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To prove :
{tex}3 \left( A B ^ { 2 } + B C ^ { 2 } + C A ^ { 2 } \right) = 4 \left( A D ^ { 2 } + B E ^ { 2 } + C F ^ { 2 } \right){/tex}

In triangle sum of squares of any two sides is equal to twice the square of half of the third side, together with twice the square of median bisecting it.
If AD is the median
{tex}A B ^ { 2 } + A C ^ { 2 } = 2 \left\{ A D ^ { 2 } + \frac { B C ^ { 2 } } { 4 } \right\}{/tex}
or, {tex}2 \left( A B ^ { 2 } + A C ^ { 2 } \right) = 4 A D ^ { 2 } + B C ^ { 2 }{/tex} ...(i)
Similarly by taking BE & CF as medians,
{tex}2 \left( A B ^ { 2 } + B C ^ { 2 } \right) = 4 B E ^ { 2 } + A C ^ { 2 }{/tex} ...(ii)
& {tex}2 \left( A C ^ { 2 } + B C ^ { 2 } \right) = 4 C F ^ { 2 } + A B ^ { 2 }{/tex} ...(iii)
By adding, (i), (ii) and (iii), we get
{tex}3 \left( A B ^ { 2 } + B C ^ { 2 } + A C ^ { 2 } \right) = 4 \left( A D ^ { 2 } + B E ^ { 2 } + C F ^ { 2 } \right){/tex}
Hence proved.

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What are the main real numbers

Posted by Shivam Jat 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 7 years, 4 months ago

Real Numbers:

All rational and all irrational number makes the collection of real number. It is denoted by the letter R
We can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line
The sum or difference of a rational number and an irrational number is an irrational number.
The product or division of a rational number with an irrational number is an irrational number.
This process of visualization of representing a decimal expansion on the number line is known as the process of successive magnification

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What are +,+=

Posted by Tanish Dhanotiya 7 years, 4 months ago

CBSE > Class 10 > Mathematics

4 answers

Nikki Kushwah 7 years, 4 months ago

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Ajay Singh 7 years, 4 months ago

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Kanika Singla 7 years, 4 months ago

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Neha Singh 7 years, 4 months ago

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All trignometry identities in class 10

Posted by Vaibhav Shukla 7 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Aryan Mishra 7 years, 4 months ago

1)sin² +cos² =1 ~ sin²= 1– cos² cos²= 1– sin² 2)1+ tan² = sec² ~ tan² = sec² – 1 1 = sec² – tan² 3)1+ cot² = cosec² ~ cot² = cosec² – 1 1 = cosec² – cot²

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State and prove pythagorous theorem

Posted by Anushka Bhatnagar 7 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sejal Barge 7 years, 4 months ago

a²+b²=c²

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out of two concentric circles ,the radius of outer circle is 5cm and the chord AC of length 8cm is a tangent to inner circle . find the radius of the inner circle

Posted by Manas Choudhari 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Since AC is a tangent to the inner circle.
{tex}\angle OBC = 90^\circ {/tex}
AC is a chord of the outer circle.
We know that, the perpendicular drawn from the centre to a chord of a circle, bisects the chord.
{tex}AC = 2BC \\ 8=2BC {/tex}
{tex}\Rightarrow{/tex} BC = 4 cm
In {tex}\Delta{/tex}OBC,
By Pythagoras theorem,
OC² = OB² + BC²
{tex}\Rightarrow{/tex} 5² = r² + 4²
{tex}\Rightarrow{/tex} r² = 5²- 4²
{tex}\Rightarrow{/tex} r² = 25 - 16
{tex}\Rightarrow{/tex} r² = 9 cm
{tex}\Rightarrow{/tex} r = 3 cm

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Show that one and only one of n;n+4;n+2 is divisible by 3

Posted by Kanha Sharma 7 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Shivay Brahmåñ 7 years, 4 months ago

Take n =3q So first n is divisible by 3 Both n+4;n+2is also divided by 3 but it leaves reminder and n does not leave any reminder so n is divisible by 3

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Syed Affan 7 years, 4 months ago

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{1+an²A\1+cot²A}= {1- tanA\cotA}²= tan ²A prove following identities

Posted by Aditya Raj 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

{tex}= \frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A } = \frac { 1 + \tan ^ { 2 } A } { 1 + \frac { 1 } { \tan ^ { 2 } A } } \cdot \because \cot A = \frac { 1 } { \tan A }{/tex}
{tex}= \frac { 1 + \tan ^ { 2 } A } { \frac { \tan ^ { 2 } A + 1 } { \tan ^ { 2 } A } } = \tan ^ { 2 } A \ldots \ldots ( 1 ){/tex}
{tex}\left( \frac { 1 - \tan A } { 1 - \cot A } \right) ^ { 2 } = \left( \frac { 1 - \tan A } { 1 - \frac { 1 } { \tan A } } \right) ^ { 2 }{/tex}
{tex}= \left\{ \frac { 1 - \tan A } { \left( \frac { \tan A - 1 } { \tan A } \right) } \right\} ^ { 2 } = ( - \tan A ) ^ { 2 } = \tan ^ { 2 } A{/tex} ....... (2)

(1) and (2) taken together given the result.

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Prove BPT . In detail?

Posted by Sakshi Mohan 7 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Elsa B 7 years, 4 months ago

You can find it in ncert textbook page no 124.

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Archit Vijay 7 years, 4 months ago

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O/O =2

Posted by Razz Yadav 7 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Kirankumar Jambenal 7 years, 4 months ago

YouTube

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Priyanshu Singh 7 years, 4 months ago

10+10÷(10)square +(10)square

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2+3=0 prove thid

Posted by Sahil Jain 7 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Priyanshu Singh 7 years, 4 months ago

(-2)+(-1)+3=0

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2/10=2

Posted by Sonu Prajapati 7 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Priyanshu Singh 7 years, 4 months ago

On multiplying numerator by 10 and denominater by 1=2

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Prove that the diagonals of trapezium intersects each other in the same ratio?

Posted by Rajdev Attri 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given in the quadrilateral ABCD
{tex}\frac{AO}{BO}=\frac{CO}{DO}{/tex}
or, {tex}\frac { A O } { C O } = \frac { B O } { D O }{/tex} ...(i)
Draw EO {tex}\parallel{/tex} AB on

In {tex}\triangle A B D,{/tex} EO {tex}\parallel{/tex} AB (By construction)
{tex}\therefore {/tex} {tex}\frac { A E } { E D } = \frac { B O } { D O }{/tex} (By BPT)...(ii)
From (i) and (ii) we get
{tex}\frac{AO}{CO}=\frac{AE}{ED}{/tex}
Hence by converse of BPT in {tex}\triangle{/tex}ADC
{tex}EO\|CD{/tex}
But {tex}EO \|AB {/tex}
So {tex}AB\|CD {/tex}
Therefore ABCD is a trapezium

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Prove the sss similarity criterion?

Posted by Rajdev Attri 7 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

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If 3sin0+5cos0=5 prove that 5sin0-3cos0=+3 and -3

Posted by Shubham Joshi 7 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Priyanshu Singh 7 years, 4 months ago

R.s.agarwal ka page no.213

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Some applications of trigonometry

Posted by Arjun Thakur 7 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Laawanya Sivakumar 7 years, 4 months ago

Comedy

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Devansh Gupta 7 years, 4 months ago

Calculator??

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Given tan A=4/3 Find the other trigonometric ratios of the angel A

Posted by Srushti Kunkolienkar 7 years, 4 months ago

CBSE > Class 10 > Mathematics

3 answers

Devansh Gupta 7 years, 4 months ago

Abhishek ir answer is wrong

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Srushti Kunkolienkar 7 years, 4 months ago

but √25 is 5 no ?

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Abhishek Varma 7 years, 4 months ago

AB=4,BC=3,AC=25(by pythagoras theorem) SinA=3/25,cosA=4/25,cotA=3/4,secA=25/4,cosecA=25/3.

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Prove √ 2 is irrational

Posted by Anju Gupta 7 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Kannu Kranti Yadav 7 years, 4 months ago

Let , us assume that √2 is rational number then it can be written in the form of a/b where b≠0 and (a,b are co prime numbers). So, √2=a/b, now squaring on both sides gives 2=a²/b²... 2b²=a² so we can say thay a² is divisible by 2 , therefore a is also divisible by 2. Then take a=2c, now putting value ofa and squaring on both sides giv gives us ...... 2b²=4c² => b²=2c² so b² is divisible by 2 therefore b is also divisible by 2. Hence our assumption is wrong , they have other common factor as 2. There fore √2 is irrational.hence, proved

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Prove that ratio of area of two similar triangle is equal to ratio of square of angle bisector

Posted by Kushagra Jaiswal 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given: {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}DEF
AM is the bisector of {tex}\angle{/tex} BAC and is the corresponding bisector of {tex}\angle{/tex} EDF
To prove:{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}
Proof: {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{B^2}}}{{D{E^2}}}{/tex}.......(i) [Area theorem]
Now {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}DEF
{tex}\Rightarrow {/tex} {tex}\angle{/tex} A = {tex}\angle{/tex} D {tex}\Rightarrow {/tex} {tex}\frac{1}{2}\angle A = \frac{1}{2}\angle D{/tex}
{tex}\Rightarrow {/tex} {tex}\angle{/tex}BAM = {tex}\angle{/tex}EDN
In {tex}\triangle {/tex}ABM and DEN {tex}\angle{/tex} B = {tex}\angle{/tex} E and {tex}\angle{/tex} BAM= {tex}\angle{/tex}EDN
{tex}\Rightarrow {/tex} {tex}\triangle {/tex}ABM {tex} \sim {/tex} {tex}\triangle {/tex}DEN
{tex}\Rightarrow {/tex} {tex}\frac{{AB}}{{DE}} = \frac{{AM}}{{DN}} \Rightarrow \frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}...................(ii)
{tex}\Rightarrow {/tex} {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex} [From (i) and (ii)] Proved.