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What is AAS congruency rule?

Posted by Muazzam Ali (Jul 17, 2017 8:54 p.m.) (Question ID: 6720)

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  • ASA      Angle Side Angle     Rule 

    Two triangles are congruent if the two angles and the included side are equal

    similarly there are 5 ways of finding if the two tringles are congruent

    SSS, SAS , AAS, ASA, HL

    Caution   AAA       Angle Angle Angle

    Triangles which have three equal angles , we cannot be sure if they are congruent without knowing at least one side    

    Answered by Hans Raj (Jul 21, 2017 8:06 a.m.)
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  • AAS Congruence Criterion: If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

    Answered by Sahdev Sharma (Jul 20, 2017 6:35 p.m.)
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  • ASA is the Angle Side Angle rule.

    In which we can say triangles are congruent by seeing their two angles and one side.

    Answered by Arpan Bhowmick (Jul 18, 2017 9:38 a.m.)
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Find the area of a right angled triangle whose hypotenuse is 25cm and base is 7cm

 

Posted by Aman Gupta (Jul 15, 2017 8:50 p.m.) (Question ID: 6666)

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  • Using Pythagoras Theorm

    {tex}Height = \sqrt {(hypotenuse)^2-(base)^2}{/tex}

    {tex}= \sqrt{625-49}{/tex}

    {tex}= \sqrt{576}= 24 {/tex} cm

    Area of ∆ = {tex}{1\over 2}\times base\times height{/tex}

    {tex}{1\over2}\times 7\times 24= 84\ cm^2{/tex}

    Answered by Sahdev Sharma (Jul 15, 2017 10:14 p.m.)
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How is 39/20 a rational number lying between 0.75 and 1.2

Posted by Latha G (Jul 13, 2017 9:18 p.m.) (Question ID: 6624)

  • Please send it as soon as possible

    Posted by Latha G (Jul 13, 2017 9:20 p.m.)
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Simplify (0.001)^1/3

Posted by Meet Sharma (Jul 11, 2017 9:59 p.m.) (Question ID: 6580)

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  • {tex}(0.001)^{1\over 3}{/tex}

    {tex}((0.1)^3)^{1\over 3}{/tex}

    {tex}(0.1)^{3\times {1\over 3}}{/tex}

    {tex}(0.1)^1= 0.1 {/tex}

    Answered by Payal Singh (Jul 12, 2017 8:48 a.m.)
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  • {tex}(0.001)^{1\over 3}{/tex}

    {tex}((0.1)^3)^{1\over 3}{/tex}

    {tex}(0.1)^{3\times {1\over 3}}{/tex}

    {tex}(0.1)^1= 0.1 {/tex}

    Answered by Sahdev Sharma (Jul 12, 2017 8:47 a.m.)
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Root 12 multiply root 5 

Posted by Poshan Sharma (Jul 11, 2017 7:49 p.m.) (Question ID: 6569)

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  • √12×√5=√60

                 =2√15

    Answered by Ashutosh Kumar (Jul 12, 2017 6:53 a.m.)
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Question no. 6maths biok chaptet 6

Posted by Aakash Kumar (Jul 11, 2017 3:14 p.m.) (Question ID: 6562)

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Solve equation for x

(i) 22x-2x+3+24=0

(ii) 32x+4+1=2.3x+2

 

Posted by Dipankar Nath (Jul 10, 2017 8:02 p.m.) (Question ID: 6543)

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  • (i) {tex}2^{2x}-2^{x+3}+2^4=0{/tex}

    {tex}=> (2^x)^{2}-2^{x}2^{3}+16=0{/tex}

    Put 2x= y

    {tex}=> y^2-8y+16=0{/tex}

    {tex}=> y^2-4y-4y+16=0{/tex}

    {tex}=> y(y-4)-4(y-4)=0{/tex}

    => (y-4)(y-4) = 0

    => y = 4

    {tex}=> 2^x= 4{/tex}

    => {tex}2^x= 2^2{/tex}

    So value of x = 2

    Answered by Sahdev Sharma (Jul 10, 2017 11:04 p.m.)
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(x+4)(x+10)

Posted by Sujal Agrawal (Jul 07, 2017 4:40 p.m.) (Question ID: 6451)

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  • (x+4)(x+10)=x(x+10)+4(x+10)

                         =x2+10x+4x+4×10

                         =x2+14x+40

    Answered by Dharmendra Kumar (Jul 21, 2017 11:38 p.m.)
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  • = (x+4)(x+10)

    = x2+10x+4x+40

    = x2+14x+40

    Answered by Ashutosh Kumar (Jul 07, 2017 6:37 p.m.)
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The equation of the bisector of the angle between the lines x+y=3 and 2x-y= 2which contains the point ( 1,1). 

Posted by Ajay Karthikeya (Jul 05, 2017 7:30 p.m.) (Question ID: 6397)

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  • (X+y-3)+k(2x-y-2)=0

    Now, at (1,1), K=-1

    Putting the value of K ,we get,

    x+y-3 -2x+y+2=0

    Or, - x+2y-1=0

    Or, x-2y=-1

    This is the equation of the bisector of the angle.

    Answered by Soumya Ghoshal (Jul 17, 2017 5:53 p.m.)
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The equation of the bisector of the angle between the lines x+y= 3 and 2x- y = 2 which contains the point (1,1). 

Posted by Ajay Karthikeya (Jul 05, 2017 7:26 p.m.) (Question ID: 6396)

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simplify: (3√3+2√3)(2√3+3√2)

Posted by Rashi Jain (Jul 05, 2017 6:22 a.m.) (Question ID: 6390)

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  • {tex}(3\sqrt 3+2\sqrt 3)(2\sqrt 3+3\sqrt 2){/tex}

    {tex}(5\sqrt 3)(2\sqrt 3+3\sqrt 2){/tex}

    {tex}10\times 3+15\sqrt 6{/tex}

    {tex}30+15\sqrt 6{/tex}

    Answered by Payal Singh (Jul 20, 2017 4:21 p.m.)
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First term of an A.P is -2 and thR 10 th term is 16. Determine the 100th term.

Posted by Ajay Karthikeya (Jul 01, 2017 6:17 p.m.) (Question ID: 6295)

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  • Given    

    1st term a1 = -2 ,  ..............10th term a10 = 16 , .................100th term  a100 = ?

    we know that  

     an = a1 + (n -1)d  ...........   a10 =a1 + (10 - 1) d...................... 16 = -2 + 9d ..................... d = 18/9 ....... .... d = 2

    a100 = a1 + (100 - 1) 2...........a100 = -2 + 99x2 ........................ a100 = -2 + 198 ............... a100 = 196      

    Answered by Hans Raj (Jul 03, 2017 3:23 p.m.)
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  • First term of AP a = -2

    Let common difference = d

    As we know

    {tex}a_n = a+(n-1)d{/tex}

    => {tex}a_{10} = a+(10-1)d{/tex}

    => 16 = -2+9d

    => 9d= 18

    => d= 2

    {tex}a_{100} = a+99d{/tex}

    => {tex}a_{100}=-2+(99\times 2){/tex}

    {tex}=>> a_{100}= -2+198=196{/tex}

    Answered by Payal Singh (Jul 01, 2017 10:39 p.m.)
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factotise:- 9x²+6x-1+25y²

Posted by Piyush Singh (Jun 28, 2017 8:27 p.m.) (Question ID: 6238)

  • each set.  contains three sets

     

    Posted by Ajay Karthikeya (Jul 01, 2017 6:43 p.m.)
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  • {tex}9x^2+6x+1-25y^2{/tex}

    {tex}= (3x)^2+2\times 3x\times 1+(1)^2-(5y)^2{/tex}

    {tex}= (3x+1)^2-(5y)^2{/tex}

    {tex}= (3x+1-5y)(3x+1+5y){/tex}

    Answered by Payal Singh (Jun 29, 2017 8:16 a.m.)
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Factorise:-  9x²+6x+1-25y²

Posted by Piyush Singh (Jun 28, 2017 3:41 p.m.) (Question ID: 6229)

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  • {tex}9x^2+6x+1-25y^2{/tex}.......................... {tex}(3x)^2 + (2x3xx1) + (1)^2 -25y^2{/tex}.................................applying Rule {tex}(a+b)^2 = a^2+b^2+2ab {/tex}

    .{tex}(3x + 1)^2 - (5y)^2{/tex} ........................... . {tex}(3x+1+5y)(3x+1-5y){/tex} .................................. .... applying Rule {tex}a^2 - b^2 = (a+b) (a-b){/tex}

    Answered by Hans Raj (Jul 03, 2017 3:44 p.m.)
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  • Ans. 

    {tex}9x^2+6x+1-25y^2{/tex}

    {tex}= (3x)^2+(2\times 3x\times 1)+(1)^2-25y^2{/tex} [using {tex}(a+b)^2 = a^2+2ab+b^2{/tex}]

    {tex}= (3x+1)^2-(5y)^2{/tex}

    {tex}= (3x+1-5y)(3x+1+5y){/tex} [using {tex}a^2-b^2 = (a+b)(a-b)]{/tex}

    Answered by Naveen Sharma (Jun 28, 2017 5:27 p.m.)
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in a triangleABC AB=AC. D is a midpoint inside triangle ABC such that BD=DC. prove that angle ABC=angle ACD

Posted by Sandeep Sood (Jun 27, 2017 7:01 p.m.) (Question ID: 6213)

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from an internal point of equilateral triangle the lengths of perpendicular drawn on three sides are respectively 14cm,10cm and 6cm in length. find the area of the triangle

Posted by Sandeep Sood (Jun 27, 2017 11:02 a.m.) (Question ID: 6204)

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  • Let the side of equilateral triangle be a and let the perpendicular drawn to sides be OP = 10 cm, OQ = 14 cm and OR = 6 cm.

    Join OA, OB and OC.

    According to question,

    Area of equilateral triangle ABC = Area of {tex}\Delta{/tex}OBC + Area of {tex}\Delta{/tex}OCA + Area of {tex}\Delta{/tex}OAB

    =>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2} \times {\rm{BC}} \times {\rm{OP}} + {1 \over 2} \times {\rm{AC}} \times {\rm{OQ}} + {1 \over 2} \times {\rm{AB}} \times {\rm{OR}}{/tex}

    =>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2} \times 10 \times a + {1 \over 2} \times 14 \times a + {1 \over 2} \times 6 \times a{/tex}

    =>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2}a\left( {10 + 14 + 6} \right){/tex}

    =>     {tex}{{\sqrt 3 } \over 4}a = {1 \over 2} \times 30{/tex}

    =>     {tex}a = {{15 \times 4} \over {\sqrt 3 }} = {{60} \over {\sqrt 3 }}{/tex} cm

    Now, Area of equilateral triangle

    {tex}{{\sqrt 3 } \over 4}{a^2} = {{\sqrt 3 } \over 4} \times {{60} \over {\sqrt 3 }} \times {{60} \over {\sqrt 3 }}{/tex}

    =>     {tex}{{\sqrt 3 } \over 4}{a^2} = 300\sqrt 3 {/tex} cm2

    Answered by Rashmi Bajpayee (Jun 27, 2017 1:27 p.m.)
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ABCD is a trapezium in which  AB||CD and AB=2CD. if its diagonal intersect each other at O,find the ratio of areas of triangles AOB and COD

Posted by Sandeep Sood (Jun 27, 2017 10:59 a.m.) (Question ID: 6203)

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How to  fatorise x^4+x^2+25

Posted by Onkar Yadav (Jun 26, 2017 11:49 p.m.) (Question ID: 6197)

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Why a^3+b^3+c^3=3abc

Posted by Onkar Yadav (Jun 26, 2017 11:48 p.m.) (Question ID: 6196)

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  • We know that

    {tex}{x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right){/tex}

    Here, if {tex}x+y+z=0{/tex}, then R.H.S. = 0

    Then,

    {tex}{x^3} + {y^3} + {z^3} - 3xyz = 0{/tex}

    =>     {tex}{x^3} + {y^3} + {z^3} = 3xyz {/tex}

    So, we can say that if {tex}a+b+c=0{/tex}, then {tex}{a^3} + {b^3} + {c^3} = 3abc{/tex}

    Answered by Rashmi Bajpayee (Jun 28, 2017 7:50 p.m.)
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if x =2 -under root 5 ,find X2 + 1/x2

Posted by Navraj Singh Johal (Jun 21, 2017 3:38 p.m.) (Question ID: 6082)

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  • Given : x = {tex}2-\sqrt 5{/tex}

    Now 

    {tex}{1\over x} = {1\over 2-\sqrt 5}{/tex}

    Rationalize the denominator,

    We get 

    {tex}{1\over x} = {1\over 2-\sqrt 5}\times {2+\sqrt 5\over 2+\sqrt 5}{/tex}

    {tex}=> {1\over x} = {2+\sqrt 5\over (2)^2-(\sqrt 5)^2} {/tex}

    {tex}=> {1\over x} = {2+\sqrt 5\over 4-5}= -2-\sqrt 5{/tex}

    Now, 

    {tex}x^2+{1\over x^2}= (2-\sqrt 5)^2+(-2-\sqrt5)^2{/tex}

    {tex}4+5-4\sqrt5+4+5+4\sqrt5{/tex}= 18

    Answered by Payal Singh (Jun 21, 2017 6:52 p.m.)
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Find the area of a parallelogram ABCD in which ABCD is 28 cm and diagonal AC=30cm.

Posted by Deepanshu Sharma (Jun 21, 2017 11:43 a.m.) (Question ID: 6073)

  • Deepanshu 28 cm is the perimeter of the parallelogram?

    Posted by Sagar Tripathi (Jun 22, 2017 8:40 p.m.)
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1) Examine whether y+3 is a factor of polynomial 2y3 + 3y2 - 7y + 6

2) Find the valude of k, if (x-1) is a factor of 4x3 + 3 x 2 - 4 x + k

3) If x + y = 5 and xy = 6, then find x3 + y3 

Posted by Madhav V (Jun 21, 2017 9:39 a.m.) (Question ID: 6070)

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  • 1) Yes, y+3 is a factor of f(x)= 2y3 + 3y2 - 7y + 6

     2) Since (x-1) is a factor of f(x)= 4x3 + 3x2 - 4x +k , x= 1 is a root of f(x). Therefore,

         f(1)= 4*(1)3 + 3*(1)2 - 4(1) +k = 0

              or, 4*1 +3*1 - 4 + k = 0

              or, 4 + 3 - 4 + k = 0

              or, 3 + k = 0

              or, k = -3

         Therefore, value of k is -3

    3)   x3 + y3 

        = (x+y)3 - 3xy(x+y)                   [ since x3 + y3 = (x + y)3  - 3xy(x+y)]

        = (5)3 - 3 * (6) * (5)                    [ x+y = 5 and xy = 6 given]

        = 125 - 90

        = 35

        Therefore,  the value of x3 + yis 35

    Answered by Sankhadip Bag (Jun 21, 2017 6:02 p.m.)
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Rationalize the followings 

3/5+2√3

 

Posted by Ankit Batra (Jun 20, 2017 1:57 p.m.) (Question ID: 6050)

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  • {tex}{3 \over 5} + {2 \over {\sqrt 3 }} = {{3\sqrt 3 + 10} \over {5\sqrt 3 }}{/tex}

    = {tex}{{3\sqrt 3 + 10} \over {5\sqrt 3 }} \times {{\sqrt 3 } \over {\sqrt 3 }}{/tex}

    = {tex}{{9 + 10\sqrt 3 } \over {15}}{/tex}

    Answered by Rashmi Bajpayee (Jun 21, 2017 8:16 p.m.)
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If x=2ab/b^2+1 Find the value of (a+x)^1/2+ (a-x)^1/2/(a+x)^1/2- (a-x)^1/2

Posted by Sandeep Sood (Jun 20, 2017 10:53 a.m.) (Question ID: 6046)

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  • Given: {tex}x = {{2ab} \over {{b^2} + 1}}{/tex}

    {tex}{{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} - \sqrt {a - x} }}{/tex}

    = {tex}{{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} - \sqrt {a - x} }} \times {{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} + \sqrt {a - x} }}{/tex}

    = {tex}{{{{\left( {\sqrt {a + x} } \right)}^2} + {{\left( {\sqrt {a - x} } \right)}^2} + 2\sqrt {a + x} .\sqrt {a - x} } \over {{{\left( {\sqrt {a + x} } \right)}^2} - {{\left( {\sqrt {a - x} } \right)}^2}}}{/tex}

    = {tex}{{a + x + a - x + 2\sqrt {{a^2} - {x^2}} } \over {a + x - a + x}}{/tex}

    = {tex}{{2a + 2\sqrt {{a^2} - {x^2}} } \over {2x}}{/tex}

    = {tex}{{a + \sqrt {{a^2} - {x^2}} } \over x}{/tex}

    = {tex}{{a + \sqrt {{a^2} - {{\left( {{{2ab} \over {{b^2} + 1}}} \right)}^2}} } \over {{{2ab} \over {{b^2} + 1}}}}{/tex}

    = {tex}a + \sqrt {{a^2} - {{4{a^2}{b^2}} \over {{{\left( {{b^2} + 1} \right)}^2}}}} \times {{{b^2} + 1} \over {2ab}}{/tex}

    = {tex}a + a\sqrt {{{{{\left( {{b^2} + 1} \right)}^2} - 4{b^2}} \over {{{\left( {{b^2} + 1} \right)}^2}}}} \times {{{b^2} + 1} \over {2ab}}{/tex}

    = {tex}a\left[ {1 + {1 \over {{b^2} + 1}}\sqrt {{b^4} + 1 + 2{b^2} - 4{b^2}} } \right] \times {{{b^2} + 1} \over {2ab}}{/tex}

    = {tex}{{{b^2} + 1 + \sqrt {{b^4} + 1 - 2{b^2}} } \over {{b^2} + 1}} \times {{{b^2} + 1} \over {2b}}{/tex}

    = {tex}{{{b^2} + 1 + \sqrt {{{\left( {{b^2} - 1} \right)}^2}} } \over {2b}}{/tex}

    = {tex}{{{b^2} + 1 + {b^2} - 1} \over {2b}}{/tex}

    = {tex}{{2{b^2}} \over {2b}}{/tex}

    = {tex}b{/tex}

    Answered by Rashmi Bajpayee (Jun 21, 2017 8:43 p.m.)
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if(0.04)^2/0.008*(0.2)^6=(0.2)^k then value of k is equal to

Posted by Sandeep Sood (Jun 20, 2017 10:45 a.m.) (Question ID: 6045)

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  • {tex}{(0.04)^{2}\over 0.008\times (0.2)^6}=(0.2)^k{/tex}

    {tex}=> {((0.2)^2)^{2}\over( 0.2)^3\times (0.2)^6}=(0.2)^k{/tex}

    {tex}=> {(0.2)^4\over( 0.2)^9}=(0.2)^k{/tex}

    {tex}=> (0.2)^{4-9}=(0.2)^k{/tex}

    {tex}=> (0.2)^{-5}=(0.2)^k{/tex}

    On comparing both sides, we get

    k = -5

    Answered by Payal Singh (Jun 21, 2017 7:47 a.m.)
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an isosceles triangle has perimeter 30 CM and each of the equal sides is 12cm.find the area of triangle 

Posted by Santosh Jha (Jun 17, 2017 8:49 a.m.) (Question ID: 5969)

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  • Perimeter of Isosceles triangle = 30 cm

    Each of equal side of given triangle = 12 cm

    Let the third side be x cm

    =>          12 + 12 + x = 30

    =>          x = 30 - 24 = 6 cm

    Now, Semi-perimeter (s) = {tex}{{30} \over 2} = 15{/tex} cm

    Using Heron's formula,

    Area of given triangle = {tex}\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} {/tex}

    = {tex}\sqrt {15\left( {15 - 12} \right)\left( {15 - 12} \right)\left( {15 - 6} \right)} {/tex}

    = {tex}\sqrt {15 \times 3 \times 3 \times 9} {/tex}

    = {tex}9\sqrt {15} {/tex} sq. cm

    = 9 x 3.87

    = 34.83 sq. cm

     

    Answered by Rashmi Bajpayee (Jun 17, 2017 11:27 a.m.)
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Factorise: 1) x4+4x2+16

                   2) x3-10x2-52x-42

Posted by Pradhyumn Soni (Jun 16, 2017 3:15 p.m.) (Question ID: 5941)

  • Dear Pradhyumn, please check the q.(2), I think there is a little mistake. There should be 53 in place of 52.

    I am giving you solution for that.

    Posted by Rashmi Bajpayee (Jun 17, 2017 1:33 p.m.)
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  • {tex}p\left( x \right) = {x^3} - 10{x^2} - 53x - 42{/tex}

    Putting x = -1 in p(x),

    {tex}p\left( { - 1} \right) = {\left( { - 1} \right)^3} - 10{\left( { - 1} \right)^2} - 53\left( { - 1} \right) - 42{/tex} = {tex} - 1 - 10 + 53 - 42 = 0{/tex}

    Since {tex}p\left( { - 1} \right) = 0{/tex}, therefore, (x + 1) is a factor of p(x)

    Now, on dividing p(x) by (x + 1), we get the quotient {tex}{x^2} - 11x - 42{/tex}

    Therefore, {tex}p\left( x \right) = {x^3} - 10{x^2} - 53x - 42{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} - 11x - 42} \right]{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} - 14x + 3x - 42} \right]{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {x\left( {x - 14} \right) + 3\left( {x - 14} \right)} \right]{/tex}

    {tex}\Rightarrow{/tex}         {tex}p\left( x \right) = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 14} \right){/tex}

     

    Answered by Rashmi Bajpayee (Jun 17, 2017 1:46 p.m.)
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  • 1) x<font size="2">4</font>+4x<font size="2">2</font>+16

    = {tex}{x^4} + 16 + 4{x^2}{/tex}

    = {tex}{\left( {{x^2}} \right)^2} + {\left( 4 \right)^2} + 2 \times {x^2} \times 4 - 2 \times {x^2} \times 4 + 4{x^2}{/tex}

    = {tex}{\left( {{x^2} + 4} \right)^2} - 8{x^2} + 4{x^2}{/tex}

    = {tex}{\left( {{x^2} + 4} \right)^2} - 4{x^2}{/tex}

    = {tex}{\left( {{x^2} + 4} \right)^2} - {\left( {2x} \right)^2}{/tex}

    = {tex}\left( {{x^2} + 4 + 2x} \right)\left( {{x^2} + 4 - 2x} \right){/tex}                 [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

    = {tex}\left( {{x^2} + 2x + 4} \right)\left( {{x^2} - 2x + 4} \right){/tex}

    Answered by Rashmi Bajpayee (Jun 17, 2017 1:10 p.m.)
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Factorise : 1) 2√2x3+5√2x2-7√2

                    2) x3+13x2+32+20

                    3) p4-81q4

 

Posted by Pradhyumn Soni (Jun 16, 2017 3:10 p.m.) (Question ID: 5940)

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  • 1) 2√2x<font size="2">3</font>+5√2x<font size="2">2</font>-7√2

    {tex}p\left( x \right) = 2\surd 2{x^3} + 5\surd 2{x^2} - 7\surd 2{/tex}

    Putting x = 1, in p(x),

    {tex}p\left( 1 \right) = 2\surd 2{\left( 1 \right)^3} + 5\surd 2{\left( 1 \right)^2} - 7\surd 2{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( 1 \right) = 2\surd 2 + 5\surd 2 - 7\surd 2{/tex} = {tex}7\surd 2 - 7\surd 2 = 0{/tex}

    Since p(1) = 0, therefore, (x - 1) is a factor of p(x)

    Dividing p(x) by (x -1), we get the quotient {tex}2\sqrt 2 {x^2} + 7\sqrt 2 {/tex}

    Therefore,

    {tex}p\left( x \right) = 2\surd 2{x^3} + 5\surd 2{x^2} - 7\surd 2{/tex}

    {tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \left( {x - 1} \right)\left[ {2\sqrt 2 {x^2} + 7\sqrt 2 } \right]{/tex}

    {tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \left( {x - 1} \right)\left[ {\sqrt 2 \left( {2{x^2} + 7} \right)} \right]{/tex}

    {tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \sqrt 2 \left( {x - 1} \right)\left( {2{x^2} + 7} \right){/tex}

    Answered by Rashmi Bajpayee (Jun 17, 2017 4:41 p.m.)
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  • 3) {tex}{p^4} - 81{q^4}{/tex}

    = {tex}{\left( {{p^2}} \right)^2} - {\left( {9{q^2}} \right)^2}{/tex}                               [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

    = {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {{p^2} - 9{q^2}} \right]{/tex}

    = {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {{{\left( p \right)}^2} - {{\left( {3q} \right)}^2}} \right]{/tex}

    = {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {\left( {p + 3q} \right)\left( {p - 3q} \right)} \right]{/tex}        [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

    = {tex}\left( {{p^2} + 9{q^2}} \right)\left( {p + 3q} \right)\left( {p - 3q} \right){/tex}

    Answered by Rashmi Bajpayee (Jun 17, 2017 1:32 p.m.)
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  • 2) p(x) = x<font size="2">3</font>+13x<font size="2">2</font>+32x+20

    putting x = -1 in given polynomial,

    p(-1) = (-1)<font size="2">3</font>+ 13(-1)<font size="2">2</font>+ 32(-1) +20

             = -1 + 13 - 32 + 20 = 0

    Since p(-1) = 0, therefore (x + 1) is a factor of p(x).

    Now, on dividing p(x) by (x + 1), we get the quotient x2 + 12x + 20

    {tex}p\left( x \right) = {x^3} + 13{x^2} + 32x + 20{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} + 12x + 20} \right]{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} + 10x + 2x + 20} \right]{/tex}

    {tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {x\left( {x + 10} \right) + 2\left( {x + 10} \right)} \right]{/tex}

    {tex}\Rightarrow{/tex}         {tex}p\left( x \right) = \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 10} \right){/tex}

    Answered by Rashmi Bajpayee (Jun 17, 2017 1:25 p.m.)
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IS zero is an even number

Posted by Sudhanshu Dash (Jun 16, 2017 8:13 a.m.) (Question ID: 5929)

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  • Yes zero is an even number.

    Answered by Ashutosh Kumar (Jun 16, 2017 8:39 a.m.)
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<hr>

​​​​​Prove that m-n=2

Posted by Sree Devi (Jun 16, 2017 7:02 a.m.) (Question ID: 5928)

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