CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

Proof: In ∆ABC AD bisects CB,So BD=CD=10cm AD=15cm AC=12cm BD=BD+CD=10+10=20cm By Pythagoras Therom AB²=BC²-AC² AB²=20²-12² AB²=400-144 AB²=256 AB=√256 AB=16cm Hence,proved😂😂😂🤣

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2+2=5 Prove it???? 🤔🤔🤔🤨🤨

Posted by Ashish Singh 2 years, 3 months ago

CBSE > Class 09 > Mathematics

5 answers

Sudhir Kumar 2 years, 3 months ago

We can start from here, 4:4=5:5 4(1:1)=5(1:1) 4=5 2+2=4 Hence, 2+2=5

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Subankar Sikdar 2 years, 3 months ago

2+2 = 4

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Ayush Kumar Jha 2 years, 3 months ago

0=0 20-20=25-25 4(5-5)=5(5-5) [ both eliminating 5-5] 4=5 We can write 2+2=4 2+2=5

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Rahul ._. 2 years, 3 months ago

2+2=4 1+1+1+1=2+1+1 Talking 1 common (1+1)(1+1)=1(2+1)+1 2+2=5

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Chanchal Choudhary 2 years, 3 months ago

2+2 =4

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Why is maths racist?

Posted by Arihan Barolia 2 years, 3 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

Expand (89999)³

Posted by Debayan Kar 2 years, 3 months ago

CBSE > Class 09 > Mathematics

3 answers

Arihan Barolia 2 years, 3 months ago

(90000)^3 + (3*(90000)*(90000)*(-1)) + (3*(-1)*(-1)*(90000)) + (-1)^3 by the identity: (a+b)^3= a^3+ 3aab+ 3bba+ b^3 So the answer after solving is >>> 728,975,700,269,999

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Hardik Saluja 2 years, 3 months ago

728,975,700,269,999

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Balwant Yadav 2 years, 3 months ago

728,975,700,269,999

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Somu have 153 mangoes . He gave 97 mangoes his friend Ram. He gave 25 mangoes his brother . Somu's mother gave 99 mangoes to somu. How many mangoes did somu have ?

Posted by Debayan Kar 2 years, 3 months ago

CBSE > Class 09 > Mathematics

5 answers

Debayan Kar 2 years, 3 months ago

137 is wrong ok

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Debayan Kar 2 years, 3 months ago

Real answer is 130 mangoes

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Debayan Kar 2 years, 3 months ago

Ans 120 mangoes

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Balwant Yadav 2 years, 3 months ago

153-99-25=29 29+99=128

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Debayan Kar 2 years, 3 months ago

137 mangoes

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ANSWER

Expand (π)³

Posted by Debayan Kar 2 years, 3 months ago

CBSE > Class 09 > Mathematics

4 answers

Akshadeep Meshram 2 years, 3 months ago

π × π × π = π³

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Debayan Kar 2 years, 3 months ago

Balwant your answer is wrong

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Debayan Kar 2 years, 3 months ago

(22/7)³=10648/343

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Debayan Kar 2 years, 3 months ago

We know that π=22/7 or 3.1472.......

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ANSWER

The equation 2x+5y=7

Posted by Abhishek Rajput 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Balwant Yadav 2 years, 3 months ago

(1,1) (-4,3)

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Debayan Kar 2 years, 3 months ago

Ans (1,1) (3.5,0)(5,0) (0,2)

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10!×20/9!= ?

Posted by Debayan Kar 2 years, 3 months ago

CBSE > Class 09 > Mathematics

3 answers

Balwant Yadav 2 years, 3 months ago

10!=3628800 20=20 9!=362880 And solve 3628800×20÷362880=200

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Debayan Kar 2 years, 3 months ago

Balwant yadav solve this question also

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Debayan Kar 2 years, 3 months ago

Ans 200

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ANSWER

How to prove any questions

Posted by Anshika Gupta 2 years, 3 months ago

CBSE > Class 09 > Mathematics

5 answers

Akshadeep Meshram 2 years, 3 months ago

First understand the concept Then start with easy question then increase your difficulty level

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Balwant Yadav 2 years, 3 months ago

Any questions prove Our eligibility

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Debayan Kar 2 years, 3 months ago

And find out key words from questions

2Thank You

Debayan Kar 2 years, 3 months ago

Circles, quadrilaterals,

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Debayan Kar 2 years, 3 months ago

Understand all questions with read properly

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Class 9 maths 8.2

Posted by Shivang Sharma 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Pooja Arya 2 years, 3 months ago

Class 9 8.2 is about mid point theorem It says .. In a triangle, the line segment that joins the mid point of any two sides is parallel to the third side also half of it

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Show that diagonals of a square that quadrilateral PQRS is a rhombus

Posted by Charu Singh 2 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 2 months ago

Given: Let ABCD is a quadrilateral.
Let its diagonal AC and BD bisect each other at right angle at point O.

$\therefore$ OA = OC, OB = OD
And $\angle$ AOB = $\angle$ BOC = $\angle$ COD = $\angle$ AOD = ${90^ \circ }$
To prove: ABCD is a rhombus.
Proof: In $\triangle$ AOD and $\triangle$ BOC,
OA = OC [Given]
$\angle$ AOD = $\angle$ BOC [Given]
OB = OD [Given]
$\therefore$ $\triangle$ AOD $\cong$ $\triangle$ COB [By SAS congruency]
$\Rightarrow$ AD = CB [By C.P.C.T.]……….(i)
Again, In $\triangle$ AOB and $\triangle$ COD,
OA = OC [Given]
$\angle$ AOB = $\angle$ COD [Given]
OB = OD [Given]
$\therefore$ $\triangle$ AOB $\cong$ $\triangle$ COD [By SAS congruency]
$\Rightarrow$ AB = CD [By C.P.C.T.]……….(ii)
Now In $\triangle$ AOB and $\triangle$ BOC,
OA = OC [Given]
$\angle$ AOB = $\angle$ BOC [Given]
OB = OB [Common]
$\therefore$ $\triangle$ AOB $\cong$ $\triangle$ COB [By SAS congruency]
$\Rightarrow$ AB = BC [By C.P.C.T.]……….(iii)
From eq. (i), (ii) and (iii),
AD = BC = CD = AB
And the diagonals of quadrilateral ABCD bisect each other at right angle.
Therefore, ABCD is a rhombus.

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A circular path with center A mathav walk at uniform speed of 0.5 m/s from gate p and reach the center of park in 150 sec what is the streat distance between the center of the path and gate M

Posted by Jaini Gandhi 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Akshadeep Meshram 2 years, 3 months ago

75m

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The value of 50- root 32×root50 is

Posted by Bhavani S. K 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Debayan Kar 2 years, 3 months ago

50-√32×√50 =50-4√2×5√2 =50-20×√2×√2 =50-20×2 =50-40 =10

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Akshadeep Meshram 2 years, 3 months ago

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ANSWER

ABCD is a rhombus and P,Q,R,S are the mid point of the sides AB,BC,CD,DA respectively. Show that the quadrilateral PQRS is a rectangle .

Posted by Sasmita Mallick 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Sehajpreet Kaur 2 years, 3 months ago

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = 1 2 AC ( Using mid-point theorem) …..(1) In Δ ADC, R and S are the mid - points of CD and AD respectively. ∴ RS || AC and RS = 1 2 AC ( Using mid-point theorem) …..(2) From equation (1) and (2) , we obtain PQ || RS and PQ = RS Since in quadrilateral PQRs, one pair of opposite sides is equal and parallel to each other , it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O. In Quadrilateral OMQN, MQ || ON ( ∵ PQ || AC) QN || OM ( ∵ QR || BD) Therefore, OMQN is a parallelogram. ∴ ∠ M Q N = ∠ N O M and ∠ P Q R = ∠ N O M However, ∠ N O M = 90 ∘ (Diagonals of a rhombus are perpendicular to each other) ∴ ∠ P Q R = 90 ∘ Clearly , PQRS is a parallelogram having one of its interior angles as 90 ∘ Hence, PQRS is a rectangle.

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Factorise : √3x²+4x+3x

Posted by Riya Singh 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Akshadeep Meshram 2 years, 3 months ago

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Mohak Yadav 2 years, 3 months ago

Question is wrong

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Which irrational number does not lie between 2/7 & 3/5

Posted by Lavesh Sahu 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Shivang Sharma 2 years, 3 months ago

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Lavesh Sahu 2 years, 3 months ago

17/35

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Solve if you are intelligent! In triangle ABC, AD is the median. AC = 12cm, DC = 10cm, AD = 15cm. Find the length AB. Hint:- AD bisects the side BC

Posted by Akshadeep Meshram 2 years, 3 months ago

CBSE > Class 09 > Mathematics

3 answers

Fatima Zafar 2 years, 3 months ago

Answer 5√13

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Akshadeep Meshram 2 years, 3 months ago

Your answer is wrong aman kumar

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Aman Kumar 2 years, 3 months ago

18cm

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Represent underroot 6 on number line

Posted by Anjali Mattu 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Debayan Kar 2 years, 3 months ago

1. Material required are ruler , notebook, pencil with compass and eraser

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Abhishek Ranjan 2 years, 3 months ago

Plz give m answer

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√21

Posted by Ayush Chaurasiya 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Ansh Raj 2 years, 3 months ago

It is irrational number.

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a³+b³= a+b (a²-ab+b²)

Posted by Aditya Raj 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Shubh Goyal 2 years, 3 months ago

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Two Parallel beams EG and EH are supported using two poles AC and BD If the beams are also parallel to ground ON.If C is 70. D is 60. Find hyd Find btg Find yxc SFX Ase

Posted by Priyanshu Aryan 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 3 months ago

Angle HYD = 120°

Angle BTG = 60°

Angle YXC = 70°

Angle SXF = 70°

Angle ASE = 70°

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x²-y² = expand

Posted by Garvit Chawla 2 years, 3 months ago

CBSE > Class 09 > Mathematics

5 answers

Shivang Sharma 2 years, 3 months ago

Don't post personal information, mobile numbers and other details.
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Shubh Goyal 2 years, 3 months ago

(x+y)(x-y)

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Garvit Sharma 2 years, 3 months ago

(x-y)whole squre

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Smriti Gupta 2 years, 3 months ago

(x+y)(x-y)

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Garvit Chawla 2 years, 3 months ago

(x+y) (x-y)

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Express y in term of x in the equation 5y-3x-10=0

Posted by Anushka Awasthi 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Preeti Dabral 2 years, 3 months ago

$y = \frac{{3x + 10}}{5}$

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Shubh Goyal 2 years, 3 months ago

y=3x+10/5

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Find the radius of sphere whose surface area is 154 cm square.

Posted by Hetvi Patel 2 years, 3 months ago

CBSE > Class 09 > Mathematics

4 answers

Preeti Dabral 2 years, 3 months ago

Let the radius of the sphere be r cm.
Surface area = 154 cm²
⇒ $4\pi r^2$ = 154
⇒ $4\times{22\over7}\times r^2=154$
⇒ $r^2={{154\times7}\over4\times22}={49\over4}$
⇒ r = ${\sqrt{49\over4}}={7\over2}$
⇒ r = 3.5 cm
∴ the radius of the sphere is 3.5 cm.

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Smriti Gupta 2 years, 3 months ago

3.5

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Siddhi Gupta 2 years, 3 months ago

S.a=154 4πr²=154 r²=154 x 7\4 x 22 r=√49/7 r=7/2 r=3.5cm Radius of sphere is 3.5 cm

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Shubh Goyal 2 years, 3 months ago

√10.5=3.24

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give four solutions x+y=13 ...

Posted by Shivanand Tiwari 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Smriti Gupta 2 years, 3 months ago

(0,13),(11,2),(10,3),(12,1)

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ANSWER

Using suitable identity evaluate 101 multiply 102

Posted by Nikunj Aggarwal 2 years, 3 months ago

CBSE > Class 09 > Mathematics

5 answers

Smriti Gupta 2 years, 3 months ago

10302

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Shubh Goyal 2 years, 3 months ago

101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x 2 +(a+b)x+ab, we have (100+1)(100+2)=(100) 2 +(1+2)100+(1)(2) =10000+(3)100+2=10000+300+2 =10302

1Thank You

Shivanand Tiwari 2 years, 3 months ago

101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x 2 +(a+b)x+ab, we have (100+1)(100+2)=(100) 2 +(1+2)100+(1)(2) =10000+(3)100+2=10000+300+2 =10302

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Rudra . 2 years, 3 months ago

101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x2+(a+b)x+ab, we have (100+1)(100+2)=(100)2+(1+2)100+(1)(2) =10000+(3)100+2 =10000+300+2 =10302

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Rudra . 2 years, 3 months ago

101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x 2 +(a+b)x+ab, we have (100+1)(100+2)=(100) 2 +(1+2)100+(1)(2) =10000+(3)100+2=10000+300+2 =10302

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Base angles of an isosceles triangle are equal

Posted by Kunal Mankar 2 years, 3 months ago

CBSE > Class 09 > Mathematics