CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Given: ABCD is a parallelogram whose diagonals are AC and BD.
To prove: AB² + BC² + CD² + DA² = AC² + BD²
Construction: Draw AM {tex} \bot {/tex} DC and BN {tex} \bot {/tex} D(Produced)

Proof: In right triangle AMD and BNC.
AD = BC ..............Opp.sides of a ||gm
AM = BN ............Both are altitudes of the same parallelogram to the same base
{tex}\therefore {/tex} {tex}\triangle {/tex} AMD {tex}\cong{/tex} {tex}\triangle {/tex} BNC ................RHS congruence criterion
{tex}\therefore {/tex} MD = NC .........(1).........CPCT
In right triangle BND,
{tex}\because {/tex} {tex}\angle{/tex} N=90°
{tex}\therefore {/tex} BD²= BN²+ DN² .............By Pythagoras theorem
= BN²+ (DC + CN)²
= BN²+ DC²+ CN²+ 2DC.CN
= (BN²+ CN²) + CN² + 2DC.CN
= BC²+ DC²+ 2DC.CN ..........(2)
In right triangle BNC with {tex}\angle{/tex}N = 90^o
BN²+CN² = BC² ......By Pythagoras theorem
In right triangle AMC
{tex}\because {/tex} {tex}\angle{/tex} M=90^o
{tex}\therefore {/tex} AC² = AM²+ MC²
= AM² = (DC - DM)²
= AM²+ DC²+ DM²- 2DC.DM
= (AM²+ DC²) + DC² - 2DC.DM
= AD²+ DC²- 2DC.DM
{tex}\because {/tex} In right triangle AMD with {tex}\angle{/tex} M=90°
AD²= AM² + DM² ..........[By Pythagoras theorem]
= AD²+ AB² - DC.CN .......From(1)
Adding (3) and (2) ,we get
AC²+ BD²= (AD²+ AB) + (BC²+ DC) = AB²+ BC²+ BC²+ CD²+ DA²

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If we have 6 class intervals then how we choose modal class

Posted by Kartik Dhiman 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Khushi Yadav 7 years, 9 months ago

Class interval that will have highest frequency will be modal class

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Plz give full table of sin cos tan cot sec cosec

Posted by Sachin Dayal 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Rutuparnna Mangal 7 years, 9 months ago

No need to know. Your teacher will teach you later.

1Thank You

Keshav Bindal 7 years, 9 months ago

See in NCERT or I have written below of sin cos tan, change there numerator with there denominator you will get value of cosec sec cot

1Thank You

Sachin Dayal 7 years, 9 months ago

Yet i have not recieved my books because i have new entered in 10th

1Thank You

Shakshi Singh 7 years, 9 months ago

See ncert book

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3 sample papers will be enough

Posted by Amanpreet Singh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Ayush Kumar Singh 7 years, 9 months ago

No at least 10

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Amanpreet Singh 7 years, 9 months ago

so how many

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Anand Kumar Thakur 7 years, 9 months ago

2Thank You

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phythagoras theorem kesa proof hota ha

Posted by Tushar Sharma 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Rutuparnna Mangal 7 years, 9 months ago

Same on you . Book nahi hai kya?

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Anand Kumar Thakur 7 years, 9 months ago

Book mein dekh le

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Plz give table of sin cos tan

Posted by Sachin Dayal 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Keshav Bindal 7 years, 9 months ago

tan0 = 0 & tan90 = un defined

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Keshav Bindal 7 years, 9 months ago

Tan0 = un defined , tan30 = 1/√3 , tan45 = 1 , tan60 = √3 , tan90 = 0

1Thank You

Keshav Bindal 7 years, 9 months ago

Cos0 = 1 , cos30 = (√3)/2 , cos45 = 1/√2 , cos60 = 1/2 , cos90 = 0

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Keshav Bindal 7 years, 9 months ago

Sin0 = 0 , sin30 = 1/2 , sin45 = 1/√2 , sin60 = (√3)/2 , sin90 = 1

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in an equilateral triangle ABC.D is a point on side BC such that BD=1/3BC.prove that 9AD2=7AB2

Posted by Harry Potter 7 years, 9 months ago

CBSE > Class 10 > Mathematics

7 answers

Nitesh Sharma 7 years, 9 months ago

Given :- An equilateral triangle ABC and D be a point on BC such that BD= 1/3 BC. To prove :- 9AD^2 = 7AB^2 Construction :- Draw AE Perpendicular to BC and Join AD. Proof :- Triangle ABC is an equilateral triangle and AE Perpendicular to BC Therefore BE = EC Thus, we have BD = 1/3 BC and DC = 2/3 BC and BE = EC = 1/3BC In Triangle AEB AE^2 + BE^2 = AB^2 [Using Pythagoras Theorem] AE^2 = AB^2 - BE^2 AD^2 - DE^2 = AB^2 - BE^2 ['.' In Triangle AED, AD^2 = AE^2 + DE^2] AD^2 = AB^2 - BE^2 + DE^2 AD^2 = AB^2 - (1/2 BC)^2 + ( BE - BD )^2 AD^2 = AN^2 - 1/4 BC ^2 + (1/2 BC - 1/3 BC )^2 AD^2 = AB^2 - 1/4 BC^2 + ( BC/6 )^2 AD^2 = AB^2 - BC^2 (1/4 - 1/36 ) AD^2 = AB^2 - BC^2 (8/36) 9AD^2 = 9AB^2 - 2BC^2 9AD^2 = 9AB^2 - 2AB^2 [ '.' AB = BC ] 9AD^2 = 7 AB^2

3Thank You

Harry Potter 7 years, 9 months ago

i know the shortest trick. i was just knowing how many here know the short ans??????

1Thank You

Anand Kumar Thakur 7 years, 9 months ago

This is too long

3Thank You

Shikhar Yadav 7 years, 9 months ago

1Thank You

Harry Potter 7 years, 9 months ago

in RD it is too long.don't you have any short sol.

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Shikhar Yadav 7 years, 9 months ago

Given in Rd Sharma

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Amir Sheikh 7 years, 9 months ago

Hi siya

1Thank You

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Kitne chapters Ho gye hai sabke?

Posted by Jasmeen Kaur 7 years, 9 months ago

CBSE > Class 10 > Mathematics

7 answers

Ayush Kumar Singh 7 years, 9 months ago

I am mathematics topper from class 10 do you have any doubt

0Thank You

Harshit Joshi 7 years, 9 months ago

I know all but I have revised 7 yet

2Thank You

Khushi Yadav 7 years, 9 months ago

All?

2Thank You

Jasmeen Kaur 7 years, 9 months ago

???

1Thank You

Amir Sheikh 7 years, 9 months ago

5 chap

1Thank You

Samaira Sen 7 years, 9 months ago

All

0Thank You

Akash Raj 7 years, 9 months ago

Kon hai

3Thank You

ANSWER

exercise 12.1 Q no. 3 anyone please i need only hint

Posted by Amanpreet Singh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Amanpreet Singh 7 years, 9 months ago

thanks a lot

0Thank You

Keshav Bindal 7 years, 9 months ago

Area of 1st region gold = πr^2 = 22/7 × 10.5 × 10.5 & Area of 2nd region red = π(R^2 - r^2) = 22/7 ×(21^2 - 10.5^2) & Area of 3rd region blue = π(R^2 - r^2) = 22/7 ×(31.5^2 - 21^2)

1Thank You

Jasmeen Kaur 7 years, 9 months ago

Radius of gold =10.5 Area of gold = 22/7 × (10.5)^2 Radius of red =10.5+10.5 Area of red =22/7×(21)^2 - area of gold Similarly add 10.5 in radius for next and for its area 22/7×(21+10.5)^2 - 22/7×(21)^2

3Thank You

ANSWER

Construct a triangle whose sides are 3 and 8

Posted by Namita Shringi 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Ayush Kumar Singh 7 years, 9 months ago

Sorry for reply ????

0Thank You

Ayush Kumar Singh 7 years, 9 months ago

Leave that????

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ANSWER

Yaar circles lesson samja do ???

Posted by Gautm S 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

P A 7 years, 9 months ago

29 ko milna Ab smjh ke ky kareg ?

1Thank You

ANSWER

3×5×7×8×6×+8+6+8_4×9 is a composite number explain

Posted by Neha S 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

♥️Bhavika Naik?? 7 years, 9 months ago

When u will solve this u will understand that the no. Has more than 2 factors so it is composite

1Thank You

Akash Raj 7 years, 9 months ago

Kon

1Thank You

ANSWER

The sum of digits of a two digits number is 9 Also nine times this number is twice the number obtained by reversing the order of the digits. Find the number

Posted by Aditya Kumar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Harry Potter 7 years, 9 months ago

the no. is 18

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ANSWER

(Polinomials) imp questionssss

Posted by Chinchu Rajesh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

4s²-4s+1

Posted by Akshat Jain 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Keshav Bindal 7 years, 9 months ago

4s^2 - 2s - 2s + 1 = 2s(2s-1)-1(2s-1) = (2s-1)(2s-1) Therefore s = 1/2

1Thank You

Harry Potter 7 years, 9 months ago

1/2,1/2

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Ch 13th ke example 4 mai tsa of bird bath lena hai toh fir csa of both cylinder and hemisphere kyu lia hai? Plz ans

Posted by ♥️Bhavika Naik?? 7 years, 9 months ago

CBSE > Class 10 > Mathematics

8 answers

Keshav Bindal 7 years, 9 months ago

And if has material but that base was joined with any other fig. base, both the base area was not calculated

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Keshav Bindal 7 years, 9 months ago

To know when in any question base was calculated or not, saw the diagram and knew that if there was any material on base, only if there was any material when it's area was calculated

1Thank You

Keshav Bindal 7 years, 9 months ago

Because there is no material at base of hemisphere

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♥️Bhavika Naik?? 7 years, 9 months ago

Samja nai ??

1Thank You

Keshav Bindal 7 years, 9 months ago

But base was jointed and it was not visible

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♥️Bhavika Naik?? 7 years, 9 months ago

Ha par base include kyu nai kiya......atlast container mai base toh hota hai na