Maths-
General
Easy

Question

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

Hint:

Find the congruence rule used and thus find whether sides are equal.

The correct answer is: 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)

    Related Questions to study

    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
    General
    Maths-

    Find the least number which must be added to 732733 to obtain a perfect square.

    Ans :- 3
    Explanation :-

    Number = 732733
    Square Root = 732733 + 3 = 732736 = (86)2 .
    We need to add 3 to the given number to make it a perfect square.

    Find the least number which must be added to 732733 to obtain a perfect square.

    Maths-General
    Ans :- 3
    Explanation :-

    Number = 732733
    Square Root = 732733 + 3 = 732736 = (86)2 .
    We need to add 3 to the given number to make it a perfect square.
    General
    Maths-

    In the given figure, PQ || RS. Check whether ΔPOQ i congruent to ΔSOR Give reason for your answer:

    In ΔPOQ and ΔSOR, we have
    straight angle P O Q equals straight angle S O R(Vertically opposite angles)
    straight angle Q P O equals straight angle R S O( since  PQ || RS, alternate angles)
    straight angle S R O equals straight angle P Q O( since  PQ || RS, alternate angles)
    By, AAA similarity criterion we can say that ΔPOQ is similar to ΔSOR, but we can’t
    conclude that ΔPOQ ⩭ ΔSOR

    In the given figure, PQ || RS. Check whether ΔPOQ i congruent to ΔSOR Give reason for your answer:

    Maths-General
    In ΔPOQ and ΔSOR, we have
    straight angle P O Q equals straight angle S O R(Vertically opposite angles)
    straight angle Q P O equals straight angle R S O( since  PQ || RS, alternate angles)
    straight angle S R O equals straight angle P Q O( since  PQ || RS, alternate angles)
    By, AAA similarity criterion we can say that ΔPOQ is similar to ΔSOR, but we can’t
    conclude that ΔPOQ ⩭ ΔSOR
    General
    Maths-

    Show that 64 is square as well as cube number

    Explanation :-
    64 = 2 to the power of 6 Here 6 is divisible by both 2 and 3 .
    so,64 is both a square and cube number.

    Show that 64 is square as well as cube number

    Maths-General
    Explanation :-
    64 = 2 to the power of 6 Here 6 is divisible by both 2 and 3 .
    so,64 is both a square and cube number.
    parallel
    General
    Maths-

    From the given figures, find by which criteria the two triangles ABC and PQR are congruent.
    In the figure XY = PZ. Prove that XZ = PY.
    In the figure XY = PZ. Prove that XZ = PY.

    In ΔABC, We have
    straight angle B A C equals 180 minus left parenthesis 70 plus 30 right parenthesis equals 80 to the power of ring operator( Sum of interior angles of a triangle is 180)
    Likewise, In ΔPQR, We have
    straight angle P Q R equals 180 minus left parenthesis 70 plus 80 right parenthesis equals 30 to the power of ring operator( Sum of interior angles of a triangle is 180)
    If we consider ΔABC and ΔPQR, we have
    AB = RQ
    straight angle C A B equals straight angle P R O equals 80 to the power of ring operator
    straight angle C B A equals straight angle P Q R equals 30 to the power of ring operator(from the figure)
    So, by ASA congruence rule we have
    ΔABC⩭ ΔRQP

    From the given figures, find by which criteria the two triangles ABC and PQR are congruent.
    In the figure XY = PZ. Prove that XZ = PY.
    In the figure XY = PZ. Prove that XZ = PY.

    Maths-General
    In ΔABC, We have
    straight angle B A C equals 180 minus left parenthesis 70 plus 30 right parenthesis equals 80 to the power of ring operator( Sum of interior angles of a triangle is 180)
    Likewise, In ΔPQR, We have
    straight angle P Q R equals 180 minus left parenthesis 70 plus 80 right parenthesis equals 30 to the power of ring operator( Sum of interior angles of a triangle is 180)
    If we consider ΔABC and ΔPQR, we have
    AB = RQ
    straight angle C A B equals straight angle P R O equals 80 to the power of ring operator
    straight angle C B A equals straight angle P Q R equals 30 to the power of ring operator(from the figure)
    So, by ASA congruence rule we have
    ΔABC⩭ ΔRQP
    General
    Maths-

    Evaluate: cube root of 3375

    Explanation :-
    3375 = 3 × 1125
    = 3 × 3 × 375
    = 3 ×3 × 3 × 125
    = 3 × 3 × 3 × 5 × 25
    = 3 × 3 × 3× 5 × 5 × 5
    3 cubed cross times 5 cubed equals 15 cubed
    cube root of 3375 equals cube root of left parenthesis 15 right parenthesis cubed end root equals 15

    Evaluate: cube root of 3375

    Maths-General
    Explanation :-
    3375 = 3 × 1125
    = 3 × 3 × 375
    = 3 ×3 × 3 × 125
    = 3 × 3 × 3 × 5 × 25
    = 3 × 3 × 3× 5 × 5 × 5
    3 cubed cross times 5 cubed equals 15 cubed
    cube root of 3375 equals cube root of left parenthesis 15 right parenthesis cubed end root equals 15
    General
    Maths-

    Evaluate: cube root of fraction numerator left parenthesis negative 1728 right parenthesis over denominator 2744 end fraction end root

    Explanation :-
    We know that 1728 equals 12 cubed not stretchy rightwards double arrow negative 1728 equals left parenthesis negative 1 right parenthesis cubed 1728 equals left parenthesis negative 12 right parenthesis cubed
    2744 equals 2 space cross times 1372 equals 2 cross times 2 cross times 686 equals 2 cross times 2 cross times 2 cross times 343 equals 2 cubed cross times 7 cubed equals left parenthesis 14 right parenthesis cubed
    Substituting respective values we get
    cube root of fraction numerator left parenthesis negative 1728 right parenthesis over denominator 2744 end fraction end root equals cube root of fraction numerator left parenthesis negative 12 right parenthesis cubed over denominator left parenthesis 14 right parenthesis cubed end fraction end root equals fraction numerator negative 12 over denominator 14 end fraction equals fraction numerator negative 6 over denominator 7 end fraction

    Evaluate: cube root of fraction numerator left parenthesis negative 1728 right parenthesis over denominator 2744 end fraction end root

    Maths-General
    Explanation :-
    We know that 1728 equals 12 cubed not stretchy rightwards double arrow negative 1728 equals left parenthesis negative 1 right parenthesis cubed 1728 equals left parenthesis negative 12 right parenthesis cubed
    2744 equals 2 space cross times 1372 equals 2 cross times 2 cross times 686 equals 2 cross times 2 cross times 2 cross times 343 equals 2 cubed cross times 7 cubed equals left parenthesis 14 right parenthesis cubed
    Substituting respective values we get
    cube root of fraction numerator left parenthesis negative 1728 right parenthesis over denominator 2744 end fraction end root equals cube root of fraction numerator left parenthesis negative 12 right parenthesis cubed over denominator left parenthesis 14 right parenthesis cubed end fraction end root equals fraction numerator negative 12 over denominator 14 end fraction equals fraction numerator negative 6 over denominator 7 end fraction
    parallel

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