Maths-
General
Easy

Question

Arrange in descending order of magnitude cube root of 2 comma root index 6 of 3 and root index 9 of 4

Hint:

Use LCM property.

The correct answer is: descending order


    Complete step by step solution:
    Here we can write, root index 8 of 2 equals 2 to the power of 1 third end exponent comma space of 1em root index 6 of 3 equals 3 to the power of 1 over 6 end exponent  and root index 9 of 4 equals 4 to the power of 1 over 9 end exponent
    On taking the LCM of 3,6,9 we have LCM as 18.
    So, not stretchy rightwards double arrow 2 to the power of 1 third end exponent equals 2 to the power of fraction numerator 1 cross times 6 over denominator 3 cross times 6 end fraction end exponent equals 2 to the power of 6 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 2 to the power of 6 close parentheses to the power of 1 over 18 end exponent equals 64 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 3 to the power of 1 over 6 end exponent equals 3 to the power of fraction numerator 1 cross times 3 over denominator 6 cross times 3 end fraction end exponent equals 3 to the power of 3 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 3 cubed close parentheses to the power of 1 over 18 end exponent equals 27 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 4 to the power of 1 over 9 end exponent equals 4 to the power of fraction numerator 1 cross times 2 over denominator 9 cross times 2 end fraction end exponent equals 4 to the power of 2 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 4 squared close parentheses to the power of 1 over 18 end exponent equals 16 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Now, we have 64 to the power of 1 over 18 end exponent comma 27 to the power of 1 over 18 end exponent and 16 to the power of 1 over 18 end exponent
    So, here descending order is 64 > 27 > 16
    not stretchy rightwards double arrow 64 to the power of 1 over 18 end exponent greater than 27 to the power of 1 over 18 end exponent greater than 16 to the power of 1 over 18 end exponent
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 2 to the power of 1 third end exponent greater than 3 to the power of 1 over 6 end exponent greater than 4 to the power of 1 over 9 end exponent end cell row cell not stretchy rightwards double arrow cube root of 2 greater than root index 6 of 3 greater than root index 9 of 4 end cell end table

    Related Questions to study

    General
    Maths-

    In △ABC, AD⊥BC Also,B D equals 3 C D Prove that 2 A B squared equals 2 A C squared plus B C squared

    Aim  :- Prove that 2 AB squared equals 2 AC squared plus BC squared
    As BD = 3 CD  ;We get BC = BD +CD = 4 CD
    Applying pythagoras theorem in ΔACD ,We get
    AC squared equals AD squared plus CD squared


    Applying pythagoras theorem in ΔABD ,We get
    AB squared equals BD squared plus AD squared not stretchy rightwards double arrow AD squared equals AB squared minus BD squared  —Eq2
    Substitute Eq2 in Eq1 ,
    AC squared equals AB squared minus BD squared plus CD squared not stretchy rightwards double arrow AC squared equals AB squared plus CD squared minus BD squared
    As BD = 3 CD  ;We get  A C squared equals A B squared plus C D squared minus 9 C D squared
    AC squared equals AB squared minus 8 CD squared not stretchy rightwards double arrow 8 CD squared plus AC squared equals AB squared
    Multiplying with 2 on both sides
    left parenthesis 4 CD right parenthesis squared plus 2 AC squared equals 2 AB squared As BC = 4 CD
    ∴ 2 AB squared equals 2 AC squared plus BC squared
    Hence proved

    In △ABC, AD⊥BC Also,B D equals 3 C D Prove that 2 A B squared equals 2 A C squared plus B C squared

    Maths-General
    Aim  :- Prove that 2 AB squared equals 2 AC squared plus BC squared
    As BD = 3 CD  ;We get BC = BD +CD = 4 CD
    Applying pythagoras theorem in ΔACD ,We get
    AC squared equals AD squared plus CD squared


    Applying pythagoras theorem in ΔABD ,We get
    AB squared equals BD squared plus AD squared not stretchy rightwards double arrow AD squared equals AB squared minus BD squared  —Eq2
    Substitute Eq2 in Eq1 ,
    AC squared equals AB squared minus BD squared plus CD squared not stretchy rightwards double arrow AC squared equals AB squared plus CD squared minus BD squared
    As BD = 3 CD  ;We get  A C squared equals A B squared plus C D squared minus 9 C D squared
    AC squared equals AB squared minus 8 CD squared not stretchy rightwards double arrow 8 CD squared plus AC squared equals AB squared
    Multiplying with 2 on both sides
    left parenthesis 4 CD right parenthesis squared plus 2 AC squared equals 2 AB squared As BC = 4 CD
    ∴ 2 AB squared equals 2 AC squared plus BC squared
    Hence proved

    General
    Maths-

    In the figure, AD is a median to BC  in straight triangle A B C and A E perpendicular B C text . If  end text B C equals a comma C A equals b comma A B equals c comma A D equals p comma A E equals h. and DE =
    x text , prove that  end text b squared plus c squared equals 2 p squared plus 1 half a squared

    Solution :-
    Aim  :- Prove that b squared plus c squared equals 2 p squared plus 1 half a squared
    Hint :- apply in pythagoras theorem on triangles AEB , AEC and AED Find b squared  and c squared  values .Now, add them and do necessary modification to get in terms of p and a.
    Explanation(proof ) :-
    As D is mid point BD = DC = BC/2 = a/2
    Applying pythagoras theorem to  triangle AEB we get ,
    A C squared equals E C squared plus A E squared not stretchy rightwards double arrow b squared equals left parenthesis x plus a divided by 2 right parenthesis squared plus h squared minus Eq 1
    Applying pythagoras theorem to  triangle AEC we get ,
    A B squared equals B E squared plus A E squared not stretchy rightwards double arrow c squared equals left parenthesis x minus a divided by 2 right parenthesis squared plus h squared minus Eq 2

    Applying pythagoras theorem to triangle AED we get ,
    A D squared equals E D squared plus A E squared not stretchy rightwards double arrow p squared equals x squared plus h squared minus Eq 3
    Adding Eq 1 and Eq 2  
    We get b squared plus c squared equals left parenthesis x minus a divided by 2 right parenthesis squared plus h squared plus left parenthesis x plus a divided by 2 right parenthesis squared plus h squared
    [By applying  open left parenthesis a minus b right parenthesis squared plus left parenthesis a plus b right parenthesis squared equals 2 open parentheses a squared plus b squared close parentheses close square brackets
    not stretchy rightwards double arrow b squared plus c squared equals 2 h squared plus 2 open parentheses x squared plus left parenthesis a divided by 2 right parenthesis squared close parentheses
    b squared plus c squared equals 2 open parentheses h squared plus x squared close parentheses plus 2 open parentheses a squared over 4 close parentheses
    Substitute Eq3 in the above condition
    b squared plus c squared equals 2 open parentheses p squared close parentheses plus 1 half a squared not stretchy rightwards double arrow b squared plus c squared equals 2 p squared plus 1 half a squared
    Hence proved

    In the figure, AD is a median to BC  in straight triangle A B C and A E perpendicular B C text . If  end text B C equals a comma C A equals b comma A B equals c comma A D equals p comma A E equals h. and DE =
    x text , prove that  end text b squared plus c squared equals 2 p squared plus 1 half a squared

    Maths-General
    Solution :-
    Aim  :- Prove that b squared plus c squared equals 2 p squared plus 1 half a squared
    Hint :- apply in pythagoras theorem on triangles AEB , AEC and AED Find b squared  and c squared  values .Now, add them and do necessary modification to get in terms of p and a.
    Explanation(proof ) :-
    As D is mid point BD = DC = BC/2 = a/2
    Applying pythagoras theorem to  triangle AEB we get ,
    A C squared equals E C squared plus A E squared not stretchy rightwards double arrow b squared equals left parenthesis x plus a divided by 2 right parenthesis squared plus h squared minus Eq 1
    Applying pythagoras theorem to  triangle AEC we get ,
    A B squared equals B E squared plus A E squared not stretchy rightwards double arrow c squared equals left parenthesis x minus a divided by 2 right parenthesis squared plus h squared minus Eq 2

    Applying pythagoras theorem to triangle AED we get ,
    A D squared equals E D squared plus A E squared not stretchy rightwards double arrow p squared equals x squared plus h squared minus Eq 3
    Adding Eq 1 and Eq 2  
    We get b squared plus c squared equals left parenthesis x minus a divided by 2 right parenthesis squared plus h squared plus left parenthesis x plus a divided by 2 right parenthesis squared plus h squared
    [By applying  open left parenthesis a minus b right parenthesis squared plus left parenthesis a plus b right parenthesis squared equals 2 open parentheses a squared plus b squared close parentheses close square brackets
    not stretchy rightwards double arrow b squared plus c squared equals 2 h squared plus 2 open parentheses x squared plus left parenthesis a divided by 2 right parenthesis squared close parentheses
    b squared plus c squared equals 2 open parentheses h squared plus x squared close parentheses plus 2 open parentheses a squared over 4 close parentheses
    Substitute Eq3 in the above condition
    b squared plus c squared equals 2 open parentheses p squared close parentheses plus 1 half a squared not stretchy rightwards double arrow b squared plus c squared equals 2 p squared plus 1 half a squared
    Hence proved

    General
    Maths-

    text  If the angles of a triangle are in the ratio  end text 5 colon 6 colon 7 text , find all the angles.  end text

    Explanation :-
    Given angles are in the ratio 5:6:7
    Step 1:- Find angles in terms of k  and find k
    The angles are 5k ,6k and 7 k respectively
    We know Sum of angles of triangle is 180°
    5k +6k+7k = 180
    18 k = 180
    K = 10°
    Step 2:- Find the actual angles of triangle
    5k = 5×10 = 50°
    6k = 6×10 = 60°
    7k = 7×10 = 70°
    Therefore 50°,60° and 70° are the angles of the triangle which satisfy the condition .

    text  If the angles of a triangle are in the ratio  end text 5 colon 6 colon 7 text , find all the angles.  end text

    Maths-General
    Explanation :-
    Given angles are in the ratio 5:6:7
    Step 1:- Find angles in terms of k  and find k
    The angles are 5k ,6k and 7 k respectively
    We know Sum of angles of triangle is 180°
    5k +6k+7k = 180
    18 k = 180
    K = 10°
    Step 2:- Find the actual angles of triangle
    5k = 5×10 = 50°
    6k = 6×10 = 60°
    7k = 7×10 = 70°
    Therefore 50°,60° and 70° are the angles of the triangle which satisfy the condition .

    parallel
    General
    Maths-

    text  In a right angled Triangle  end text A B C comma less than B equals 90 to the power of ring operator text  and  end text D text  is the mid point of  end text B C text . Prove that  end text A C squared equals A D squared plus 3 C D

    Aim  :- Prove that A C squared equals A D squared plus 3 C D squared
    Explanation(proof ) :-
    Applying pythagoras theorem in triangle ABC,we get
    A C squared equals A B squared plus B C squared                —- Eq1
    Applying pythagoras theorem in triangle ABD ,we get
    A D squared equals A B squared plus B D squared                 — Eq2
    As AD is median , D is mid point then BC = 2BD

    Subtract Eq1 - Eq2
    We get ,A C squared minus A D squared equals A B squared plus B C squared minus open parentheses A B squared plus B D squared close parentheses
    A C squared minus A D squared equals B C squared minus B D squared text  substitute  end text BC equals 2 BD
    A C squared minus A D squared equals 4 B D squared minus B D squared
    As D is mid point BD = CD ,
    substitute BD = CD in the above equation
    A C squared minus A D squared equals 3 C D squared
    A C squared equals A D squared plus 3 C D squared
    Hence proved

    text  In a right angled Triangle  end text A B C comma less than B equals 90 to the power of ring operator text  and  end text D text  is the mid point of  end text B C text . Prove that  end text A C squared equals A D squared plus 3 C D

    Maths-General
    Aim  :- Prove that A C squared equals A D squared plus 3 C D squared
    Explanation(proof ) :-
    Applying pythagoras theorem in triangle ABC,we get
    A C squared equals A B squared plus B C squared                —- Eq1
    Applying pythagoras theorem in triangle ABD ,we get
    A D squared equals A B squared plus B D squared                 — Eq2
    As AD is median , D is mid point then BC = 2BD

    Subtract Eq1 - Eq2
    We get ,A C squared minus A D squared equals A B squared plus B C squared minus open parentheses A B squared plus B D squared close parentheses
    A C squared minus A D squared equals B C squared minus B D squared text  substitute  end text BC equals 2 BD
    A C squared minus A D squared equals 4 B D squared minus B D squared
    As D is mid point BD = CD ,
    substitute BD = CD in the above equation
    A C squared minus A D squared equals 3 C D squared
    A C squared equals A D squared plus 3 C D squared
    Hence proved
    General
    Maths-

    The Quadrilateral PQRS has angles at  S,Q right angles and the diagonals PR, QS are perpendicular. Prove that SR = QR.

    Solution :-
    Aim  :- Prove that SR = QR
    Explanation(proof ) :-
    We know that if a quadrilateral has a sum of opposite angles 180°
    ∠S  + ∠ Q = 90 + 90 = 180°(S and Q are opposite angle)
    Here PQRS is cyclic and ∠S = 90°
    As the angle angle in a semicircle is  90° then PSR lies in a semi circle
    and PR is diameter.
    As PR passes through centre and perpendicular to chord SQ
    The  perpendicular drawn from centre to chord bisects the chord
    So, OS = OQ as PR bisects chord SQ
    In  straight capital delta OQR and  straight capital delta OSR
    OS = OQ (side)
    ∠SOR = ∠QOR =90° (Angle)
    OP =OP (common side )
    By SAS rule straight capital delta OQR ≅  straight capital delta OSR
    By congruence we get , SR= QR
    Hence proved

    The Quadrilateral PQRS has angles at  S,Q right angles and the diagonals PR, QS are perpendicular. Prove that SR = QR.

    Maths-General
    Solution :-
    Aim  :- Prove that SR = QR
    Explanation(proof ) :-
    We know that if a quadrilateral has a sum of opposite angles 180°
    ∠S  + ∠ Q = 90 + 90 = 180°(S and Q are opposite angle)
    Here PQRS is cyclic and ∠S = 90°
    As the angle angle in a semicircle is  90° then PSR lies in a semi circle
    and PR is diameter.
    As PR passes through centre and perpendicular to chord SQ
    The  perpendicular drawn from centre to chord bisects the chord
    So, OS = OQ as PR bisects chord SQ
    In  straight capital delta OQR and  straight capital delta OSR
    OS = OQ (side)
    ∠SOR = ∠QOR =90° (Angle)
    OP =OP (common side )
    By SAS rule straight capital delta OQR ≅  straight capital delta OSR
    By congruence we get , SR= QR
    Hence proved
    General
    Maths-

    P and Q are points on the sides CA and CB  respectively of a straight triangle ABC  right angled at  c. Prove that AQ squared plus BP squared equalsA B squared plus P Q squared

    Aim  :- Prove that A Q squared plus B P squared equals A B squared plus P Q squared
    Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB
    find the equation, add them and substitute proper conditions to prove the condition
     

    Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB
    find the equation, add them and substitute proper conditions to prove the condition
    Explanation(proof ) :-
    Applying pythagoras theorem to  triangle AQC we get ,
    A Q squared equals A C squared plus Q C squared minus Eq 1
    Applying pythagoras theorem to  triangle PBC we get ,
    B P squared equals P C squared plus B C squared minus Eq 2
    Applying pythagoras theorem to  triangle ABC we get ,
    A B squared equals A C squared plus B C squared minus Eq 3
    Applying pythagoras theorem to  triangle PCQ we get ,
    P Q squared equals P C squared plus Q C squared minus Eq 4
    Adding Eq1 and Eq2 we get ,
    A Q squared plus B P squared equals A C squared plus Q C squared plus P C squared plus B C squared
    A Q squared plus B P squared equals open parentheses A C squared plus B C squared close parentheses plus open parentheses Q C squared plus P C squared close parentheses left curly bracket text  substitute Eq3 and Eq4}  end text
    A Q squared plus B P squared equals A B squared plus P Q squared
    Hence proved 

    P and Q are points on the sides CA and CB  respectively of a straight triangle ABC  right angled at  c. Prove that AQ squared plus BP squared equalsA B squared plus P Q squared

    Maths-General
    Aim  :- Prove that A Q squared plus B P squared equals A B squared plus P Q squared
    Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB
    find the equation, add them and substitute proper conditions to prove the condition
     

    Hint :- Applying pythagoras theorem to both triangles ABC, ACQ,PCQ and PCB
    find the equation, add them and substitute proper conditions to prove the condition
    Explanation(proof ) :-
    Applying pythagoras theorem to  triangle AQC we get ,
    A Q squared equals A C squared plus Q C squared minus Eq 1
    Applying pythagoras theorem to  triangle PBC we get ,
    B P squared equals P C squared plus B C squared minus Eq 2
    Applying pythagoras theorem to  triangle ABC we get ,
    A B squared equals A C squared plus B C squared minus Eq 3
    Applying pythagoras theorem to  triangle PCQ we get ,
    P Q squared equals P C squared plus Q C squared minus Eq 4
    Adding Eq1 and Eq2 we get ,
    A Q squared plus B P squared equals A C squared plus Q C squared plus P C squared plus B C squared
    A Q squared plus B P squared equals open parentheses A C squared plus B C squared close parentheses plus open parentheses Q C squared plus P C squared close parentheses left curly bracket text  substitute Eq3 and Eq4}  end text
    A Q squared plus B P squared equals A B squared plus P Q squared
    Hence proved 
    parallel
    General
    Maths-

    In △ABC, ∠B=90 and  is the mid point of BC. Prove that A C squared minus A D squared equals 3 B D squared
     

    Hint :- use pythagoras theorem in triangle ABD and ABC

    Hint :- use pythagoras theorem in triangle ABD and ABC
    We get AC squared and AD squared subtract them and substitute BC = 2BD
    In the equation to get the result.
    Explanation(proof ) :-
    Applying pythagoras theorem in triangle ABC,we get
    A C squared equals A B squared plus B C squared  —- Eq1
    Applying pythagoras theorem in triangle ABD ,we get
    AD squared equals AB squared plus BD squared Eq2
    As AD is median , D is mid point then BC = 2BD
    Subtract Eq1 - Eq2
    We get, AC squared minus AD squared equals AB squared plus BC squared minus open parentheses AB squared plus BD squared close parentheses
    AC squared minus AD squared equals BC squared minus BD squared text  substitute  end text BC equals 2 BD
    AC squared minus AD squared equals 4 BD squared minus BD squared
    AC squared minus AD squared equals 3 BD squared
    Hence proved

    In △ABC, ∠B=90 and  is the mid point of BC. Prove that A C squared minus A D squared equals 3 B D squared
     

    Maths-General
    Hint :- use pythagoras theorem in triangle ABD and ABC

    Hint :- use pythagoras theorem in triangle ABD and ABC
    We get AC squared and AD squared subtract them and substitute BC = 2BD
    In the equation to get the result.
    Explanation(proof ) :-
    Applying pythagoras theorem in triangle ABC,we get
    A C squared equals A B squared plus B C squared  —- Eq1
    Applying pythagoras theorem in triangle ABD ,we get
    AD squared equals AB squared plus BD squared Eq2
    As AD is median , D is mid point then BC = 2BD
    Subtract Eq1 - Eq2
    We get, AC squared minus AD squared equals AB squared plus BC squared minus open parentheses AB squared plus BD squared close parentheses
    AC squared minus AD squared equals BC squared minus BD squared text  substitute  end text BC equals 2 BD
    AC squared minus AD squared equals 4 BD squared minus BD squared
    AC squared minus AD squared equals 3 BD squared
    Hence proved
    General
    Maths-

    ABC text  is an equilateral triangle  end text. straight K text  is a point on  end text BC text  such that  end text BK colon KC equals 2 colon 1 text . Prove that  end text 9 AK squared equals 7 AC squared

    Aim  :- Prove that 9 A K squared equals 7 A C squared
    Explanation(proof ) :-
    BK:KC = 2:1 KC :( KC+BK) = 1:(1+2)  KC :CB = 1:3
    In equilateral triangle all sides are equal
    Let , AB = BC =  CA  = x
    Given CK =1 third BC = X over 3 ( BC is trisected at K)
    In equilateral triangle ,the altitudes and medians coincide
    so,AE is median also i.e E is midpoint of BC
    CE = X over 2
    From diagram ,KE = CE - CK =x over 2 minus x over 3 equals x over 6
    We get height of equilateral triangle when side x is fraction numerator x square root of 3 over denominator 2 end fraction

    Applying pythagoras theorem on triangle ADE
    We get, A K squared equals E K squared plus A E squared equals open parentheses x over 6 close parentheses squared plus open parentheses fraction numerator x square root of 3 over denominator 2 end fraction close parentheses squared
    A K squared equals x squared over 36 plus fraction numerator 3 x squared over denominator 4 end fraction equals fraction numerator 28 x squared over denominator 36 end fraction equals 7 over 9 x squared
    Substitute x by AC as x = AC
    A K squared equals 7 over 9 A C squared not stretchy rightwards double arrow 9 A K squared equals 7 A C squared
    Hence proved

    ABC text  is an equilateral triangle  end text. straight K text  is a point on  end text BC text  such that  end text BK colon KC equals 2 colon 1 text . Prove that  end text 9 AK squared equals 7 AC squared

    Maths-General
    Aim  :- Prove that 9 A K squared equals 7 A C squared
    Explanation(proof ) :-
    BK:KC = 2:1 KC :( KC+BK) = 1:(1+2)  KC :CB = 1:3
    In equilateral triangle all sides are equal
    Let , AB = BC =  CA  = x
    Given CK =1 third BC = X over 3 ( BC is trisected at K)
    In equilateral triangle ,the altitudes and medians coincide
    so,AE is median also i.e E is midpoint of BC
    CE = X over 2
    From diagram ,KE = CE - CK =x over 2 minus x over 3 equals x over 6
    We get height of equilateral triangle when side x is fraction numerator x square root of 3 over denominator 2 end fraction

    Applying pythagoras theorem on triangle ADE
    We get, A K squared equals E K squared plus A E squared equals open parentheses x over 6 close parentheses squared plus open parentheses fraction numerator x square root of 3 over denominator 2 end fraction close parentheses squared
    A K squared equals x squared over 36 plus fraction numerator 3 x squared over denominator 4 end fraction equals fraction numerator 28 x squared over denominator 36 end fraction equals 7 over 9 x squared
    Substitute x by AC as x = AC
    A K squared equals 7 over 9 A C squared not stretchy rightwards double arrow 9 A K squared equals 7 A C squared
    Hence proved
    General
    Maths-

    The perpendicular AD on the base BC of Triangle ABC intersects BC in D such that BD = 3CD. Prove that
    2 A B squared equals 2 A C squared plus B C squared

    Aim  :- Prove that 2 A B squared equals 2 A C squared plus B C squared
    Explanation(proof ) :-
    As BD = 3 CD  ;We get BC = BD +CD = 4 CD
    Applying pythagoras theorem in ΔACD ,We get
    A C squared equals A D squared plus C D squared                                            — Eq1
    Applying pythagoras theorem in ΔABD ,We get
    A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared            —Eq2

    Substitute Eq2 in Eq1 ,

     A C squared equals A B squared minus B D squared plus C D squared not stretchy rightwards double arrow A C squared equals A B squared plus C D squared minus B D squared

    As BD = 3 CD  ;We get A C squared equals A B squared plus C D squared minus 9 C D squared
    A C squared equals A B squared minus 8 C D squared not stretchy rightwards double arrow 8 C D squared plus A C squared equals A B squared
    Multiplying with 2 on both sides

    left parenthesis 4 C D right parenthesis squared plus 2 A C squared equals 2 A B squared AsBC equals 4 CD
    therefore 2 A B squared equals 2 A C squared plus B C squared
    Hence proved


    The perpendicular AD on the base BC of Triangle ABC intersects BC in D such that BD = 3CD. Prove that
    2 A B squared equals 2 A C squared plus B C squared

    Maths-General
    Aim  :- Prove that 2 A B squared equals 2 A C squared plus B C squared
    Explanation(proof ) :-
    As BD = 3 CD  ;We get BC = BD +CD = 4 CD
    Applying pythagoras theorem in ΔACD ,We get
    A C squared equals A D squared plus C D squared                                            — Eq1
    Applying pythagoras theorem in ΔABD ,We get
    A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared            —Eq2

    Substitute Eq2 in Eq1 ,

     A C squared equals A B squared minus B D squared plus C D squared not stretchy rightwards double arrow A C squared equals A B squared plus C D squared minus B D squared

    As BD = 3 CD  ;We get A C squared equals A B squared plus C D squared minus 9 C D squared
    A C squared equals A B squared minus 8 C D squared not stretchy rightwards double arrow 8 C D squared plus A C squared equals A B squared
    Multiplying with 2 on both sides

    left parenthesis 4 C D right parenthesis squared plus 2 A C squared equals 2 A B squared AsBC equals 4 CD
    therefore 2 A B squared equals 2 A C squared plus B C squared
    Hence proved


    parallel
    General
    Maths-

    In an equilateral straight triangle ABC text ,  end textthe side BC  is trisected at D . Prove that 9 A D squared equals 7 A B squared
    .

    Solution :-
    Aim  :- Prove that 9 A D squared equals 7 A B squared
    Hint :- let the side of equilateral triangle be x , then we get height of equilateral
    triangle as  left parenthesis x square root of 3 right parenthesis divided by 2 .Now , Find length of DE and apply pythagoras theorem in
    Right angle triangle ADE to find length of AD in terms of AB.
    Explanation(proof ) :-
    In equilateral triangle all sides are equal
    Let , AB = BC =  CA  = x
    Given BD = ⅓ BC = x/3 ( BC is trisected at D)
    n equilateral triangle ,the altitudes and medians coincide
    so,AE is median also i.e E is midpoint of BC
    BE = x/2
    From diagram ,DE = BE - BD equals x over 2 minus x over 3 equals x over 6
    We get height of equilateral triangle when side x isfraction numerator x square root of 3 over denominator 2 end fraction
    Applying pythagoras theorem on triangle ADE
    We get, A D squared equals D E squared plus A E squared equals open parentheses x over 6 close parentheses squared plus open parentheses fraction numerator x square root of 3 over denominator 2 end fraction close parentheses squared
    A D squared equals x squared over 36 plus fraction numerator 3 x squared over denominator 4 end fraction equals fraction numerator 28 x squared over denominator 36 end fraction equals 7 over 9 x squared
    Substitute x by AB as x = AB
    A D squared equals 7 over 9 A B squared not stretchy rightwards double arrow 9 A D squared equals 7 A B squared

    Hence proved

    In an equilateral straight triangle ABC text ,  end textthe side BC  is trisected at D . Prove that 9 A D squared equals 7 A B squared
    .

    Maths-General
    Solution :-
    Aim  :- Prove that 9 A D squared equals 7 A B squared
    Hint :- let the side of equilateral triangle be x , then we get height of equilateral
    triangle as  left parenthesis x square root of 3 right parenthesis divided by 2 .Now , Find length of DE and apply pythagoras theorem in
    Right angle triangle ADE to find length of AD in terms of AB.
    Explanation(proof ) :-
    In equilateral triangle all sides are equal
    Let , AB = BC =  CA  = x
    Given BD = ⅓ BC = x/3 ( BC is trisected at D)
    n equilateral triangle ,the altitudes and medians coincide
    so,AE is median also i.e E is midpoint of BC
    BE = x/2
    From diagram ,DE = BE - BD equals x over 2 minus x over 3 equals x over 6
    We get height of equilateral triangle when side x isfraction numerator x square root of 3 over denominator 2 end fraction
    Applying pythagoras theorem on triangle ADE
    We get, A D squared equals D E squared plus A E squared equals open parentheses x over 6 close parentheses squared plus open parentheses fraction numerator x square root of 3 over denominator 2 end fraction close parentheses squared
    A D squared equals x squared over 36 plus fraction numerator 3 x squared over denominator 4 end fraction equals fraction numerator 28 x squared over denominator 36 end fraction equals 7 over 9 x squared
    Substitute x by AB as x = AB
    A D squared equals 7 over 9 A B squared not stretchy rightwards double arrow 9 A D squared equals 7 A B squared

    Hence proved

    General
    Maths-


    ABC and DBC are two isosceles triangles on the same base BC. Show
    that straight angle ABD equals straight angle ACD

    Aim :- Prove ∠ ABD = ∠ ACD
    Explanation :-
    In triangle ABC, AB = AC (isosceles triangle )
    so, ∠ ABC = ∠ ACB
    In triangle BDC, BD = DC (isosceles triangle )
    ∠ DBC =∠ DCB
    Adding above results we get,
    ∠ABC  + ∠ DBC = ∠ ACB + ∠ DCB
    [ from diagram we get ∠ ABC  + ∠ DBC = ∠ ABD and ∠ ACB + ∠ DCB = ACD
     ∠ ABD = ∠ACD
    Hence proved


    ABC and DBC are two isosceles triangles on the same base BC. Show
    that straight angle ABD equals straight angle ACD

    Maths-General
    Aim :- Prove ∠ ABD = ∠ ACD
    Explanation :-
    In triangle ABC, AB = AC (isosceles triangle )
    so, ∠ ABC = ∠ ACB
    In triangle BDC, BD = DC (isosceles triangle )
    ∠ DBC =∠ DCB
    Adding above results we get,
    ∠ABC  + ∠ DBC = ∠ ACB + ∠ DCB
    [ from diagram we get ∠ ABC  + ∠ DBC = ∠ ABD and ∠ ACB + ∠ DCB = ACD
     ∠ ABD = ∠ACD
    Hence proved
    General
    Maths-

    PQ and RS are respectively the smallest and longest sides of a quadrilateral PQRS. Show that ∠ P > ∠ R

    Aim :- Prove ∠P  ∠R


     
    In triangle PQR,
    PQ < QR (PQ is smallest)
    so, ∠PRQ ∠QPR
    In triangle PSR,
    PS < SR  (RS is longest side)
    ∠ PRS ∠ RPS
    Adding above results we get,
    ∠PRQ + ∠PRS ∠QPR + ∠RPS
    ∠R  ∠P
    Hence proved

    PQ and RS are respectively the smallest and longest sides of a quadrilateral PQRS. Show that ∠ P > ∠ R

    Maths-General
    Aim :- Prove ∠P  ∠R


     
    In triangle PQR,
    PQ < QR (PQ is smallest)
    so, ∠PRQ ∠QPR
    In triangle PSR,
    PS < SR  (RS is longest side)
    ∠ PRS ∠ RPS
    Adding above results we get,
    ∠PRQ + ∠PRS ∠QPR + ∠RPS
    ∠R  ∠P
    Hence proved

    parallel
    General
    Maths-

    Classify whether the following Surd is rational or irrational , justify. 3√24− 5√5
     

    Complete step by step solution:
    3 square root of 24 equals 6 square root of 6 (It is an irrational number)
    5 square root of 5 (It is an irrational number)
    Here both the numbers are irrational.
    On subtraction, we have 6 square root of 6 minus 5 square root of 5 , we get an irrational number.
    This cant be written in the form p over q, where p and q are integers.
    3 square root of 24 minus 5 square root of 5 is an irrational number.

    Classify whether the following Surd is rational or irrational , justify. 3√24− 5√5
     

    Maths-General
    Complete step by step solution:
    3 square root of 24 equals 6 square root of 6 (It is an irrational number)
    5 square root of 5 (It is an irrational number)
    Here both the numbers are irrational.
    On subtraction, we have 6 square root of 6 minus 5 square root of 5 , we get an irrational number.
    This cant be written in the form p over q, where p and q are integers.
    3 square root of 24 minus 5 square root of 5 is an irrational number.
    General
    Maths-

    In △ABC, AD and  are the bisectors of ∠A and ∠C respectively. AD and CE intersect at . If ∠ABC = 90^∘, then find ∠AOC.

    Hint :-As we know AD and CE are angular bisectors we find ∠OAC and  ∠OCA in terms of  ∠A and ∠C  .We find the value of  ∠AOC by using the sum of angles in the triangle is 180°

    Explanation :-
    Step 1:-  find∠OAC and  ∠OCA
    In ABC ,
    ∠A+∠B+∠C = 180°(sum of angles in the triangle)
    ∠A+∠C = 180°- 90°
    ∠A+∠C = 90°

    ∠OAC = ½ ∠A (AD is angular bisector )
    ∠OCA = ½ ∠C (CE is angular bisector )
    Step 2 :- Find ∠AOC
    In AOC ,
    ∠AOC + ∠OAC + ∠OCA = 180°(sum of angles in the triangle)
    ∠AOC + ½ ∠A  + ½ ∠C = 180°
    ∠AOC + ½ (∠A  + ∠C) = 180° (Substitute ∠A+∠C = 90°)
    ∠AOC + ½ (90°) = 180° ∠AOC  = 180°- 45° = 135°.
    ∴ ∠AOC   = 135°

    In △ABC, AD and  are the bisectors of ∠A and ∠C respectively. AD and CE intersect at . If ∠ABC = 90^∘, then find ∠AOC.

    Maths-General
    Hint :-As we know AD and CE are angular bisectors we find ∠OAC and  ∠OCA in terms of  ∠A and ∠C  .We find the value of  ∠AOC by using the sum of angles in the triangle is 180°

    Explanation :-
    Step 1:-  find∠OAC and  ∠OCA
    In ABC ,
    ∠A+∠B+∠C = 180°(sum of angles in the triangle)
    ∠A+∠C = 180°- 90°
    ∠A+∠C = 90°

    ∠OAC = ½ ∠A (AD is angular bisector )
    ∠OCA = ½ ∠C (CE is angular bisector )
    Step 2 :- Find ∠AOC
    In AOC ,
    ∠AOC + ∠OAC + ∠OCA = 180°(sum of angles in the triangle)
    ∠AOC + ½ ∠A  + ½ ∠C = 180°
    ∠AOC + ½ (∠A  + ∠C) = 180° (Substitute ∠A+∠C = 90°)
    ∠AOC + ½ (90°) = 180° ∠AOC  = 180°- 45° = 135°.
    ∴ ∠AOC   = 135°
    General
    Maths-

    A B C text  is a right triangle with  end text straight angle A equals 90 to the power of ring operator BL text  and  end text CM text  are its medians. Prove that  end text 4 open parentheses BL squared plus CM squared close parentheses equals 5 BC squared text .  end text

    Solution :-
    Aim  :- Prove that 4 open parentheses B L squared plus C M squared close parentheses equals 5 B C squared
    Hint :- Applying pythagoras theorem to both triangles ACM and BLA. find the BL2
    and CM2  add them and substitute proper conditions to prove the condition .
    Explanation(proof ) :-
    Given,   BL and CM are the medians of a triangle ABC
    M is midpoint of AB i.e AM = ½ AB
    L is mid point of AC i.e AL = ½ AC
    Applying pythagoras theorem to  triangle BLA we get ,
    B L squared equals A L squared plus A B squared
    As we know AL = ½ AC we getB L squared equals A B squared plus fraction numerator A C squared over denominator 4 end fraction minus Eq 1
    Applying pythagoras theorem to  triangle ACM we get ,
    C M squared equals A C squared plus A M squared
    As we know AM = ½ AB we get C M squared equals A C squared plus fraction numerator A B squared over denominator 4 end fraction minus Eq 2— Eq2
    Applying pythagoras theorem to  triangle ABC we get ,
    A B squared plus A C squared equals B C squared minus Eq 3


    Adding Eq1 and Eq2 we get ,
    B L squared plus C M squared equals A B squared plus fraction numerator A C squared over denominator 4 end fraction plus A C squared plus fraction numerator A B squared over denominator 4 end fraction
    not stretchy rightwards double arrow B L squared plus C M squared equals 5 over 4 open parentheses A B squared plus A C squared close parentheses
    Substitute Eq 3 in the above , we get 
    B L squared plus C M squared equals 5 over 4 open parentheses B C squared close parentheses not stretchy rightwards double arrow 4 open parentheses B L squared plus C M squared close parentheses equals 5 B C squared
    Hence proved

    A B C text  is a right triangle with  end text straight angle A equals 90 to the power of ring operator BL text  and  end text CM text  are its medians. Prove that  end text 4 open parentheses BL squared plus CM squared close parentheses equals 5 BC squared text .  end text

    Maths-General
    Solution :-
    Aim  :- Prove that 4 open parentheses B L squared plus C M squared close parentheses equals 5 B C squared
    Hint :- Applying pythagoras theorem to both triangles ACM and BLA. find the BL2
    and CM2  add them and substitute proper conditions to prove the condition .
    Explanation(proof ) :-
    Given,   BL and CM are the medians of a triangle ABC
    M is midpoint of AB i.e AM = ½ AB
    L is mid point of AC i.e AL = ½ AC
    Applying pythagoras theorem to  triangle BLA we get ,
    B L squared equals A L squared plus A B squared
    As we know AL = ½ AC we getB L squared equals A B squared plus fraction numerator A C squared over denominator 4 end fraction minus Eq 1
    Applying pythagoras theorem to  triangle ACM we get ,
    C M squared equals A C squared plus A M squared
    As we know AM = ½ AB we get C M squared equals A C squared plus fraction numerator A B squared over denominator 4 end fraction minus Eq 2— Eq2
    Applying pythagoras theorem to  triangle ABC we get ,
    A B squared plus A C squared equals B C squared minus Eq 3


    Adding Eq1 and Eq2 we get ,
    B L squared plus C M squared equals A B squared plus fraction numerator A C squared over denominator 4 end fraction plus A C squared plus fraction numerator A B squared over denominator 4 end fraction
    not stretchy rightwards double arrow B L squared plus C M squared equals 5 over 4 open parentheses A B squared plus A C squared close parentheses
    Substitute Eq 3 in the above , we get 
    B L squared plus C M squared equals 5 over 4 open parentheses B C squared close parentheses not stretchy rightwards double arrow 4 open parentheses B L squared plus C M squared close parentheses equals 5 B C squared
    Hence proved
    parallel

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