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

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Tangets from external point to the circle are equal

 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.

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Class 10 BPT Basic Proportionality Theorem

 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.

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Class 10 Board Exam

 

Is enough time left for Board Exam preparation? 

My answer is yes. Time is enough even for those who did not start preparation till today, till date. 

If you want to perform better then you have enough time. If you are careless and not serious enough, you don't have any wish regarding the Board Exam, then you don't have any time. 

First Plan

First plan your Study Plan, think about your target and resources you have, then prepare your Time Table and adhere tobit. 

Priority Plan

Make a list of the subjects you are good at and the subjects you thik you are not good enough at. It means you will have to give extra time for those subjects where you want to improve. 

After selecting the subjects you will have to choose the topic you feel comfortable with . First prepare that topic entirely.. After that focus on some other difficult topics, to enhance your knowledge and increase your marks. 

Follow The Plan

Generally it is seen that after following some days, one stops following the plan and retreat later. So don't commit this mistake. 

Remind Yourself Daily

It is your duty towards yourself to remind yourself about your aim and target. 

This what I believe in , If you agree then try this, I think it will be fruitful for you. 

Thanks. 

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Class 10 -Prove √3 is a Irrational number.

 

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.

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