Maths-
General
Easy

Question

Ray OQ bisects ∠𝑃𝑂𝑅. Find 𝑚∠𝑃𝑂𝑄 if 𝑚∠𝑃𝑂𝑅 = 42°.

Hint:

  • Angle bisector bisects the angle in two equal parts.
  • If the measure of angle is 2xo then the angle bisector will bisect it in two parts measuring xo each.

The correct answer is: 21 digres


     
    • Step by step explanation: 
      • Given: 
    Ray OQ bisects ∠𝑃𝑂𝑅

     𝑚∠POR = 42°

    • Step 1:
    • Let 𝑚∠POQ = xo
    • As we know angle bisectors bisects the angle in two equal parts.
    Hence,

    𝑚∠POQ = 𝑚∠QOR = xo

    • Step 2:
    From figure it is clear that:

    ∠POQ + ∠QOR = ∠POR

    x + x = ∠POR

    2x = ∠POR
    We know that m ∠POR = 42o

    ∴ 2x = 42

    x = 42 over 2

    x = 21o

    m ∠POQ = 21o.

    • Final Answer: 
    Hence, 𝑚∠POQ is 21o.

    Related Questions to study

    General
    Maths-

    Ray OQ bisects ∠𝑃𝑂𝑅. Find 𝑚∠𝑃𝑂𝑅 if 𝑚∠𝑃𝑂𝑄 = 24°.

    • Hint:
      • Angle bisector bisects the angle in two equal parts.
      • If the measure of angle is 2xo then the angle bisector will bisect it in two parts measuring xo each.
    • Step by step explanation: 
      • Given: 
    Ray OQ bisects ∠𝑃𝑂𝑅
    𝑚∠POQ = 24°
    • Step 1:
    • Let 𝑚∠POQ = xo
    • As we know angle bisectors bisects the angle in two equal parts.
    Hence,
    𝑚∠POQ = 𝑚∠QOR = xo
    • Step 2:
    From figure it is clear that:
    ∠POQ + ∠QOR = ∠POR
    x + x = ∠POR
    2x = ∠POR
    We know that x = 24o
    ∴ ∠POR = 2x = 2(24)
    ∠POR = 48o.
    • Final Answer: 
    Hence, 𝑚∠𝑃𝑂𝑅 is 48o.

    Ray OQ bisects ∠𝑃𝑂𝑅. Find 𝑚∠𝑃𝑂𝑅 if 𝑚∠𝑃𝑂𝑄 = 24°.

    Maths-General
    • Hint:
      • Angle bisector bisects the angle in two equal parts.
      • If the measure of angle is 2xo then the angle bisector will bisect it in two parts measuring xo each.
    • Step by step explanation: 
      • Given: 
    Ray OQ bisects ∠𝑃𝑂𝑅
    𝑚∠POQ = 24°
    • Step 1:
    • Let 𝑚∠POQ = xo
    • As we know angle bisectors bisects the angle in two equal parts.
    Hence,
    𝑚∠POQ = 𝑚∠QOR = xo
    • Step 2:
    From figure it is clear that:
    ∠POQ + ∠QOR = ∠POR
    x + x = ∠POR
    2x = ∠POR
    We know that x = 24o
    ∴ ∠POR = 2x = 2(24)
    ∠POR = 48o.
    • Final Answer: 
    Hence, 𝑚∠𝑃𝑂𝑅 is 48o.
    General
    Maths-

    Find: a) QR
    b) 𝑚∠𝑃𝑄𝑇

    HINT – Observe the figure carefully and notice which sides/ angles are equal
    SOL – (a) In the figure, it is shown that PQ = QR and
    PQ = 22 in.
    rightwards double arrow QR = 22 inches

    (b) PR is a straight line rightwards double arrow ∠ PQR = 180°
    rightwards double arrow ∠PQT + ∠ RQT = 180°    ---- (1)
    It is given that ∠ RQT = 90°
    Substituting in (1)
    We get, ∠PQT + 90° = 180°
    ∠PQT = 180° - 90°
    ∠PQT = 90°.

    Find: a) QR
    b) 𝑚∠𝑃𝑄𝑇

    Maths-General
    HINT – Observe the figure carefully and notice which sides/ angles are equal
    SOL – (a) In the figure, it is shown that PQ = QR and
    PQ = 22 in.
    rightwards double arrow QR = 22 inches

    (b) PR is a straight line rightwards double arrow ∠ PQR = 180°
    rightwards double arrow ∠PQT + ∠ RQT = 180°    ---- (1)
    It is given that ∠ RQT = 90°
    Substituting in (1)
    We get, ∠PQT + 90° = 180°
    ∠PQT = 180° - 90°
    ∠PQT = 90°.
    General
    Maths-

    A sports store sells a total of 70- Soccer balls in one month and collects a total of $2,400. Write and Solve a System of equations to determine how many of each type of soccer ball were sold.

    Solution :-
    Hint :- Given, Total income of the store for 70 soccer balls is $2,400.
    There are two types of soccer balls and the cost of each type is different .
    Frame equation considering no.of limited edition soccer balls sold be x
    And no.of Pro NSL soccer ball sold be y and solve them to find  x and y.
    Ans :- The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.
    Explanation :-
    Let no.of limited edition soccer balls sold be x ,no.of Pro NSL soccer ball sold be y.
    Step 1:- Frame equations
    Total no.of ball is 70
    I.e x + y = 70             —Eq1
    Total cost of balls is $2,400
    Cost of x  limited edition soccer balls is 65x (as per ball cost is given in diagram)
    And Cost of y Pro NSL soccer ball is 15x(as per ball cost is given in diagram)
    I.e 65x + 15y = 2,400            —Eq2
    Step 2:- Eliminate y to find x
    Do Eq2 -15(Eq1) to eliminate y
    65x + 15y - 15(x+y) = 2400 - 15(70)
    65x - 15x = 1350
    50x = 1350 ⇒ x = 27
    Step 3:- substitute value of x to find y
    x + y = 70 ⇒ 27 + y = 70
    ⇒ y = 70 - 27
    ∴y = 43
    ∴The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

    A sports store sells a total of 70- Soccer balls in one month and collects a total of $2,400. Write and Solve a System of equations to determine how many of each type of soccer ball were sold.

    Maths-General
    Solution :-
    Hint :- Given, Total income of the store for 70 soccer balls is $2,400.
    There are two types of soccer balls and the cost of each type is different .
    Frame equation considering no.of limited edition soccer balls sold be x
    And no.of Pro NSL soccer ball sold be y and solve them to find  x and y.
    Ans :- The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.
    Explanation :-
    Let no.of limited edition soccer balls sold be x ,no.of Pro NSL soccer ball sold be y.
    Step 1:- Frame equations
    Total no.of ball is 70
    I.e x + y = 70             —Eq1
    Total cost of balls is $2,400
    Cost of x  limited edition soccer balls is 65x (as per ball cost is given in diagram)
    And Cost of y Pro NSL soccer ball is 15x(as per ball cost is given in diagram)
    I.e 65x + 15y = 2,400            —Eq2
    Step 2:- Eliminate y to find x
    Do Eq2 -15(Eq1) to eliminate y
    65x + 15y - 15(x+y) = 2400 - 15(70)
    65x - 15x = 1350
    50x = 1350 ⇒ x = 27
    Step 3:- substitute value of x to find y
    x + y = 70 ⇒ 27 + y = 70
    ⇒ y = 70 - 27
    ∴y = 43
    ∴The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.
    parallel
    General
    Maths-

    Ray OQ bisects ∠𝑃𝑂𝑅. Find the value of x.

    • Step by step explanation: 
      • Given:

    𝑚∠POQ = (x - 4) °  

    𝑚∠QOR = (5x - 20) °.

    • Step 1:
    • From the figure it is clear that,

    POR is right angle hence POR = 90o.
    and 

    POR = POQ + QOR 

    • Step 2:
    • Put values of ∠COD and ∠COP 

    ∠POR = ∠POQ + ∠QOR  

    90 = (x - 4) + (5x - 20)

    90 = 5x + x - (20 + 4)

    90 = 6x - 24

    6x = 90 - 24

    6x = 66

    x = 66 over 6

    x =  11

    • Final Answer:
    Hence, the x is 11.

    Ray OQ bisects ∠𝑃𝑂𝑅. Find the value of x.

    Maths-General
    • Step by step explanation: 
      • Given:

    𝑚∠POQ = (x - 4) °  

    𝑚∠QOR = (5x - 20) °.

    • Step 1:
    • From the figure it is clear that,

    POR is right angle hence POR = 90o.
    and 

    POR = POQ + QOR 

    • Step 2:
    • Put values of ∠COD and ∠COP 

    ∠POR = ∠POQ + ∠QOR  

    90 = (x - 4) + (5x - 20)

    90 = 5x + x - (20 + 4)

    90 = 6x - 24

    6x = 90 - 24

    6x = 66

    x = 66 over 6

    x =  11

    • Final Answer:
    Hence, the x is 11.
    General
    Maths-

    Identify all pairs of congruent angles and congruent segments.

    • Step by step explanation: 
    • Step 1:
    Find pair of congruent angles
    From the figure it is clear that

    ∠BAC = ∠QPR

    ∠BCA = ∠QRP

    ∠ABC = ∠PQR
    This are pair of congruent angles.

    • Step 2:
    Find pair of congruent segments
    From the figure it is clear that

    BA = QP                          
    This is pair of congruent segments.

    Identify all pairs of congruent angles and congruent segments.

    Maths-General
    • Step by step explanation: 
    • Step 1:
    Find pair of congruent angles
    From the figure it is clear that

    ∠BAC = ∠QPR

    ∠BCA = ∠QRP

    ∠ABC = ∠PQR
    This are pair of congruent angles.

    • Step 2:
    Find pair of congruent segments
    From the figure it is clear that

    BA = QP                          
    This is pair of congruent segments.

    General
    Maths-

    ∠𝐴 & ∠𝐵 are complementary.  ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅  ∠𝐶

    HINT – If sum of two angles is 90°, we say they are complementary angles
    SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
    rightwards double arrow ∠A + ∠B = 90°               ---- (1)
    Also, 𝐵 & ∠𝐶 are complementary
    rightwards double arrow ∠B + ∠C = 90°              ---- (2)
    From (1) and (2)
    We get, ∠A + ∠B = ∠B + ∠C
    rightwards double arrow ∠A = ∠C
    rightwards double arrow ∠𝐴 ≅  ∠𝐶
    Hence Proved.

    ∠𝐴 & ∠𝐵 are complementary.  ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅  ∠𝐶

    Maths-General
    HINT – If sum of two angles is 90°, we say they are complementary angles
    SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
    rightwards double arrow ∠A + ∠B = 90°               ---- (1)
    Also, 𝐵 & ∠𝐶 are complementary
    rightwards double arrow ∠B + ∠C = 90°              ---- (2)
    From (1) and (2)
    We get, ∠A + ∠B = ∠B + ∠C
    rightwards double arrow ∠A = ∠C
    rightwards double arrow ∠𝐴 ≅  ∠𝐶
    Hence Proved.
    parallel
    General
    Maths-

    Given: Ray OR bisects ∠𝑃𝑂𝑆.
    Prove: 𝑚∠1 = 𝑚∠2

    SOL – It is given that ray OR bisects ∠𝑃𝑂𝑆 i.e. OR is an angle bisector.
    We know that an angle bisector divides an angle into two congruent angles.
    rightwards double arrow ∠ POR ≅ ∠ ROS
    rightwards double arrow 𝑚∠1 = 𝑚∠2
    Hence Proved.

    Given: Ray OR bisects ∠𝑃𝑂𝑆.
    Prove: 𝑚∠1 = 𝑚∠2

    Maths-General
    SOL – It is given that ray OR bisects ∠𝑃𝑂𝑆 i.e. OR is an angle bisector.
    We know that an angle bisector divides an angle into two congruent angles.
    rightwards double arrow ∠ POR ≅ ∠ ROS
    rightwards double arrow 𝑚∠1 = 𝑚∠2
    Hence Proved.
    General
    Maths-

    Solve the following by using the method of substitution
    Y = - 2X-3
    Y = - X-4

    Ans :- x = 1 ; y = -5
    Explanation :-
    ⇒  y = -2x - 3 — eq 1
    ⇒  y = -x - 4—- eq 2
    Step 1 :- find x by substituting y = 4x + 2 in eq 2.
    -2x - 3 =   -x – 4  ⇒  2x + 3x + 4
    ⇒  2x – x = 4 - 3 ⇒ x  = 4 - 3
    ⇒  x  = 1
    Step 2 :- substitute value of x and find y
    ⇒  y =  - x – 4  ⇒  y = -1 - 4
    ∴  y =  - 5
    ∴   x = 1 and y = - 5  is the solution of the given pair of equations.

    Solve the following by using the method of substitution
    Y = - 2X-3
    Y = - X-4

    Maths-General
    Ans :- x = 1 ; y = -5
    Explanation :-
    ⇒  y = -2x - 3 — eq 1
    ⇒  y = -x - 4—- eq 2
    Step 1 :- find x by substituting y = 4x + 2 in eq 2.
    -2x - 3 =   -x – 4  ⇒  2x + 3x + 4
    ⇒  2x – x = 4 - 3 ⇒ x  = 4 - 3
    ⇒  x  = 1
    Step 2 :- substitute value of x and find y
    ⇒  y =  - x – 4  ⇒  y = -1 - 4
    ∴  y =  - 5
    ∴   x = 1 and y = - 5  is the solution of the given pair of equations.
    General
    Maths-

    Solve the system of equations by elimination :
    X - 2Y = 1
    2X + 3Y= - 12

    SOLUTION:
    HINT: Perform any arithmetic operation and then find.
    Complete step by step solution:
    Let x - 2y = 1…(i)
    and 2x + 3y=-12….(ii)
    On multiplying (i) with 2, we get 2(x - 2y = 1)
    ⇒ 2x - 4y = 2…(iii)
    Now, we have the coefficients of x in (ii) and (iii) to be the same.
    On subtracting (ii) from (iii),
    we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y
    and RHS to be 2 - (- 12) = 14
    On equating LHS and RHS, we have - 7y = 14
    ⇒ y = - 2
    On substituting the value of y in (i), we get x - 2 × - 2 =1
    ⇒ x + 4 = 1
    ⇒ x = 1-4
    ⇒ x = - 3
    Hence we get x = - 3 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations

    Solve the system of equations by elimination :
    X - 2Y = 1
    2X + 3Y= - 12

    Maths-General
    SOLUTION:
    HINT: Perform any arithmetic operation and then find.
    Complete step by step solution:
    Let x - 2y = 1…(i)
    and 2x + 3y=-12….(ii)
    On multiplying (i) with 2, we get 2(x - 2y = 1)
    ⇒ 2x - 4y = 2…(iii)
    Now, we have the coefficients of x in (ii) and (iii) to be the same.
    On subtracting (ii) from (iii),
    we get LHS to be 2x - 4y - (2x + 3y) = - 4y - 3y = - 7y
    and RHS to be 2 - (- 12) = 14
    On equating LHS and RHS, we have - 7y = 14
    ⇒ y = - 2
    On substituting the value of y in (i), we get x - 2 × - 2 =1
    ⇒ x + 4 = 1
    ⇒ x = 1-4
    ⇒ x = - 3
    Hence we get x = - 3 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations
    parallel
    General
    Maths-

    Solve the equation. Write a reason for each step.
    𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

    HINT – Open the brackets
    SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
    Opening the brackets
    We get, x – 2 + 3x + 6 = 3x + 10
    rightwards double arrow 4x + 4 = 3x + 10      ( Adding similar terms )
    rightwards double arrow 4x – 3x = 10 – 4
    rightwards double arrow x = 6.

    Solve the equation. Write a reason for each step.
    𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10

    Maths-General
    HINT – Open the brackets
    SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
    Opening the brackets
    We get, x – 2 + 3x + 6 = 3x + 10
    rightwards double arrow 4x + 4 = 3x + 10      ( Adding similar terms )
    rightwards double arrow 4x – 3x = 10 – 4
    rightwards double arrow x = 6.
    General
    Maths-

    Use Substitution to solve each system of equations :
    6X - 3Y = -6
    Y = 2X + 2

    Solution :-
    Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations  .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.
    Ans :- infinite no.of solutions .
    Explanation :-
    y = 2x + 2— eq 1
    6x - 3y = -6—- eq 2
    Step 1 :- find x by substituting y = 2x + 2 in eq 2.
    6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6
    -6 = -6
    Here we get -6 = -6 which is always true i.e always having a root .
    They coincide with each other and have infinite no.of solutions
    They have infinite no.of solutions for the given system of equations

    Use Substitution to solve each system of equations :
    6X - 3Y = -6
    Y = 2X + 2

    Maths-General
    Solution :-
    Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value of  x in the equations  .If we get a true statement we say they have infinite solutions .If we get the false statement we say they have no solution.
    Ans :- infinite no.of solutions .
    Explanation :-
    y = 2x + 2— eq 1
    6x - 3y = -6—- eq 2
    Step 1 :- find x by substituting y = 2x + 2 in eq 2.
    6x - 3 (2x + 2) = -6 ⇒ 6 x -6x - 6 = -6
    -6 = -6
    Here we get -6 = -6 which is always true i.e always having a root .
    They coincide with each other and have infinite no.of solutions
    They have infinite no.of solutions for the given system of equations

    General
    Maths-

    Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅
    Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

    SOL – It is given that 𝑚∠𝑃 = 30° and  𝑚∠𝑄 = 30°

    rightwards double arrow 𝑚∠𝑃 = 𝑚∠𝑄           ---- (1)
    Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅    ---- (1)
    Using transitive property which states that if A = B and B = C then A = C
    We get from (1) and (2),

    𝑚∠P = 𝑚∠R = 30°
    Hence Proved

    Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅
    Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

    Maths-General
    SOL – It is given that 𝑚∠𝑃 = 30° and  𝑚∠𝑄 = 30°

    rightwards double arrow 𝑚∠𝑃 = 𝑚∠𝑄           ---- (1)
    Further, it is given that 𝑚∠𝑄 = 𝑚∠𝑅    ---- (1)
    Using transitive property which states that if A = B and B = C then A = C
    We get from (1) and (2),

    𝑚∠P = 𝑚∠R = 30°
    Hence Proved

    parallel
    General
    Maths-

    Use the given information and the diagram to prove the statement.
    Given: 𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180° and 𝑚∠𝑃𝑀𝐶 = 150°
    Prove: 𝑚∠𝑃𝑀𝐷 = 30°

    HINT – Use the information given.
    SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150°    ---- (1)
    𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°
    150° + 𝑚∠𝑃𝑀𝐷 = 180°             ( - From (1) )
    𝑚∠𝑃𝑀𝐷 = 180° - 150°
    = 30°
    Hence Proved.

    Use the given information and the diagram to prove the statement.
    Given: 𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180° and 𝑚∠𝑃𝑀𝐶 = 150°
    Prove: 𝑚∠𝑃𝑀𝐷 = 30°

    Maths-General
    HINT – Use the information given.
    SOL – It is given that 𝑚∠𝑃𝑀𝐶 = 150°    ---- (1)
    𝑚∠𝑃𝑀𝐶 + 𝑚∠𝑃𝑀𝐷 = 180°
    150° + 𝑚∠𝑃𝑀𝐷 = 180°             ( - From (1) )
    𝑚∠𝑃𝑀𝐷 = 180° - 150°
    = 30°
    Hence Proved.
    General
    Maths-

    Solve the system of equations by elimination :
    3X + 2Y = 8
    X + 4Y = - 4

    Complete step by step solution:
    Let 3x + 2y = 8…(i)
    and x + 4y = - 4….(ii)
    On multiplying (ii) with 3, we get 3(x + 4y=-4)
    ⇒3x + 12y = - 12…(iii)
    Now, we have the coefficients of  in (i) and (iii) to be the same.
    On subtracting (i) from (iii),
    we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y
    and RHS to be - 12 - 8 = - 20
    On equating LHS and RHS, we have 10y = - 20
    ⇒y = - 2
    On substituting the value of y in (i), we get 3x + 2× - 2 = 8
    ⇒ 3x - 4 = 8
    ⇒ 3x = 8 + 4
    ⇒ 3x = 12
    ⇒x = 4
    Hence we get x = 4 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations

    Solve the system of equations by elimination :
    3X + 2Y = 8
    X + 4Y = - 4

    Maths-General
    Complete step by step solution:
    Let 3x + 2y = 8…(i)
    and x + 4y = - 4….(ii)
    On multiplying (ii) with 3, we get 3(x + 4y=-4)
    ⇒3x + 12y = - 12…(iii)
    Now, we have the coefficients of  in (i) and (iii) to be the same.
    On subtracting (i) from (iii),
    we get LHS to be 3x + 12y - (3x + 2y) = 12y - 2y = 10y
    and RHS to be - 12 - 8 = - 20
    On equating LHS and RHS, we have 10y = - 20
    ⇒y = - 2
    On substituting the value of y in (i), we get 3x + 2× - 2 = 8
    ⇒ 3x - 4 = 8
    ⇒ 3x = 8 + 4
    ⇒ 3x = 12
    ⇒x = 4
    Hence we get x = 4 and y = - 2
    Note: We can also solve these system of equations by making the coefficients of y
    to be the same in both the equations
    General
    Maths-

    Give a two-column proof.
    Given:

    Prove: PR = 25 in

    HINT – Use Segment Addition Postulate
    SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
    rightwards double arrow PR = PQ + QR
    rightwards double arrow PR = 12 + 13
    rightwards double arrow PR = 25 in.
    Hence Proved.

    Give a two-column proof.
    Given:

    Prove: PR = 25 in

    Maths-General
    HINT – Use Segment Addition Postulate
    SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
    rightwards double arrow PR = PQ + QR
    rightwards double arrow PR = 12 + 13
    rightwards double arrow PR = 25 in.
    Hence Proved.
    parallel

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