Question

# Find: a) QR

b) πβ πππ

Hint:

### Observe the figure carefully and notice which sides/ angles are equal

## The correct answer is: a) QR = 22 inches, b) β πππ = 90

### SOL β (a) In the figure, it is shown that PQ = QR and

PQ = 22 in.

Β QR = 22 inches

(b) PR is a straight line β PQR = 180Β°

β PQT + β RQT = 180Β°Β Β Β ---- (1)

It is given that β RQT = 90Β°

Substituting in (1)

We get, β PQT + 90Β° = 180Β°

β PQT = 180Β° - 90Β°

β PQT = 90Β°.

### Related Questions to study

### 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 findΒ 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 = 70Β Β Β Β Β Β Β βEq1

Total cost of balls is $2,400

Cost of xΒ 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 = 2,400Β Β Β Β Β Β βEq2

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 β x = 27

Step 3:- substitute value of x to find y

x + y = 70 β 27 + y = 70

β y = 70 - 27

β΄y = 43

β΄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 findΒ 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 = 70Β Β Β Β Β Β Β βEq1

Total cost of balls is $2,400

Cost of xΒ 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 = 2,400Β Β Β Β Β Β βEq2

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 β x = 27

Step 3:- substitute value of x to find y

x + y = 70 β 27 + y = 70

β y = 70 - 27

β΄y = 43

β΄The no.of limited edition soccer balls sold be 27 and no.of Pro NSL soccer ball sold are 43.

### Ray OQ bisects β πππ
.Β Find the value of x.

- Step by step explanation:Β
- Given:

πβ POQ = (x - 4) Β°Β Β

πβ QOR = (5x - 20) Β°.

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

β POR is right angle hence β POR = 90^{o}.

andΒ

β POR = β POQ + β QORΒ

- Step 2:
- Put values of β COD and β COPΒ

β POR = β POQ + β QOR Β

90 = (x - 4) +Β (5x - 20)

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

90 = 6x -Β 24

6x = 90 - 24

6x = 66

x =Β

x = Β 11

- Final Answer:

### Ray OQ bisects β πππ
.Β Find the value of x.

- Step by step explanation:Β
- Given:

πβ POQ = (x - 4) Β°Β Β

πβ QOR = (5x - 20) Β°.

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

β POR is right angle hence β POR = 90^{o}.

andΒ

β POR = β POQ + β QORΒ

- Step 2:
- Put values of β COD and β COPΒ

β POR = β POQ + β QOR Β

90 = (x - 4) +Β (5x - 20)

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

90 = 6x -Β 24

6x = 90 - 24

6x = 66

x =Β

x = Β 11

- Final Answer:

### Identify all pairs of congruent angles and congruent segments.

- Step by step explanation:Β

- Step 1:

From the figure it is clear that

β BAC = β QPR

β BCA = β QRP

β ABC = β PQR

This are pair of congruent angles.

- Step 2:

From the figure it is clear that

BA = QPΒ Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β

This is pair of congruent segments.

### Identify all pairs of congruent angles and congruent segments.

- Step by step explanation:Β

- Step 1:

From the figure it is clear that

β BAC = β QPR

β BCA = β QRP

β ABC = β PQR

This are pair of congruent angles.

- Step 2:

From the figure it is clear that

BA = QPΒ Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β

This is pair of congruent segments.

### β π΄ &ββ π΅ are complementary. Β β π΅ &ββ πΆ are complementary. Prove: β π΄ β ββ πΆ

SOL β It is given that β π΄ &ββ π΅ are complementary

β A + β B = 90Β°Β Β Β Β Β Β Β Β Β Β Β Β Β Β ---- (1)

Also, π΅ &ββ πΆ are complementary

Β β B + β C = 90Β°Β Β Β Β Β Β Β Β Β Β Β Β Β ---- (2)

From (1) and (2)

We get, β A + β B = β B + β C

Β β A = β C

Β β π΄ β ββ πΆ

Hence Proved.

### β π΄ &ββ π΅ are complementary. Β β π΅ &ββ πΆ are complementary. Prove: β π΄ β ββ πΆ

SOL β It is given that β π΄ &ββ π΅ are complementary

β A + β B = 90Β°Β Β Β Β Β Β Β Β Β Β Β Β Β Β ---- (1)

Also, π΅ &ββ πΆ are complementary

Β β B + β C = 90Β°Β Β Β Β Β Β Β Β Β Β Β Β Β ---- (2)

From (1) and (2)

We get, β A + β B = β B + β C

Β β A = β C

Β β π΄ β ββ πΆ

Hence Proved.

### Given: Ray OR bisects β πππ.

Prove: πβ 1 = πβ 2

We know that anΒ angle bisector divides an angle into two congruent angles.

β PORΒ β β ROS

Β πβ 1 = πβ 2

Hence Proved.

### Given: Ray OR bisects β πππ.

Prove: πβ 1 = πβ 2

We know that anΒ angle bisector divides an angle into two congruent angles.

β PORΒ β β ROS

Β πβ 1 = πβ 2

Hence Proved.

### Solve the following by using the method of substitution

Y = - 2X-3

Y = - X-4

Explanation :-

β Β y = -2x - 3Β β eq 1

βΒ y = -x - 4β- eq 2

Step 1 :- find x by substituting y = 4x + 2 in eq 2.

-2x - 3 = Β Β -x β 4 Β β Β 2x + 3x + 4

β Β 2x β x = 4 - 3 β x Β = 4 - 3

β Β x Β = 1

Step 2 :- substitute value of x and find y

βΒ y = Β - x β 4 Β β Β y = -1 - 4

β΄ Β y = Β - 5

β΄ Β Β x = 1 and y = - 5Β 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 :-

β Β y = -2x - 3Β β eq 1

βΒ y = -x - 4β- eq 2

Step 1 :- find x by substituting y = 4x + 2 in eq 2.

-2x - 3 = Β Β -x β 4 Β β Β 2x + 3x + 4

β Β 2x β x = 4 - 3 β x Β = 4 - 3

β Β x Β = 1

Step 2 :- substitute value of x and find y

βΒ y = Β - x β 4 Β β Β y = -1 - 4

β΄ Β y = Β - 5

β΄ Β Β x = 1 and y = - 5Β 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 = 1β¦(i)

and 2x + 3y=-12β¦.(ii)

On multiplying (i) with 2, we get 2(x - 2y = 1)

β 2x - 4y = 2β¦(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

β y = - 2

On substituting the value of y in (i), we get x - 2 Γ - 2 =1

β x + 4 = 1

β x = 1-4

β 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 = 1β¦(i)

and 2x + 3y=-12β¦.(ii)

On multiplying (i) with 2, we get 2(x - 2y = 1)

β 2x - 4y = 2β¦(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

β y = - 2

On substituting the value of y in (i), we get x - 2 Γ - 2 =1

β x + 4 = 1

β x = 1-4

β 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.

π₯ β 2 + 3(π₯ + 2) = 3π₯ + 10

SOL β It is given that π₯ β 2 + 3(π₯ + 2) = 3π₯ + 10

Opening the brackets

We get, x β 2 + 3x + 6 = 3x + 10

4x + 4 = 3x + 10Β Β Β Β Β ( Adding similar terms )

Β 4x β 3x = 10 β 4

Β x = 6.

### Solve the equation. Write a reason for each step.

π₯ β 2 + 3(π₯ + 2) = 3π₯ + 10

SOL β It is given that π₯ β 2 + 3(π₯ + 2) = 3π₯ + 10

Opening the brackets

We get, x β 2 + 3x + 6 = 3x + 10

4x + 4 = 3x + 10Β Β Β Β Β ( Adding similar terms )

Β 4x β 3x = 10 β 4

Β 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 ofΒ x in the equationsΒ .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 + 2β eq 1

6x - 3y = -6β- eq 2

Step 1 :- find x by substituting y = 2x + 2 in eq 2.

6x - 3 (2x + 2) = -6 β 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 ofΒ x in the equationsΒ .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 + 2β eq 1

6x - 3y = -6β- eq 2

Step 1 :- find x by substituting y = 2x + 2 in eq 2.

6x - 3 (2x + 2) = -6 β 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: πβ π = 30Β°, πβ π = 30Β° , πβ π = πβ π

Prove: πβ π β
πβ π
= 30Β°

πβ π = πβ πΒ Β Β Β Β Β Β Β Β Β ---- (1)

Further, it is given that πβ π = πβ π
Β Β Β ---- (1)

Using transitive property which states that if A = B and B = C then A = C

We get from (1) and (2),

πβ P = πβ R = 30Β°

Hence Proved

### Given: πβ π = 30Β°, πβ π = 30Β° , πβ π = πβ π

Prove: πβ π β
πβ π
= 30Β°

πβ π = πβ πΒ Β Β Β Β Β Β Β Β Β ---- (1)

Further, it is given that πβ π = πβ π
Β Β Β ---- (1)

Using transitive property which states that if A = B and B = C then A = C

We get from (1) and (2),

πβ P = πβ R = 30Β°

Hence Proved

### Use the given information and the diagram to prove the statement.

Given: πβ πππΆ + πβ πππ· = 180Β° and πβ πππΆ = 150Β°

Prove: πβ πππ· = 30Β°

SOL β It is given that πβ πππΆ = 150Β° Β Β Β ---- (1)

πβ πππΆ + πβ πππ· = 180Β°

150Β° + πβ πππ· = 180Β°Β Β Β Β Β Β Β Β Β Β Β Β ( - From (1) )

πβ πππ· = 180Β° - 150Β°

= 30Β°

Hence Proved.

### Use the given information and the diagram to prove the statement.

Given: πβ πππΆ + πβ πππ· = 180Β° and πβ πππΆ = 150Β°

Prove: πβ πππ· = 30Β°

SOL β It is given that πβ πππΆ = 150Β° Β Β Β ---- (1)

πβ πππΆ + πβ πππ· = 180Β°

150Β° + πβ πππ· = 180Β°Β Β Β Β Β Β Β Β Β Β Β Β ( - From (1) )

πβ πππ· = 180Β° - 150Β°

= 30Β°

Hence Proved.

### Solve the system of equations by elimination :

3X + 2Y = 8

X + 4Y = - 4

Let 3x + 2y = 8β¦(i)

and x + 4y = - 4β¦.(ii)

On multiplying (ii) with 3, we get 3(x + 4y=-4)

β3x + 12y = - 12β¦(iii)

Now, we have the coefficients of Β 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

βy = - 2

On substituting the value of y in (i), we get 3x + 2Γ - 2 = 8

β 3x - 4 = 8

β 3x = 8 + 4

β 3x = 12

βx = 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 = 8β¦(i)

and x + 4y = - 4β¦.(ii)

On multiplying (ii) with 3, we get 3(x + 4y=-4)

β3x + 12y = - 12β¦(iii)

Now, we have the coefficients of Β 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

βy = - 2

On substituting the value of y in (i), we get 3x + 2Γ - 2 = 8

β 3x - 4 = 8

β 3x = 8 + 4

β 3x = 12

βx = 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 β 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

Β PR = 12 + 13

Β PR = 25 in.

Hence Proved.

### Give a two-column proof.

Given:

Prove: PR = 25 in

SOL β 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

Β PR = 12 + 13

Β PR = 25 in.

Hence Proved.

### Use Substitution to solve each system of equations :

- 3X - Y = 7

X + 2Y = 6

Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value ofΒ x in the equations.

Ans :- x = -4; y = 5

Explanation :-

-Β 3x - y = 7 β - 3x - 7 = yΒ Β Β Β Β Β Β Β Β Β Β β eq 1

x + 2y = 6Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β β- eq 2

x + 2 ( - 3x - 7) = 6 β x - 6x - 14 = 6

β - 5x = 14 + 6 β - 5x = 20

β x = -4

Step 2 :- substitute value of x and find y

β - 3x - 7 = y β y = - 3 (- 4) - 7

β y = 12 - 7

β΄ y = 5

x = -4 and y = 5 is the solution of the given pair of equations.

### Use Substitution to solve each system of equations :

- 3X - Y = 7

X + 2Y = 6

Hint :- find x by substituting y (in terms of x) in the equation and find y by substituting value ofΒ x in the equations.

Ans :- x = -4; y = 5

Explanation :-

-Β 3x - y = 7 β - 3x - 7 = yΒ Β Β Β Β Β Β Β Β Β Β β eq 1

x + 2y = 6Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β β- eq 2

x + 2 ( - 3x - 7) = 6 β x - 6x - 14 = 6

β - 5x = 14 + 6 β - 5x = 20

β x = -4

Step 2 :- substitute value of x and find y

β - 3x - 7 = y β y = - 3 (- 4) - 7

β y = 12 - 7

β΄ y = 5

x = -4 and y = 5 is the solution of the given pair of equations.

### If Q is the midpoint of PR, prove that PR = 2 PQ.

Give a two-column proof.

SOL β If Q is mid β pointΒ Γ PQ = QR Β Β Β Β Β ---- (1)

Using Segment Addition Postulate,

We get, Β PR = PQ + QR

PR = PQ + PQ

PR = 2 PQ

Hence Proved.

### If Q is the midpoint of PR, prove that PR = 2 PQ.

Give a two-column proof.

SOL β If Q is mid β pointΒ Γ PQ = QR Β Β Β Β Β ---- (1)

Using Segment Addition Postulate,

We get, Β PR = PQ + QR

PR = PQ + PQ

PR = 2 PQ

Hence Proved.