast two times to understand better. 

Share your doubts and concerns with your teacher. 

Follow above suggestions and  gain expertise in maths. 

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How to pass in Maths class 10

 To Pass in Maths

Comparatively other subjects it is very easy to pass in the Maths because here 

1) You don't have to remember things only there is need to understand. 

2) You don't have to worry about language some grammatically mistakes don't affect your performance you can still achieve good marks. 

3) Generally questions come from the NCERT Book. 

4) Exemplar is also very good book try to solve examples of this. 

5) Choose Five chapters of your choice and solve all types of questions of those chapters. 

6) Go through previous years papers and solve those questions which are from your chosen chapters. 

I think if you follow these suggestions then you can pass easily in maths with flying colors. 

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Basic Maths

Basic Maths? 

Is Basic Maths easy for  the Board  Exam? Answer is it is said but generally sometime it becomes little difficult also. 

First after adopting the subject don't analysis ver much about the difference between Basic Maths and Standard Maths. 

You will have to practise all the questions given in the text book because iin the book there is no differentiation between questions that thi for Standard and that is for Basic, any question may come in any paper either in Standard Or Basic. 

Don't have have such mindset that you will have easy question paper because you have adopted Maths Basic. 

Some Important points one should keep in his mind. 

1) Don't be in illusion the Basic Paper will be easy. 

2) Don't think that the students who have chosen Basic Maths they will have to work less. 

3) If  come across difficult question then solve it prepare it don't think that that difficult question is for those students who have taken Maths Standard? 

4) Don't demean yourself because you have taken Maths Basic. 

5) When solve any sample paper don't search first that is it Basic OR Standard. 

6) Try to solve standard  and basic paper of maths both. 

7) Make target of numbers you want to gain work for that don't think that you can gain those marks without efforts because you have taken maths Basic. 

You will have to work hard for both ie Maths Basic & Maths Standard, so don't loose time in analysing the differentiate between Maths Basic and Maths Standard. 

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How to benefits from CBSE previous years papers

 CBSE previous years papers, are very useful but when they used in right manner then they become more useful. 

Here you srg being some  suggestions that how you should use the previous years papers generally types of questions are repeated. 

First choose the topic ie Probably

Just open the previous year paper in descending manners it means first solve 2024 paper later take 2023....and so on

Solve all the questions of probability a available in the paper then take other year's paper and solve all the questions related to the probability. 

In such manner you can change the chapter and and  again follow suit. 

After this you can focus on Case Study questions came in the previous years papers. 

After doing such exercise you can have a rough idea that how  cbse is framing the questions. Then you  will become confident and your mind will be ready to attempt all the questions

It will be better to keep previous years papers in the hard copy going on the website and search online will distract your mind  from study. 

It will be better to gonna on CBSE OFFICIAL ACADEMIC site to download the question papers and marking scheme insted of going on some other websites. 

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Q 1 Find relationship between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). so

∴ AP = BP

         AP 2 = BP 2 

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

x2 – 4x + 4 + y2 – 10y + 25

                 =

x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y +29 =  6x - 14y +58 

⇒ -10x + 4y = 29

 is the required relation.

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How To Prepare For The Board Exam

 

Today I want  to share some tips for the coming Board Exam. What I am shareing , it is based  on the my experience. 

Some Tips must be kept in the mind for the  Board Exam. 

1. Attempt all the questions in the exam. 

2. First attempt all the easy questions. 

3. In objective questions write only options like A, B, C, D  no need to write the complete answer. 

4. Read the question carefully and adopt the right  strategy to solve the question. 

5 .Learn the formula correctly. 

6. Learn the trigonometry table correctly and also learn the identities. 

7. Draw proper diagram. 

8. Write property Given, To prove, Construction, proof, Conclusion. 

9  Select the topic which you like most  , first focus on them

10. Understand the pattern of the paper clearly, ie for Maths Class 10  Board Exam. 

    20  MCQ of  1 marks                  20   marks

    5 questions of  2 marks           10   marks

    6 questions of  3 marks           18  marks

   4  questions of  5  marks          20 marks

   3  questions of  case study       12 marks

  Total  38 questions   80 Marks

Follow if you like, depends on you

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Prove that the parallelogram circumscribing a circle is a rhombus. 

Since ABCD is a parallelogram circumscribed in a circle

$AB=CD........\left(1\right)$

$BC=AD........\left(2\right)$

$DR=DS$ (Tangents on the circle from same point $D$)

$CR=CQ$(Tangent on the circle from same point $C$)

$BP=BQ$ (Tangent on the circle from same point $B$ )

$AP=AS$ (Tangents on the circle from same point $A$)

Adding all these equations we get

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(DR+CR)+(BP+AP)=(CQ+BQ)+(DS+AS)$

$CD+AB=AD+BC$

Putting the value of equation $1$ and $2$ in the above equation we get

$2AB=2BC$

$AB=BC...........\left(3\right)$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$ we get

$AB=BC=CD=DA$

$\therefore ABCD$ is a Rhombus

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1-IMP 2025 CLASS 10 MATHS CIRCLE CHAPTER 10

 

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

Proof:

∠1 = 90° …(i)
∠2 = 90° …(ii)
∠1 = ∠2----by (1) &(2) (AIA) 
∴PQ || RS

Q 2. Prove that the length of the tangents drawn from an external point to a circle are equal.. 

Answer:

Given :

Let AP and BP be the two tangents drawn from external point P to the circle with centre O.

To Prove : AP = BP

Proof :

In $\Delta$ AOP and $\Delta$BOP

OA = OB (radii) 

$\angle OAP=\angle OBP={90}^{\circ }$ 

(Line drawn from from center to the tangent through the point of contact is perpendicular) 

OP = OP (common)

$\therefore \Delta AOP\cong \Delta BOP$ (R.H.S.) 

$\therefore$ AP = BP ( C P C T ) 

Therefore the length of the

 tangents drawn from an

 external point to a circle are

 equal.

Q 3. Statement: 

BPT (Basic Proportionality Theorem), 

If a line is drawn parallel to one of the  triangle to intersects the other two sides in distinct points, 

then the other two sides of 

the  triangle are divided  into the same ratio. 

Proof:

Given:

In ∆ABC, DE || BC and AB and AC are intersected by DE at points D and E respectively.

To prove:

AD / DB = AE / EC

Construction:

Join BE and CD.and

Draw:

EG⊥AB and DF⊥AC

Proof:

We know that

ar( Δ ADE) = 1 / 2 × AD × EG   

ar( Δ DBE) = 1 / 2 × DB × EG   

So

ar(Δ ADE) / ar(Δ DBE) =

 AD / DB   ------- (1)

Similarly,

ar(Δ ADE) / ar(Δ ECD) =

 AE / EC   ----------(2)

Now, 

Δ DBE and Δ ECD

are the on the same base DE and also between the same parallels i.e. DE and BC,

So

ar(Δ DBE) = ar(Δ ECD)  ---(3)

By (1), (2) , (3) 

AD / DB = AE / EC  

Hence  proved.

Q 4.

Prove √3 is a Irrational number.

Proof:

Let √3 be a rational number.

So, $\sqrt{3}=\frac{\mathrm{p\; \_\_\_\_\; (1)}}{q}$

On squaring both sides

$3=\frac{p^2}{q^2}$
$q^2=\frac{p^2}{3}$

$\Rightarrow \mathrm{3 is\; a\; factor\; of}$$p^2$

$\Rightarrow \mathrm{3\; is\; a\; factor\; of\; p.}$

Now, again let p = 3 c.

So, $\sqrt{3}=\frac{\mathrm{3\; c}}{q}$

On squaring both sides
$3=\frac{\mathrm{9\; c}{}^{}}{q^2}$
$q^2=\mathrm{3\; c}{}^{}$
$c^2=\frac{q^2}{3}$
$\Rightarrow \mathrm{3\; is\; factor\; of}$$q^2$
$\Rightarrow 3$ is a factor of q.

Here 3 is a common factor of p, q  both

 So p, q are not co-prime.

Therefore our assumption is wrong. 

√$\sqrt{3}$ is an irrational number.

Q 5.

Prove that the parallelogram circumscribing a circle is a rhombus. 

 ABCD is a parallelogram  , So opposite sides of a parallelogram are equal. 

$AB=CD.$

$BC=AD.-----++++(1)$

$DR=DS$ --------------(2) 

 (Tangents on the circle from same point $D$)

Similarly

$CR=C\mathrm{Q\; ----------(3)}$

AP = AS ------------(4) 

PB = BQ  -------++(5) 

Adding all these equations we get

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(DR+CR)+(BP+AP)=(DS+AS)$ +( BQ + CQ) 

$CD+AB=AD+BC$

But   AB =CD   &   BC=AD

     AB + AB = BC + BC

$2AB=2BC$

$\mathrm{Therefore\; A}B=BC-------(6)$

From equation $\left(1\right)$, $\mathrm{\&\; (6)}$we get

$AB=BC=CD=DA$

$\therefore ABCD$ is a Rhombus

Q 6. Find relationship between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). so

∴ AP = BP

         AP 2 = BP 2 

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

x2 – 4x + 4 + y2 – 10y + 25

                 =

x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y +29 =  6x - 14y +58 

⇒ -10x + 4y = 29

 is the required relation.

Q 6.

 

 Find relationship between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). so

∴ AP = BP

         AP 2 = BP 2 

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

x2 – 4x + 4 + y2 – 10y + 25

                 =

x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y +29 =  6x - 14y +58 

⇒ -10x + 4y = 29

 is the required relation.

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