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 = x^{o}
- As we know angle bisectors bisects the angle in two equal parts.
Hence,
𝑚∠POQ = 𝑚∠QOR = x^{o}
- Step 2:
From figure it is clear that:
∠POQ + ∠QOR = ∠POR
x + x = ∠POR
2x = ∠POR
We know that m ∠POR = 42^{o}
∴ 2x = 42
x =
x = 21^{o}
m ∠POQ = 21^{o}.
- Final Answer:
Hence, 𝑚∠POQ is 21^{o}.
- Given:
We know that m ∠POR = 42^{o}
Related Questions to study
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 2x^{o} then the angle bisector will bisect it in two parts measuring x^{o} each.
- Step by step explanation:
- Given:
𝑚∠POQ = 24°
- Step 1:
- Let 𝑚∠POQ = x^{o}
- As we know angle bisectors bisects the angle in two equal parts.
𝑚∠POQ = 𝑚∠QOR = x^{o}
- Step 2:
∠POQ + ∠QOR = ∠POR
x + x = ∠POR
2x = ∠POR
We know that x = 24^{o}
∴ ∠POR = 2x = 2(24)
∠POR = 48^{o}.
- Final Answer:
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 2x^{o} then the angle bisector will bisect it in two parts measuring x^{o} each.
- Step by step explanation:
- Given:
𝑚∠POQ = 24°
- Step 1:
- Let 𝑚∠POQ = x^{o}
- As we know angle bisectors bisects the angle in two equal parts.
𝑚∠POQ = 𝑚∠QOR = x^{o}
- Step 2:
∠POQ + ∠QOR = ∠POR
x + x = ∠POR
2x = ∠POR
We know that x = 24^{o}
∴ ∠POR = 2x = 2(24)
∠POR = 48^{o}.
- Final Answer:
Find: a) QR
b) 𝑚∠𝑃𝑄𝑇
SOL – (a) In the figure, it is shown that PQ = QR and
PQ = 22 in.
QR = 22 inches
(b) PR is a straight line ∠ PQR = 180°
∠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
SOL – (a) In the figure, it is shown that PQ = QR and
PQ = 22 in.
QR = 22 inches
(b) PR is a straight line ∠ PQR = 180°
∠PQT + ∠ RQT = 180° ---- (1)
It is given that ∠ RQT = 90°
Substituting in (1)
We get, ∠PQT + 90° = 180°
∠PQT = 180° - 90°
∠PQT = 90°.
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.
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.
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.
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 = 90^{o}.
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 =
x = 11
- Final Answer:
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 = 90^{o}.
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 =
x = 11
- Final Answer:
Identify all pairs of congruent angles and congruent segments.
- Step by step explanation:
- Step 1:
From the figure it is clear that
∠BAC = ∠QPR
∠BCA = ∠QRP
∠ABC = ∠PQR
This are pair of congruent angles.
- Step 2:
From the figure it is clear that
BA = QP
This is pair of congruent segments.
Identify all pairs of congruent angles and congruent segments.
- Step by step explanation:
- Step 1:
From the figure it is clear that
∠BAC = ∠QPR
∠BCA = ∠QRP
∠ABC = ∠PQR
This are pair of congruent angles.
- Step 2:
From the figure it is clear that
BA = QP
This is pair of congruent segments.
∠𝐴 & ∠𝐵 are complementary. ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅ ∠𝐶
SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
∠A + ∠B = 90° ---- (1)
Also, 𝐵 & ∠𝐶 are complementary
∠B + ∠C = 90° ---- (2)
From (1) and (2)
We get, ∠A + ∠B = ∠B + ∠C
∠A = ∠C
∠𝐴 ≅ ∠𝐶
Hence Proved.
∠𝐴 & ∠𝐵 are complementary. ∠𝐵 & ∠𝐶 are complementary. Prove: ∠𝐴 ≅ ∠𝐶
SOL – It is given that ∠𝐴 & ∠𝐵 are complementary
∠A + ∠B = 90° ---- (1)
Also, 𝐵 & ∠𝐶 are complementary
∠B + ∠C = 90° ---- (2)
From (1) and (2)
We get, ∠A + ∠B = ∠B + ∠C
∠A = ∠C
∠𝐴 ≅ ∠𝐶
Hence Proved.
Given: Ray OR bisects ∠𝑃𝑂𝑆.
Prove: 𝑚∠1 = 𝑚∠2
We know that an angle bisector divides an angle into two congruent angles.
∠ POR ≅ ∠ ROS
𝑚∠1 = 𝑚∠2
Hence Proved.
Given: Ray OR bisects ∠𝑃𝑂𝑆.
Prove: 𝑚∠1 = 𝑚∠2
We know that an angle bisector divides an angle into two congruent angles.
∠ POR ≅ ∠ ROS
𝑚∠1 = 𝑚∠2
Hence Proved.
Solve the following by using the method of substitution
Y = - 2X-3
Y = - X-4
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
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 system of equations by elimination :
X - 2Y = 1
2X + 3Y= - 12
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
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 equation. Write a reason for each step.
𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
Opening the brackets
We get, x – 2 + 3x + 6 = 3x + 10
4x + 4 = 3x + 10 ( Adding similar terms )
4x – 3x = 10 – 4
x = 6.
Solve the equation. Write a reason for each step.
𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
SOL – It is given that 𝑥 − 2 + 3(𝑥 + 2) = 3𝑥 + 10
Opening the brackets
We get, x – 2 + 3x + 6 = 3x + 10
4x + 4 = 3x + 10 ( Adding similar terms )
4x – 3x = 10 – 4
x = 6.
Use Substitution to solve each system of equations :
6X - 3Y = -6
Y = 2X + 2
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
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
Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅
Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°
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°
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
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°
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°
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.
Solve the system of equations by elimination :
3X + 2Y = 8
X + 4Y = - 4
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
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
Give a two-column proof.
Given:
Prove: PR = 25 in
SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
PR = PQ + QR
PR = 12 + 13
PR = 25 in.
Hence Proved.
Give a two-column proof.
Given:
Prove: PR = 25 in
SOL – In the given figure, line segment has two end points namely P and R and Q is lying on the line segment PR
PR = PQ + QR
PR = 12 + 13
PR = 25 in.
Hence Proved.