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There is no important questuons at all , you will trt to understand 1)how to construct a similar triangle in given ratio .2)how to cunstruct tangents of a circle.

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Find one rational and irarional no. Between √3 and √5.

Posted by Juhi Kumari 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Juhi Kumari 6 years, 11 months ago

Thanks

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Yogita Ingle 6 years, 11 months ago

√3 = 1.732......
√5 = 2.236.....
The numbers which can be expressed in the form of p/q, where q ≠ 0 are called rational numbers. A terminating decimal and a non-terminating non-repeating decimal can be represented as a rational number.
Therefore, a rational number between √3 and √5 is 2.
So, three irrational numbers between √3 and √5 are: 1.79877985647984123564........ , 2.0100100010000100001110001.......... , 2.1212121212123121234.......

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Solve for x :(a-b) x^2+(b-c)x+(c-a)=0

Posted by Santosh Lamani 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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The radii of the bases of two right circular solid cones of same height r1 and r2 respectively . The cones are melted and recasted into a solid sphere of radius R. Show that the height of each cone is given by h = 4R^3/r1^3+r2^3

Posted by Santosh Lamani 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let h be the height of each cone. Then,
Sum of the volumes of two cones = Volume of the sphere
{tex}\Rightarrow{/tex}{tex}\frac { 1 } { 3 } \pi r _ { 1 } ^ { 2 } h + \frac { 1 } { 3 } \pi r _ { 2 } ^ { 2 } h = \frac { 4 } { 3 } \pi R ^ { 3 }{/tex}
{tex}\Rightarrow{/tex} {tex}\left( r _ { 1 } ^ { 2 } + r _ { 2 } ^ { 2 } \right) h = 4 R ^ { 3 }{/tex}
{tex}\Rightarrow{/tex} {tex}h = \frac { 4 R ^ { 3 } } { r _ { 1 } ^ { 2 } + r _ { 2 } ^ { 2 } }{/tex}

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If tanA = ntanB and sinA = msinB , then prove that cos square A = m square - 1 upon n square + 1

Posted by Anjali Singh 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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tanA/1-tanA - cotA/1-cotA = CosA+sinA/ cosA- sinA

Posted by Kanishka Maurya 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Rahul Singh 6 years, 11 months ago

Change the tanA and cotA into sinA and CosA Then solve

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Sums of the first p,q,r terms of an AP are a,b,c respectively. Prove that a/p(q-r) +b/q(r-p) +c/r (p-q) = 0 .

Posted by Geetanjali Ahirwar 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,
a = Sum of p terms
{tex}\Rightarrow \quad a = \frac { p } { 2 } \{ 2 A + ( p - 1 ) D \}{/tex}
{tex}\Rightarrow \quad \frac { 2 a } { p } = \{ 2 A + ( p - 1 ) D \}{/tex}...(i)
b = Sum of q terms
{tex}\Rightarrow \quad b = \frac { q } { 2 } \{ 2 A + ( q - 1 ) D \}{/tex}
{tex}\Rightarrow \quad \frac { 2 b } { q } = \{ 2 A + ( q - 1 ) D \}{/tex}...(ii)
and, c = Sum of r terms
{tex}\Rightarrow \quad c = \frac { r } { 2 } \{ 2 A + ( r - 1 ) D \}{/tex}
{tex}\Rightarrow \quad \frac { 2 c } { r } = \{ 2 A + ( r - 1 ) D \}{/tex}...(iii)
Multiplying equation (i), (ii) and (iii) by (q - r), (r - p) and (p - q) respectively and adding, we get
{tex}\frac { 2 a } { p } ( q - r ) + \frac { 2 b } { q } ( r - p ) + \frac { 2 c } { r } ( p - q ){/tex}
= {2A + (p - 1)D} (q - r) + {2A + (q - 1)D} (r - p)+ {2A+(r - 1)D} (p-q)
= 2Aq - 2Ar + (p-1)(q -r) D + 2Ar - 2AP + (q - 1) (r -p)D + 2Ap - 2Aq + (r -1)(p-q)A
= 2Aq - 2Ar + 2Ar - 2AP+ 2Ap - 2Aq + (p-1)(q -r) D + (q - 1) (r -p)D + (r -1)(p-q)A
= 2A (q - r + r - p + p - q) + D {(p-1) (q - r) + (q - 1) (r - p) + (r - 1) (p-q)}
= 2A × 0 + D × 0
= 0

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SinA-2sin 2A = tanA

Posted by Tushar Srivastava 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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if p(x, y) is any point joining the points A( a,0) and B (0, b) then show that x/a + y/b= 1

Posted by Shriya ?? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Shriya ?? 6 years, 11 months ago

thank u soo so much??

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Naman Jain 6 years, 11 months ago

These points are collinear so using the formula of area of triangle the area will be equal to 0 0=1/2|a(y-b)+x(b-0)+0(0-y)| |ay-ab+xb-0+0| ay-ab+xb=0 ay+xb=ab ay+xb/ab=1 Taking lcm x/a+y/b = 1 Hence proved

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Shriya ?? 6 years, 11 months ago

plz answer bata do

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a number 'x' is selected from (-3, -2, -1,0 ,1 , 2,3) find the probability that (1.) 1x1_<2 (2.) x^2>4

Posted by Shriya ?? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Rahul Singh 6 years, 11 months ago

1.(1) 2.(2/7)

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Shriya ?? 6 years, 11 months ago

a n s w e r please

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2 dice are roled once find the probability of getting such nos. on dice whose product is a perfect square

Posted by Shriya ?? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Rahul Singh 6 years, 11 months ago

2/9

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If nth term of an A. P is(2n+1) then what is the sum of its three termd

Posted by Sandhya Chaurasia 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Chetna Pandey 6 years, 11 months ago

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Is the HCF of a and b is 15 and a into B is 4500 then find the LCM of a and b

Posted by Khushi Pali 6 years, 11 months ago

CBSE > Class 10 > Mathematics

6 answers

Shriya ?? 6 years, 11 months ago

lcm ×15 = 4500 => lcm = 4500/15 => lcm = 300

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Shriya ?? 6 years, 11 months ago

in this ques appyl (hcf × lcm = product of hcf and lcm ) this formula

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Shriya ?? 6 years, 11 months ago

this ques is right

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Divanshu Rawat 6 years, 11 months ago

300

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Mamta Singh 6 years, 11 months ago

The Question is incorrect Because there given only b and to find LCM ,first we want to find a which is not possible in this question

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Shriya ?? 6 years, 11 months ago

lcm = 500

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Exercise 3.1 ke tino ke Solutions bajo koyi

Posted by Aman Kumar Jha 6 years, 11 months ago

CBSE > Class 10 > Mathematics

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Draw a pair of tangents to circle of radius 4.5 cm , which are inclined to each other at an angle of 45 degree

Posted by Tamanna Dhiman 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Anushka Jugran ? 6 years, 11 months ago

How can I draw here???

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If points R (x,y) lies on line segment joining points P(a,b)and (b,a). Then prove that x+y=a+b

Posted by Reet Brar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

According to the question, R(x, y) is a point on the line segment joining the points P(a, b) and Q(b, a)
Let point R(x + y) divides the line joining P(a,b) and Q(b,a) in the ratio {tex}\lambda{/tex} : 1.
{tex}\therefore x = \frac { \lambda b + a } { \lambda + 1 }{/tex}
{tex}y = \frac { \lambda a + b } { \lambda + 1 }{/tex}
Adding, {tex}x+y=\frac { \lambda b + a + \lambda a + b } { \lambda + 1 }{/tex}
{tex}= \frac { \lambda ( a + b ) + 1 \times ( a + b ) } { \lambda + 1 }{/tex}
{tex}= \frac { ( \lambda + 1 ) \times ( a + b ) } { \lambda + 1 } = a + b{/tex}
{tex}\Rightarrow x+y=a+b{/tex}
Hence Proved.