Maths-
General
Easy

Question

Simplify 4√12+5√27 - 3√75 +√300

Hint:

Use prime factorization.

The correct answer is: 18√3


    Complete step by step solution:
    On prime factorization, we can write
    4 square root of 12 equals 4 cross times square root of 2 cross times 2 cross times 3 end root comma 5 square root of 27 equals 5 cross times square root of 3 cross times 3 cross times 3 end root comma 3 square root of 75 equals 3 cross times square root of 3 cross times 5 cross times 5 end root
    and square root of 300 equals square root of 2 cross times 2 cross times 3 cross times 5 cross times 5 end root
    This can be written as 8 square root of 3 comma 15 square root of 3 comma 15 square root of 3 and 10 square root of 3 respectively.
    So, 4 square root of 12 plus 5 square root of 27 minus 3 square root of 75 plus square root of 300 equals 8 square root of 3 plus 15 square root of 3 minus 15 square root of 3 plus 10 square root of 3 equals 18 square root of 3

    Related Questions to study

    General
    Maths-

    Simplify √45 - 3√20 +4√5

    Complete step by step solution:
    On prime factorization, we can write
    square root of 45 equals square root of 3 cross times 3 cross times 5 end root comma 3 square root of 20 equals 3 cross times square root of 2 cross times 2 cross times 5 end root text  and  end text 4 square root of 5 equals 4 square root of 5
    This can be written as 3 square root of 5 comma 6 square root of 5 and 4 square root of 5 respectively.
    So, square root of 45 minus 3 square root of 20 plus 4 square root of 5 equals 3 square root of 5 minus 6 square root of 5 plus 4 square root of 5 equals 1 square root of 5 equals square root of 5

    Simplify √45 - 3√20 +4√5

    Maths-General
    Complete step by step solution:
    On prime factorization, we can write
    square root of 45 equals square root of 3 cross times 3 cross times 5 end root comma 3 square root of 20 equals 3 cross times square root of 2 cross times 2 cross times 5 end root text  and  end text 4 square root of 5 equals 4 square root of 5
    This can be written as 3 square root of 5 comma 6 square root of 5 and 4 square root of 5 respectively.
    So, square root of 45 minus 3 square root of 20 plus 4 square root of 5 equals 3 square root of 5 minus 6 square root of 5 plus 4 square root of 5 equals 1 square root of 5 equals square root of 5
    General
    Maths-

    Arrange in ascending order of magnitude square root of 5 comma root index 8 of 11 and 2 root index 6 of 3

    Complete step by step solution:
    Here we can write, square root of 5 equals 5 to the power of 1 half end exponent comma root index 8 of 11 equals 11 to the power of 1 third end exponent text  and  end text 2 root index 6 of 3 equals root index 6 of 12 equals 12 to the power of 1 over 6 end exponent
    On taking the LCM of 2,3,6 we have LCM as 6.
    So, not stretchy rightwards double arrow 5 to the power of 1 half end exponent equals 5 to the power of fraction numerator 1 cross times 3 over denominator 2 cross times 3 end fraction end exponent equals 5 to the power of 3 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 5 cubed close parentheses to the power of 1 over 6 end exponent equals 125 to the power of 1 over 6 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 11 to the power of 1 third end exponent equals 11 to the power of fraction numerator 1 cross times 2 over denominator 3 cross times 2 end fraction end exponent equals 11 to the power of 2 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 11 squared close parentheses to the power of 1 over 6 end exponent equals 121 to the power of 1 over 6 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 12 to the power of 1 over 6 end exponent
    Now, we have 125 to the power of 1 over 6 end exponent comma 121 to the power of 1 over 6 end exponent and 12 to the power of 1 over 6 end exponent
    So, here ascending order is 12 < 121 < 125
    not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 121 to the power of 1 over 6 end exponent less than 125 to the power of 1 over 6 end exponent
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 11 to the power of 1 third end exponent less than 5 to the power of 1 half end exponent end cell row cell not stretchy rightwards double arrow root index 6 of 12 less than cube root of 11 less than square root of 5 end cell end table

    Arrange in ascending order of magnitude square root of 5 comma root index 8 of 11 and 2 root index 6 of 3

    Maths-General
    Complete step by step solution:
    Here we can write, square root of 5 equals 5 to the power of 1 half end exponent comma root index 8 of 11 equals 11 to the power of 1 third end exponent text  and  end text 2 root index 6 of 3 equals root index 6 of 12 equals 12 to the power of 1 over 6 end exponent
    On taking the LCM of 2,3,6 we have LCM as 6.
    So, not stretchy rightwards double arrow 5 to the power of 1 half end exponent equals 5 to the power of fraction numerator 1 cross times 3 over denominator 2 cross times 3 end fraction end exponent equals 5 to the power of 3 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 5 cubed close parentheses to the power of 1 over 6 end exponent equals 125 to the power of 1 over 6 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 11 to the power of 1 third end exponent equals 11 to the power of fraction numerator 1 cross times 2 over denominator 3 cross times 2 end fraction end exponent equals 11 to the power of 2 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 11 squared close parentheses to the power of 1 over 6 end exponent equals 121 to the power of 1 over 6 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 12 to the power of 1 over 6 end exponent
    Now, we have 125 to the power of 1 over 6 end exponent comma 121 to the power of 1 over 6 end exponent and 12 to the power of 1 over 6 end exponent
    So, here ascending order is 12 < 121 < 125
    not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 121 to the power of 1 over 6 end exponent less than 125 to the power of 1 over 6 end exponent
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 11 to the power of 1 third end exponent less than 5 to the power of 1 half end exponent end cell row cell not stretchy rightwards double arrow root index 6 of 12 less than cube root of 11 less than square root of 5 end cell end table
    General
    Maths-

    The Diagonals of a square ABCD meet at O. Prove that A B squared equals 2 A O squared


    Aim  :- Prove that AB squared equals 2 AO squared


    Let length of side of square be a
    Then ,applying using pythagoras theorem in triangle ADC
    We get A C squared equals A D squared plus D C squared not stretchy rightwards double arrow A C squared equals a squared plus a squared
    not stretchy rightwards double arrow AC squared equals 2 straight a squared
    As diagonals bisect each other AC = 2OA
    20 straight A squared equals straight a squared where a =AB
    We get AB squared equals 2 AO squared
    Hence proved

    The Diagonals of a square ABCD meet at O. Prove that A B squared equals 2 A O squared

    Maths-General

    Aim  :- Prove that AB squared equals 2 AO squared


    Let length of side of square be a
    Then ,applying using pythagoras theorem in triangle ADC
    We get A C squared equals A D squared plus D C squared not stretchy rightwards double arrow A C squared equals a squared plus a squared
    not stretchy rightwards double arrow AC squared equals 2 straight a squared
    As diagonals bisect each other AC = 2OA
    20 straight A squared equals straight a squared where a =AB
    We get AB squared equals 2 AO squared
    Hence proved

    parallel
    General
    Maths-

    △ PQR is a right triangle with ∠Q=90 M is the midpoint of QR. Prove that

    P M squared equals P R squared minus 3 Q M squared


    As M is mid point of QR then, QM = ½ RQ
    We get PR squared equals PQ squared plus left parenthesis 2 QM right parenthesis squared not stretchy rightwards double arrow PR squared equals PQ squared plus 4 Q straight Q squared —Eq1
    Applying Pythagoras theorem For triangle PQM
    PM squared equals PQ squared plus QM squared —Eq2
    Substitute Eq2 in Eq1
    We get
    PR squared equals PQ squared plus QM squared plus 3 Q straight Q squared
    not stretchy rightwards double arrow PR squared equals PM squared plus 3 QM squared not stretchy rightwards double arrow PM squared equals PR squared minus 3 QM squared
    Hence proved

    △ PQR is a right triangle with ∠Q=90 M is the midpoint of QR. Prove that

    Maths-General
    P M squared equals P R squared minus 3 Q M squared


    As M is mid point of QR then, QM = ½ RQ
    We get PR squared equals PQ squared plus left parenthesis 2 QM right parenthesis squared not stretchy rightwards double arrow PR squared equals PQ squared plus 4 Q straight Q squared —Eq1
    Applying Pythagoras theorem For triangle PQM
    PM squared equals PQ squared plus QM squared —Eq2
    Substitute Eq2 in Eq1
    We get
    PR squared equals PQ squared plus QM squared plus 3 Q straight Q squared
    not stretchy rightwards double arrow PR squared equals PM squared plus 3 QM squared not stretchy rightwards double arrow PM squared equals PR squared minus 3 QM squared
    Hence proved

    General
    Maths-

    In Triangle A B C comma straight angle A D B equals 90 to the power of ring operator. Prove thatA B squared equals A C squared plus B C squared minus 2 B C times C D

    Aim  :- Prove that  A B squared equals A C squared plus B C squared minus 2 B C times C D
    Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get
    A C squared equals A D squared plus C D squared — Eq1
    Applying pythagoras theorem in Δ ABD ,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 minus Eq 2
    Substitute Eq1 in Eq2 ,

    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

    text  By using  end text a squared minus b squared equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis text  and  end text B C equals B D plus C D text  we get,  end text

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

                   not stretchy rightwards double arrow A C squared equals A B squared plus B C left parenthesis 2 C D minus B C right parenthesis

                      A C squared equals A B squared plus 2 B C. C D minus B C squared

                     A C squared plus B C squared minus 2 B C times C D equals A B squared
    therefore space A B squared equals A C squared plus B C squared minus 2 B C times C D
    Hence proved

    In Triangle A B C comma straight angle A D B equals 90 to the power of ring operator. Prove thatA B squared equals A C squared plus B C squared minus 2 B C times C D

    Maths-General
    Aim  :- Prove that  A B squared equals A C squared plus B C squared minus 2 B C times C D
    Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get
    A C squared equals A D squared plus C D squared — Eq1
    Applying pythagoras theorem in Δ ABD ,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 minus Eq 2
    Substitute Eq1 in Eq2 ,

    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

    text  By using  end text a squared minus b squared equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis text  and  end text B C equals B D plus C D text  we get,  end text

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

                   not stretchy rightwards double arrow A C squared equals A B squared plus B C left parenthesis 2 C D minus B C right parenthesis

                      A C squared equals A B squared plus 2 B C. C D minus B C squared

                     A C squared plus B C squared minus 2 B C times C D equals A B squared
    therefore space A B squared equals A C squared plus B C squared minus 2 B C times C D
    Hence proved

    General
    Maths-

    In straight triangle P Q R comma Q M perpendicular R P and P R squared minus P Q squared equals Q R squared prove that  QM squared equals PM cross times MR.

    Solution :-
    Aim  :- Prove that Q M squared equals P M cross times M R

    Hint :- using inverse of pythagoras theorem, we get right angle at Q,
    Now using angles prove the similarity of QMR and PMQ triangles. And find the
    ratios to get the desired property
    Explanation(proof ) :-
    Given , P R squared minus P Q squared equals Q R squared not stretchy rightwards double arrow P R squared equals P Q squared plus Q R squared
    Here ,PR is hypotenuse and angle Q is right angle.
    Then straight angle P equals 90 minus straight angle R
    straight angle P equals 90 minus straight angle R text  So we get  end text straight angle P equals straight angle M Q R
    We know straight angle M equals straight angle M equals 90 left parenthesis text  degrees  end text right parenthesis
    By AA similarity

    straight capital delta Q M R tilde straight triangle P M Q

    By similarity we get fraction numerator Q M over denominator P M end fraction equals fraction numerator R M over denominator M Q end fraction not stretchy rightwards double arrow Q M squared equals P M cross times M R
    Hence proved.

     

    In straight triangle P Q R comma Q M perpendicular R P and P R squared minus P Q squared equals Q R squared prove that  QM squared equals PM cross times MR.

    Maths-General
    Solution :-
    Aim  :- Prove that Q M squared equals P M cross times M R

    Hint :- using inverse of pythagoras theorem, we get right angle at Q,
    Now using angles prove the similarity of QMR and PMQ triangles. And find the
    ratios to get the desired property
    Explanation(proof ) :-
    Given , P R squared minus P Q squared equals Q R squared not stretchy rightwards double arrow P R squared equals P Q squared plus Q R squared
    Here ,PR is hypotenuse and angle Q is right angle.
    Then straight angle P equals 90 minus straight angle R
    straight angle P equals 90 minus straight angle R text  So we get  end text straight angle P equals straight angle M Q R
    We know straight angle M equals straight angle M equals 90 left parenthesis text  degrees  end text right parenthesis
    By AA similarity

    straight capital delta Q M R tilde straight triangle P M Q

    By similarity we get fraction numerator Q M over denominator P M end fraction equals fraction numerator R M over denominator M Q end fraction not stretchy rightwards double arrow Q M squared equals P M cross times M R
    Hence proved.

     
    parallel
    General
    Maths-

    In the figure,P Q perpendicular P S comma P Q vertical line vertical line S R comma straight angle S Q R equals 28 to the power of ring operator text  and  end text straight angle Q R T equals 65 to the power of ring operator text ,  end textthen find the value of x and y.

    Explanation :-
    Step 1:-Find ∠QSR
    We  know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.
    ∠QSR +∠SQR = ∠QRT
    ∠QSR +28° = 65°
    ∠QSR = 65° -28° = 37°
    Step 2:- Find value of x
    PQ || SR and QS intersects both the lines
    We get angle x =  ∠QSR ( alternate interior angles)
    x =37°
    Step 3:- Find value of y
    In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)
    90°+37° +y =180 °
    y = 180-127
    y = 53 °

    In the figure,P Q perpendicular P S comma P Q vertical line vertical line S R comma straight angle S Q R equals 28 to the power of ring operator text  and  end text straight angle Q R T equals 65 to the power of ring operator text ,  end textthen find the value of x and y.

    Maths-General
    Explanation :-
    Step 1:-Find ∠QSR
    We  know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.
    ∠QSR +∠SQR = ∠QRT
    ∠QSR +28° = 65°
    ∠QSR = 65° -28° = 37°
    Step 2:- Find value of x
    PQ || SR and QS intersects both the lines
    We get angle x =  ∠QSR ( alternate interior angles)
    x =37°
    Step 3:- Find value of y
    In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)
    90°+37° +y =180 °
    y = 180-127
    y = 53 °

    General
    Maths-

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

    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

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

    Maths-General
    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
    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

    parallel
    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 .

    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
    parallel
    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 
    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
    parallel

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