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

First find length of each sides of ∆ Let A( 4 , 0) B(0, 0) and C (0 , 3) use distance formula, AB =√(4²+0) =4 BC= √(0+3²) = 3 CA =√(4²+3²) =5 now , perimeter of ∆ = 3 + 4 + 5= 12 unit........, hope it will help you.....??

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Puja Sahoo? 6 years, 10 months ago

12 unit

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Can you prove collinearity of three points by Section formula?explaine with an example?

Posted by Gurudhan Sathish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Gurudhan Sathish 6 years, 10 months ago

Yes mam,You are correct but by taking m:n u can prove mam

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Puja Sahoo? 6 years, 10 months ago

We need to prove the points (3,-2),(5,2) and(8,8) are collinear. A=(3,-2) B=(5,2) C=(8,8) Let The points B divides AC in the ratio of k+1 Then the coordinates will be, (8k+2/k+1,8k+5/k+1) Coordinates of B are (4,6) Comparing we get, 8k+2/k+1=4 and 8k+5/k+1=6 8k+2=4k+4 and 8k+5=6k+6 4k=2 and 2k=1 k=2/4 and k=1/2 k=1/2 and k=1/2 Value of k is same in both x and y directions.Therefore Points A,B,C are collinears. hope it helps u....

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How to derive divisibility rules of 7,13,17,19,23,27,31,37,41,43......etc?

Posted by Gurudhan Sathish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gurudhan Sathish 6 years, 10 months ago

This can derive by the concept of Osculator.

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Can you apply mean by Step deviation method of grouped data with un equal class sizes ?explaine with an example?

Posted by Gurudhan Sathish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gurudhan Sathish 6 years, 10 months ago

Yes...We can find ,by setting h value maximum divisor of all deviation.

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What is the draw back of Remainder and Factors theorem?How to over come it?

Posted by Gurudhan Sathish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gurudhan Sathish 6 years, 10 months ago

If the divisor is not linear polynomial we are unable to apply R.T and F.T ,so we can convert given polynomial into linear factors

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What is the draw back of Basic propornality theorem ?How it is Overcome?

Posted by Gurudhan Sathish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Gurudhan Sathish 6 years, 10 months ago

BPT does not explain lengths of parallel sides in a triangle.To over come this we have concept of Criteria for similarity b/w Triangles

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Puja Sahoo? 6 years, 10 months ago

Kya, BPT theorem ka drawback bhi hai.....?

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ANSWER

A parallelogram has vortices (1,4),(7,11),(k,4)and(1,-3).

Posted by Ispita Biswal 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

D.J Alok 6 years, 10 months ago

Complete your question.....

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ANSWER

Sin6A -cos6A

Posted by Vashvi Kashyap 5 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

(X+1) (x+2) (x+3) (x+4)-8=0

Posted by Krishna Bansal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Maths sare chapter kitne marks ke aate hai

Posted by Kapil Phalswal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Chirayu . 6 years, 10 months ago

Ch- 1 (6 marks) 3 question total Ch- 2,3,4,5 (20marks) 8 question total Ch-7 (6 marks) 3 queston total Ch- 8,9 (12 marks) 4 question total Ch- 6,10,11 (15 marks) 5 question total Ch 12,13 (10 marks) 3 question total Ch- 14, 15 (11 marks) 4 question total

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25x^4-20x^2+4=0

Posted by Krishna Bansal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Paulo Dybala 6 years, 10 months ago

25x^4 - 20x^2 + 4=0 25x^4 - 10x^2 - 10x^2 +4 =0 5x^2(5x^2 - 2) - 2(5x^2 - 2) = 0 (5x^2 -2 ) , (5x^2 - 2 ) = 0 x^2 = 2/5 , x^2 = 2/5 x = √2/5 ,. x= √2/5

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Z^4-13z^2+36=0

Posted by Krishna Bansal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Yogita Ingle 6 years, 10 months ago

(z -3)(z + 3) (z - 2)(z + 2) = 0
z = 3,-3,2 or -2

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Yogita Ingle 6 years, 10 months ago

z⁴ - 9z² - 4z² + 36 = 0
z² (z² - 9) - 4(z² - 9) = 0
(z² - 9) (z² - 4) = 0
(z - 9) (z + 9) (z - 4)(z + 4) = 0
z = 9,-9, 4 and -4

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Z^4-10z^2+9=0

Posted by Krishna Bansal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 6 years, 10 months ago

z4 - 10z2 + 9 = 0
z4 - 9z2 - 1z2 + 9 = 0
z2 (z2 - 9) -1(Z2 - 9) = 0
(z2 - 9) (z2 - 1) = 0
(z -3) (z+3) (z - 1) (z + 1) = 0
z = -3, 3, 1, -1

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If x=3 in one root of the quadrate equation xsquare_2kx_6=0 then find the value of k

Posted by Sourobh Shende 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Anshika / U Can Call Me Alien ???? 6 years, 10 months ago

Hey... I think this can be ur answer!! Polynomial - x2 - 2kx - 6 =0 P (3)=(3)2 - 2(3)k =6 =9 - 6k =6 =-6k =-3 =k =1/2 Hope it helps you dear ☺️☺️

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Dhirendra Behera 6 years, 10 months ago

The value of k will be 1

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If in an A.P. the sum of m terms is equal to n and the sum sum of n terms is equal to m, then proove that the sum of (m+n) term is -(m+n).

Posted by Sõhìt Páñdéy 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let a be the first term and d be the common difference of the given A.P. Then,
S_m= n
{tex} \Rightarrow \quad \frac { m } { 2 } \{ 2 a + ( m - 1 ) d \} = n{/tex}
{tex}\Rightarrow{/tex} 2am + m (m - 1) d = 2n ...(i)
and, S_n = m
{tex}\Rightarrow \quad \frac { n } { 2 } \{ 2 a + ( n - 1 ) d \} = m{/tex}
{tex}\Rightarrow{/tex} 2an + n (n -1)d = 2m ...(ii)
Subtracting equation (ii) from equation (i), we get
2a (m - n) + {m (m -1) - n (n - 1)} d = 2n - 2m
{tex}\Rightarrow{/tex} 2a (m - n) + {(m² - n²) - (m - n)} d = -2 (m - n)
{tex}\Rightarrow{/tex} 2a + (m + n -1) d = - 2 [On dividing both sides by (m - n)] ...(iii)
Now, {tex}S _ { m + n } = \frac { m + n } { 2 } \{ 2 a + ( m + n - 1 ) d \}{/tex}
{tex}\Rightarrow \quad S _ { m + n } = \frac { ( m + n ) } { 2 } ( - 2 ){/tex} [Using (iii)]
{tex}\Rightarrow{/tex} S_m+n= - (m + n)

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Prove that in a right angles triangle square of the hypotenuse is equal to sum of the squares of other two sides.

Posted by Nirbhay Mehra Ishu 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given : {tex}\triangle \mathrm{ABC}{/tex} is right angle at B
To prove: {tex}A C^{2}=A B^{2}+B C^{2}{/tex}

Construction: Draw {tex}B D \perp A C{/tex}

Proof:In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}ABD

{tex}\angle{/tex}ABC={tex}\angle{/tex}ADB=90

{tex}\angle{/tex}A is common

{tex}\angle{/tex}ABC{tex}\sim{/tex}{tex}\angle{/tex}ADB
{tex}\Rightarrow \frac{A D}{A B}=\frac{A B}{A C}{/tex}
{tex}\Rightarrow{/tex} AD. AC = AB² ...(i)
Similarly {tex}\Delta \mathrm{BDC} \sim \Delta \mathrm{ABC}{/tex}
{tex}\Rightarrow \frac{C D}{B C}=\frac{B C}{A C}{/tex}
{tex}\Rightarrow{/tex} CD.AC = BC² ...(ii)
Adding (i) and (ii)
AD.AC + CD. AC= AB² + BC²
AC(AD + CD) = AB² + BC²
AC {tex}\times{/tex} AC = AB² + BC²
AC² = AB² + BC²
Hence Proved

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Student observe the following spinner what is the probability of obtaining red colour

Posted by Furkan Mohmmad 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

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A sector is cut from a circle of radius 21 cm. The angle of a sector is 150. Find the length of the arc and area of the sector.

Posted by Abhishek Kumar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

4 answers

Dhirendra Behera 6 years, 10 months ago

Length of the sector will be 55cm. Area of the sector will be 577.5cm square.

1Thank You

Gaurav Seth 6 years, 10 months ago

Here, we have
r = 21 cm and ө = 120°
Let OACBO is the given sector and is the given length of arc, then

(i) The length of the arc (ACB)

(ii) Area of sector (OACBO)

4Thank You

Krishna Goyal 6 years, 10 months ago

Area of sector =577.5

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Krishna Goyal 6 years, 10 months ago

Length of sector =55.00004

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Can (x-2) be the remainder on division of a polynomial p(x) by (2x+3) ? Justify your answer.

Posted by Vinay Sahu 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

No , (x - 2) cannot be the remainder on division of a polynomial p(x) by(2x +3) Because degree of remainder must be less than degree of divisor. Here degree of x-2 and 2x-3 is same that is 1.

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A tent consist of a frustum of a cone capped by a cone if the radii of the ends of the frustum be 13 and 7 m the height of the frustum be 8m and slant height of the conical cap be 12m find the canvas required for the tent Please solve this question step by a step

Posted by Pratibha Sahu 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Radius of base of frustum = 13 m (R)
Radius of base of conical cap = 7 m (r)
Slant height of conical cap = 12 m (l)
height of frustum = 8 m (h)
Slant height of frustum
{tex}L = \sqrt { n ^ { 2 } + ( R - r ) ^ { 2 } }{/tex}
{tex}\Rightarrow \quad L = \sqrt { ( 8 ) ^ { 2 } + ( 13 - 7 ) ^ { 2 } }{/tex}
{tex}\Rightarrow \quad L = \sqrt { 64 + 36 } = \sqrt { 100 } = 10 \mathrm { m }{/tex}
therefore, Canvas required for the tent = Curved Surface area of cone + Curved Surface area of the frustum
{tex}= \pi rl + \pi L ( R + r ){/tex}
{tex}= \frac { 22 } { 7 } \times 7 \times 12 + \frac { 22 } { 7 } \times 10 ( 13 + 7 ){/tex}
{tex}= \frac { 22 } { 7 } \times 84 + \frac { 22 } { 7 } \times 10 \times 20{/tex}
{tex}= \frac { 22 } { 7 } ( 84 + 200 ){/tex}
{tex}= \frac { 22 } { 7 } \times 284{/tex}
= 892.57 m²

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If three coins are tossed together then find the probability of getting two head

Posted by Jhalashya Choudhary 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Pratibha Sahu 6 years, 10 months ago

5/8

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Akshaya Chandrasekar 6 years, 10 months ago

Possible outcomes= 8 (H,H,H) (H,T,T) (T,T,T) (T,T,H) (T,H,T) (H,T,H) (H,H,T) (T,H,H) Probability of getting two heads = 4 (H,H,H) (H,H,T) (H,T,H) (T,H,H) Therefore, probability of getting two heads = 4÷8 = 1÷2

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Anshika / U Can Call Me Alien ???? 6 years, 10 months ago

1/2

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CosA+SinA=√2CosA then prove that CosA-SinA=√2SinA

Posted by Yogesh Kumar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Paulo Dybala 6 years, 10 months ago

From given eqn 1 SinA = √2 cosA - cos A SinA = cosA(√2 - 1) SinA /√2-1 = cosA Rationalising SinA × √2 + 1 ----------------------- = cosA √2 - 1 × √2 +1 √2 sinA + sinA. = CosA ------------------------- 2 - 1 √ 2 sinA =. CosA - SinA Hence proved?

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Nishihoney Lodha 6 years, 10 months ago

By squaring both sides: (cosA+sinA)2=(√2cosA)2 By adding (cosA +sinA)2 +( cosA -sinA)2----------->cos2A+sin2A+2cosAsinA+cos2A+sin2A-2cosAsinA---------->cos2A+sin2A+cos2A+sin2A------>1+1=2. Now,(cosA-sinA)2=2-(cosA+sinA)2-------->so, (cosA-sinA)2=2-2cos2A------>(cosA-sinA)2=2(1-cos2A)------>(cosA-sinA)2=2sin2A-------->(cosA-sinA)=√2sinA.... Hence,proved. (Here 2 is the power.)

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If secA + tanA = p, then find the value of cosecA

Posted by Jiya Rana 6 years, 10 months ago

CBSE > Class 10 > Mathematics