Maths-
General
Easy

Question

The perpendicular bisectors of the sides of a triangle ABC meet at point I. prove that IA=IB=IC

Hint:

Find the congruence rules used and thus find the equal sides.

The correct answer is: IA = IB = IC




    In ΔABC, in which AD is perpendicular bisector of BC
    BE is perpendicular bisector of CA
    CF is perpendicular bisector of AB
    AD, BE and CF meet at I
    To prove IA=IB=IC
    In ΔBID and ΔCID
    BD = DC  (given)
    straight angle B D I equals straight angle C D I equals 90 to the power of ring operator(AD is perpendicular bisector of BC)
    ID = ID (common)
    Hence by SAS congruence rule, we have
    ΔBID  ΔCID
    So, IB = IC (Corresponding part of congruent triangle)
    Similarly, In ΔCIE and ΔAIE
    CE = AE (given)
    straight angle C E I equals straight angle A E I equals 90 to the power of ring operator(BE is perpendicular bisector of AC)
    IE = IE (common)
    By SAS congruence rule, ΔCIE  ΔAIE
    IC = IA (Corresponding part of congruent triangle)
    Thus, IA = IB = IC

    Related Questions to study

    General
    Maths-

    Use prime factorization method to find the square root of each of the following :
    c) 1089

    c) Let's find the prime factorization of the number.
    1089
    = 3 × 363
    = 3 × 3 × 121
    = 3 × 3 × 11 ×1 1
    equals 3 squared cross times 11 squared equals left parenthesis 3 cross times 11 right parenthesis squared equals 33 squared
    The square root of 1089 is 33.

    Use prime factorization method to find the square root of each of the following :
    c) 1089

    Maths-General
    c) Let's find the prime factorization of the number.
    1089
    = 3 × 363
    = 3 × 3 × 121
    = 3 × 3 × 11 ×1 1
    equals 3 squared cross times 11 squared equals left parenthesis 3 cross times 11 right parenthesis squared equals 33 squared
    The square root of 1089 is 33.
    General
    Maths-

    Use prime factorization method to find the square root of each of the following :
    b)175

    Let's find the prime factorization of the number
    30625  = 5 × 6125
    = 5 × 5 × 1225
    = 5 × 5 × 5 × 245
    = 5 × 5 × 5  5 × 49
    = 5  ×5 × 5 × 5 × 7 × 7
    equals 5 to the power of 4 cross times 7 squared equals 25 squared cross times 7 squared equals left parenthesis 25 cross times 7 right parenthesis squared equals left parenthesis 175 right parenthesis squared The square root of 30625 is 175.

    Use prime factorization method to find the square root of each of the following :
    b)175

    Maths-General
    Let's find the prime factorization of the number
    30625  = 5 × 6125
    = 5 × 5 × 1225
    = 5 × 5 × 5 × 245
    = 5 × 5 × 5  5 × 49
    = 5  ×5 × 5 × 5 × 7 × 7
    equals 5 to the power of 4 cross times 7 squared equals 25 squared cross times 7 squared equals left parenthesis 25 cross times 7 right parenthesis squared equals left parenthesis 175 right parenthesis squared The square root of 30625 is 175.
    General
    Maths-

    Use prime factorization method to find the square root of each of the following :
    a)11025

    Explanation :-
    a) Let's find the prime factorization of the  number.
    11025  =3 × 3675
    = 3 × 3 × 1225
    = 3 × 3 × 5*245
    = 3 × 3 × 5 × 5 × 49
    = 3 × 3 × 5 × 5 × 7 × 7
    3 squared cross times 5 squared cross times 7 squared equals left parenthesis 3 cross times 5 cross times 7 right parenthesis squared equals left parenthesis 105 right parenthesis squared
    The square root of 11236 is 105.

    Use prime factorization method to find the square root of each of the following :
    a)11025

    Maths-General
    Explanation :-
    a) Let's find the prime factorization of the  number.
    11025  =3 × 3675
    = 3 × 3 × 1225
    = 3 × 3 × 5*245
    = 3 × 3 × 5 × 5 × 49
    = 3 × 3 × 5 × 5 × 7 × 7
    3 squared cross times 5 squared cross times 7 squared equals left parenthesis 3 cross times 5 cross times 7 right parenthesis squared equals left parenthesis 105 right parenthesis squared
    The square root of 11236 is 105.
    parallel
    General
    Maths-

    In Triangle ABC , AB= AC, and D is a point on AB. Such that AD=DC=BC. Show that straight angle B A C equals 36 to the power of ring operator



    In ΔABC, we have
    CD = BC (given)
    Let straight angle D B C equals straight angle B D C equals x to the power of ring operator (since angles opposite to the equal sides are always equal)
    Since AB = AC, straight angle A B C equals straight angle A C B equals x to the power of ring operator
    In ΔABC,
    straight angle A B C plus straight angle A C B plus straight angle B A C equals 180 to the power of ring operator(By angle sum property of the triangle)
    table attributes columnspacing 1em end attributes row cell straight angle B A C plus x plus x equals 180 to the power of ring operator end cell row cell straight angle B A C equals 180 to the power of ring operator minus 2 x end cell end table
    Since AD = CD (given),
    straight angle A C D equals straight angle D A C
    So,straight angle D A C equals 180 to the power of ring operator minus 2 x
    Now,straight angle B C D equals straight angle A C B minus straight angle A C D equals x minus open parentheses 180 to the power of ring operator minus 2 x close parentheses equals 3 x minus 180 to the power of ring operator
    straight angle B D C plus straight angle D B C plus straight angle B C D equals 180 to the power of ring operator(angle sum property)
    x plus x plus 3 x minus 180 equals 180 to the power of ring operator
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 5 x equals 360 to the power of ring operator end cell row cell not stretchy rightwards double arrow x equals 72 to the power of ring operator end cell end table
    Now, straight angle B A C equals 180 minus 2 x equals 180 minus 2 left parenthesis 72 right parenthesis equals 36 to the power of ring operator.

    In Triangle ABC , AB= AC, and D is a point on AB. Such that AD=DC=BC. Show that straight angle B A C equals 36 to the power of ring operator

    Maths-General


    In ΔABC, we have
    CD = BC (given)
    Let straight angle D B C equals straight angle B D C equals x to the power of ring operator (since angles opposite to the equal sides are always equal)
    Since AB = AC, straight angle A B C equals straight angle A C B equals x to the power of ring operator
    In ΔABC,
    straight angle A B C plus straight angle A C B plus straight angle B A C equals 180 to the power of ring operator(By angle sum property of the triangle)
    table attributes columnspacing 1em end attributes row cell straight angle B A C plus x plus x equals 180 to the power of ring operator end cell row cell straight angle B A C equals 180 to the power of ring operator minus 2 x end cell end table
    Since AD = CD (given),
    straight angle A C D equals straight angle D A C
    So,straight angle D A C equals 180 to the power of ring operator minus 2 x
    Now,straight angle B C D equals straight angle A C B minus straight angle A C D equals x minus open parentheses 180 to the power of ring operator minus 2 x close parentheses equals 3 x minus 180 to the power of ring operator
    straight angle B D C plus straight angle D B C plus straight angle B C D equals 180 to the power of ring operator(angle sum property)
    x plus x plus 3 x minus 180 equals 180 to the power of ring operator
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 5 x equals 360 to the power of ring operator end cell row cell not stretchy rightwards double arrow x equals 72 to the power of ring operator end cell end table
    Now, straight angle B A C equals 180 minus 2 x equals 180 minus 2 left parenthesis 72 right parenthesis equals 36 to the power of ring operator.
    General
    Maths-

    In the triangle ABC, the angles at A and C are 40° and 95° respectively. If N is the foot of the perpendicular from C to AB , prove that CN = BN.



    In ΔABC, We have
    straight angle C A B equals 40 to the power of ring operator (given)
    straight angle A C B equals 95 to the power of ring operator (given)
    straight angle C B A equals 180 minus left parenthesis 95 plus 40 right parenthesis equals 45 to the power of ring operator(sum of interior angles of a triangle is 180)
    We know that, straight angle C N B equals 90 to the power of ring operator ( CN AB)
    Hence, straight angle N C B equals 180 minus left parenthesis 90 plus 45 right parenthesis equals 45 to the power of ring operator
    So, In ΔCNB, the base angles are equal (straight angle N C B equals straight angle C B N equals 45 to the power of ring operator) hence sides
    opposite to equal base angles are also equal. Hence we get, CN = NB

    In the triangle ABC, the angles at A and C are 40° and 95° respectively. If N is the foot of the perpendicular from C to AB , prove that CN = BN.

    Maths-General


    In ΔABC, We have
    straight angle C A B equals 40 to the power of ring operator (given)
    straight angle A C B equals 95 to the power of ring operator (given)
    straight angle C B A equals 180 minus left parenthesis 95 plus 40 right parenthesis equals 45 to the power of ring operator(sum of interior angles of a triangle is 180)
    We know that, straight angle C N B equals 90 to the power of ring operator ( CN AB)
    Hence, straight angle N C B equals 180 minus left parenthesis 90 plus 45 right parenthesis equals 45 to the power of ring operator
    So, In ΔCNB, the base angles are equal (straight angle N C B equals straight angle C B N equals 45 to the power of ring operator) hence sides
    opposite to equal base angles are also equal. Hence we get, CN = NB
    General
    Maths-

    Given XYZ is a triangle in which XW is the bisector of the angle x and XW ⊥ yz. Prove that xy= xz.



    In ΔXYW  and  ΔXZW
    straight angle W X Y equals straight angle W X Z(since XW is the bisector of angle X)
    straight angle X W Y equals straight angle X W Z(XW YZ)
    XW = XW (common side)
    Hence by ASA congruence rule, we have ΔXYW  ⩭   ΔXZW
    So, XY = XZ (Corresponding part of congruence triangles)

    Given XYZ is a triangle in which XW is the bisector of the angle x and XW ⊥ yz. Prove that xy= xz.

    Maths-General


    In ΔXYW  and  ΔXZW
    straight angle W X Y equals straight angle W X Z(since XW is the bisector of angle X)
    straight angle X W Y equals straight angle X W Z(XW YZ)
    XW = XW (common side)
    Hence by ASA congruence rule, we have ΔXYW  ⩭   ΔXZW
    So, XY = XZ (Corresponding part of congruence triangles)
    parallel
    General
    Maths-

    ABC is a triangle in which altitudes BE and CF to sides AC and AB respectively are equal. Show that
    (i) ΔABE ⩭ ΔACF
    (ii)  AB = AC



    (i) In ΔABE and ΔACF, we have
    straight angle B A E equals straight angle C A F ( Common angle)
    straight angle A E B equals straight angle A F C (Since BE ⟂ AC and CF ⟂  AB)
    BE = CF (given)
    Hence, by AAS criterion ΔABE  ⩭ ΔACF
    (ii) Since ΔABE  ⩭  ΔACF, we have
    AB = AC (Corresponding part of congruent triangles)

    ABC is a triangle in which altitudes BE and CF to sides AC and AB respectively are equal. Show that
    (i) ΔABE ⩭ ΔACF
    (ii)  AB = AC

    Maths-General


    (i) In ΔABE and ΔACF, we have
    straight angle B A E equals straight angle C A F ( Common angle)
    straight angle A E B equals straight angle A F C (Since BE ⟂ AC and CF ⟂  AB)
    BE = CF (given)
    Hence, by AAS criterion ΔABE  ⩭ ΔACF
    (ii) Since ΔABE  ⩭  ΔACF, we have
    AB = AC (Corresponding part of congruent triangles)
    General
    Maths-

    In the figure, OA = OD and OB = OC. Show that:
    (i) AOB ⩭ DOC
    (ii)  AB || CD

    (i) In ΔAOB and ΔDOC, we have
    AO = OD (given)
    BO = OC (given)
    Since CB and AD intersects, we have
    straight angle A O B equals straight angle D O C (vertically opposite angles)
    So by SAS congruence rule, we have
    ΔAOB ⩭ ΔDOC
    (ii)Since ΔAOB ⩭ ΔDOC, we have
    straight angle O A B equals straight angle O D C ( Corresponding part of congruent triangles)
    But they form alternate interior angles and since they are equal , AB || CD.

    In the figure, OA = OD and OB = OC. Show that:
    (i) AOB ⩭ DOC
    (ii)  AB || CD

    Maths-General
    (i) In ΔAOB and ΔDOC, we have
    AO = OD (given)
    BO = OC (given)
    Since CB and AD intersects, we have
    straight angle A O B equals straight angle D O C (vertically opposite angles)
    So by SAS congruence rule, we have
    ΔAOB ⩭ ΔDOC
    (ii)Since ΔAOB ⩭ ΔDOC, we have
    straight angle O A B equals straight angle O D C ( Corresponding part of congruent triangles)
    But they form alternate interior angles and since they are equal , AB || CD.
    General
    Maths-

    If AD ⟂ CD and BC ⟂ CD , AQ = PB and DP = CQ. Prove that straight angle D A Q equals straight angle C B P

    AD ⊥ CD (given)
    and BC ⊥ CD (given)
    AQ = BP and DP = CQ (given)
    To prove : ∠ DAQ = ∠ CBP
    AD ⊥ CD (given)
    and BC ⊥ CD (given)
    ∴ ∠ D = ∠ C (each 90°)
    ∵ DP = CQ (given)
    On adding PQ to both sides. we get
    DP + PQ = PQ + CQ…(i)
    ⇒ DQ = CP
    Now, in right angles ADQ and BCP
    ∴ AQ = BP (hypotenuse)
    DQ = CP (from (i))
    ∴ Δ ADQ ≡ Δ BCP (Right angle hypotenuse side (RHS congruence rule))
    ∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)

    If AD ⟂ CD and BC ⟂ CD , AQ = PB and DP = CQ. Prove that straight angle D A Q equals straight angle C B P

    Maths-General
    AD ⊥ CD (given)
    and BC ⊥ CD (given)
    AQ = BP and DP = CQ (given)
    To prove : ∠ DAQ = ∠ CBP
    AD ⊥ CD (given)
    and BC ⊥ CD (given)
    ∴ ∠ D = ∠ C (each 90°)
    ∵ DP = CQ (given)
    On adding PQ to both sides. we get
    DP + PQ = PQ + CQ…(i)
    ⇒ DQ = CP
    Now, in right angles ADQ and BCP
    ∴ AQ = BP (hypotenuse)
    DQ = CP (from (i))
    ∴ Δ ADQ ≡ Δ BCP (Right angle hypotenuse side (RHS congruence rule))
    ∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)
    parallel
    General
    Maths-

    Prove that the medians bisecting the equal sides of an isosceles triangle are also equal. If OA = OB and OD = OC, show that:
    (i) AOD ⩭ BOC
    (ii)  AD || BC

    (i) In ΔAOD and ΔBOC, we have
    AO = OB (given)
    OD = OC (given)
    Since CD and AB intersects, we have
    straight angle A O D equals straight angle B O C(vertically opposite angles)
    So by SAS congruence rule, we have
    ΔAOD ⩭ ΔBOC
    (ii) SinceΔAOD ⩭ ΔBOC, we have
    straight angle O A D equals straight angle O B C( Corresponding part of congruent triangles)
    But they form alternate interior angles and since they are equal,  AD || BC.

    Prove that the medians bisecting the equal sides of an isosceles triangle are also equal. If OA = OB and OD = OC, show that:
    (i) AOD ⩭ BOC
    (ii)  AD || BC

    Maths-General
    (i) In ΔAOD and ΔBOC, we have
    AO = OB (given)
    OD = OC (given)
    Since CD and AB intersects, we have
    straight angle A O D equals straight angle B O C(vertically opposite angles)
    So by SAS congruence rule, we have
    ΔAOD ⩭ ΔBOC
    (ii) SinceΔAOD ⩭ ΔBOC, we have
    straight angle O A D equals straight angle O B C( Corresponding part of congruent triangles)
    But they form alternate interior angles and since they are equal,  AD || BC.
    General
    Maths-

    Kumar wants to spread a carpet on the floor of his classroom. The floor is a square
    with an area of 4160.25 square feet. What is the length of carpet on each side?

    Ans :- 64.5 feet
    Explanation :-
    Let the length of the classroom (square)  be a cm .
    The we get area of class room =
    So we get , a squared equals 4160.25 not stretchy rightwards double arrow a equals square root of 4160.25 end root
    We know that the length of  side of square is length of the carpet to be used to cover it
    I.e length of side of square = the required length of square carpet.

    Using the long division method to find value of square root of 4160.25

    a equals square root of 4160.25 end root equals 64.5 text  (feet)  end text
    The length of carpet be on each side is 64.5 text  (feet)  end text.

    Kumar wants to spread a carpet on the floor of his classroom. The floor is a square
    with an area of 4160.25 square feet. What is the length of carpet on each side?

    Maths-General
    Ans :- 64.5 feet
    Explanation :-
    Let the length of the classroom (square)  be a cm .
    The we get area of class room =
    So we get , a squared equals 4160.25 not stretchy rightwards double arrow a equals square root of 4160.25 end root
    We know that the length of  side of square is length of the carpet to be used to cover it
    I.e length of side of square = the required length of square carpet.

    Using the long division method to find value of square root of 4160.25

    a equals square root of 4160.25 end root equals 64.5 text  (feet)  end text
    The length of carpet be on each side is 64.5 text  (feet)  end text.
    General
    Maths-

    The volume of a cubical box is 12167 cm3. Find the side of the box and its 6 sides area?

    Length of the side of the box is 23 cm. Area of each side is 529 .
    Area of 6 sides combined = 3174 cm2.
    Explanation :-
    Step1:- find the length of side of the cube
    Let the length of the side of the cube be a .
    The volume of cube = a cubed
    Then, a cubed equals 3174 equals 23 cross times 529 equals 23 cross times 23 cross times 23 equals 23 cubed
    The length of the side of the square is 23 cm .
    Step2:- find the  side area of the cube
    The side are  of the cube = a squared equals 23 squared equals 529 cm squared
    Step 3:- find the side area of 6 sides .
    In the cube the side area of all sides is equal so, we get as the area of  6 sides.
    The side area of 6 sides = .6 straight a squared equals 6 cross times 23 squared equals 6 cross times 529 equals 3174 cm squared
    We get that Length of the side of the box is 23 cm. Area of 6 sides combined = 3174 cm2

    The volume of a cubical box is 12167 cm3. Find the side of the box and its 6 sides area?

    Maths-General
    Length of the side of the box is 23 cm. Area of each side is 529 .
    Area of 6 sides combined = 3174 cm2.
    Explanation :-
    Step1:- find the length of side of the cube
    Let the length of the side of the cube be a .
    The volume of cube = a cubed
    Then, a cubed equals 3174 equals 23 cross times 529 equals 23 cross times 23 cross times 23 equals 23 cubed
    The length of the side of the square is 23 cm .
    Step2:- find the  side area of the cube
    The side are  of the cube = a squared equals 23 squared equals 529 cm squared
    Step 3:- find the side area of 6 sides .
    In the cube the side area of all sides is equal so, we get as the area of  6 sides.
    The side area of 6 sides = .6 straight a squared equals 6 cross times 23 squared equals 6 cross times 529 equals 3174 cm squared
    We get that Length of the side of the box is 23 cm. Area of 6 sides combined = 3174 cm2
    parallel
    General
    Maths-

    The area of a square is 289 m2, find its perimeter?

    Ans :- 68
    Explanation :-
    Step 1:-
    Let the length of the square be a
    The area of square = (length of side)2  equals a squared
    Then, a squared equals 289 equals 17 squared not stretchy rightwards double arrow a equals 17 m
    Step 2:- find the perimeter of square by using length
    We find the perimeter of the square = 4a = 4 × 17 = 68m.
    The perimeter of square whose area is 289 (m2) is 68 m.

    The area of a square is 289 m2, find its perimeter?

    Maths-General
    Ans :- 68
    Explanation :-
    Step 1:-
    Let the length of the square be a
    The area of square = (length of side)2  equals a squared
    Then, a squared equals 289 equals 17 squared not stretchy rightwards double arrow a equals 17 m
    Step 2:- find the perimeter of square by using length
    We find the perimeter of the square = 4a = 4 × 17 = 68m.
    The perimeter of square whose area is 289 (m2) is 68 m.
    General
    Maths-

    What is the least seven digits which is a perfect square. Also find the square root of the number so obtained

    Ans :- 1 00 00 00 (106)
    Explanation :-
    The least seven digit number is 1 00 00 00(106) which is already a perfect square as the given number is the perfect square of 1000

    What is the least seven digits which is a perfect square. Also find the square root of the number so obtained

    Maths-General
    Ans :- 1 00 00 00 (106)
    Explanation :-
    The least seven digit number is 1 00 00 00(106) which is already a perfect square as the given number is the perfect square of 1000
    General
    Maths-

    Find the least number which must be subtracted from 23416 to make it a perfect square.

    Ans :- 7
    Explanation :-

    We need to subtract the 7 from 23416 to make it a perfect square.

    Find the least number which must be subtracted from 23416 to make it a perfect square.

    Maths-General
    Ans :- 7
    Explanation :-

    We need to subtract the 7 from 23416 to make it a perfect square.
    parallel

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