Question

# Given: 𝑚∠𝑃 = 30°, 𝑚∠𝑄 = 30° , 𝑚∠𝑄 = 𝑚∠𝑅

Prove: 𝑚∠𝑃 ≅ 𝑚∠𝑅 = 30°

Hint:

### Use Transitive Property

## The correct answer is:

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

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),

Hence Proved

### Related Questions to study

### 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.

### 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 of x in the equations.

Ans :- x = -4; y = 5

Explanation :-

- 3x - y = 7 ⇒ - 3x - 7 = y — eq 1

x + 2y = 6 —- eq 2

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

⇒ - 5x = 14 + 6 ⇒ - 5x = 20

⇒ x = -4

Step 2 :- substitute value of x and find y

⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7

⇒ y = 12 - 7

∴ 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 of x in the equations.

Ans :- x = -4; y = 5

Explanation :-

- 3x - y = 7 ⇒ - 3x - 7 = y — eq 1

x + 2y = 6 —- eq 2

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

⇒ - 5x = 14 + 6 ⇒ - 5x = 20

⇒ x = -4

Step 2 :- substitute value of x and find y

⇒ - 3x - 7 = y ⇒ y = - 3 (- 4) - 7

⇒ y = 12 - 7

∴ 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 – If Q is mid – point Þ PQ = QR ---- (1)

Using Segment Addition Postulate,

We get, 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 – If Q is mid – point Þ PQ = QR ---- (1)

Using Segment Addition Postulate,

We get, PR = PQ + QR

PR = PQ + PQ

PR = 2 PQ

Hence Proved.

### Compare 𝑚∠𝑃𝑂𝑄𝑎𝑛𝑑∠𝑆𝑂𝑇.

- Step by step explanation:
- Step 1:
- Find ∠POQ:

- Align the protractor with the ray OP on 0
^{o}as shown above. - Start reading the inner scale from the 0°.
- Step 2:
- From the figure we can see that ray OP is aligned on mark 0
^{o}. And ray OQ is aligned on mark 30^{o}.

^{o}- 0

^{o}).

∠POQ = 30

^{o}.

- Step 3:
- Find ∠SOT:
- Align the protractor with the ray OT on 0
^{o}as shown above. - Start reading the outer scale from the 0°.
- Step 4:
- From the figure we can see that ray OT is aligned on mark 0
^{o}. And the ray OS is aligned on mark 20^{o}.

- Align the protractor with the ray OT on 0

^{o}.

∠SOT = 20^{o}.

- Step 5:
- Compare ∠POQ and ∠SOT

^{o}.

- Final Answer:

^{o}.

### Compare 𝑚∠𝑃𝑂𝑄𝑎𝑛𝑑∠𝑆𝑂𝑇.

- Step by step explanation:
- Step 1:
- Find ∠POQ:

- Align the protractor with the ray OP on 0
^{o}as shown above. - Start reading the inner scale from the 0°.
- Step 2:
- From the figure we can see that ray OP is aligned on mark 0
^{o}. And ray OQ is aligned on mark 30^{o}.

^{o}- 0

^{o}).

∠POQ = 30

^{o}.

- Step 3:
- Find ∠SOT:
- Align the protractor with the ray OT on 0
^{o}as shown above. - Start reading the outer scale from the 0°.
- Step 4:
- From the figure we can see that ray OT is aligned on mark 0
^{o}. And the ray OS is aligned on mark 20^{o}.

- Align the protractor with the ray OT on 0

^{o}.

∠SOT = 20^{o}.

- Step 5:
- Compare ∠POQ and ∠SOT

^{o}.

- Final Answer:

^{o}.

### Which postulate will you use to prove that PR = PQ + QR?

SOL – Option (b)

Segment addition postulate – It states that if a line segment has two endpoints, A and C, a third point B lies on the line segment AC if and only if this equation AB + BC = AC

Since in figure given above, line segment has two end points namely P and R and third point Q lies on the line segment PR Þ PR = PQ + QR

### Which postulate will you use to prove that PR = PQ + QR?

SOL – Option (b)

Segment addition postulate – It states that if a line segment has two endpoints, A and C, a third point B lies on the line segment AC if and only if this equation AB + BC = AC

Since in figure given above, line segment has two end points namely P and R and third point Q lies on the line segment PR Þ PR = PQ + QR

### Solve the system of equations by elimination :

X - 2Y = - 2

3X + 2Y = 30

HINT: Perform any arithmetic operation and then find.

Complete step by step solution:

Let x - 2y = - 2…(i)

and 3x + 2y = 30….(ii)

On adding (i) and (ii),

we get LHS to be x - 2y + 3x + 2y = 4x

and RHS to be -2+ 30 = 28

On equating LHS and RHS, we get 4x = 28

⇒ x = 7

On substituting the value of x in (i), we get 7 - 2y = - 2

⇒ - 2y = - 2 - 7

⇒ - 2y = - 9

⇒ y =

Hence we get x = 7 and

Note: We can also solve these system of equations by making the coefficients of x

to be the same in both the equations.

### Solve the system of equations by elimination :

X - 2Y = - 2

3X + 2Y = 30

HINT: Perform any arithmetic operation and then find.

Complete step by step solution:

Let x - 2y = - 2…(i)

and 3x + 2y = 30….(ii)

On adding (i) and (ii),

we get LHS to be x - 2y + 3x + 2y = 4x

and RHS to be -2+ 30 = 28

On equating LHS and RHS, we get 4x = 28

⇒ x = 7

On substituting the value of x in (i), we get 7 - 2y = - 2

⇒ - 2y = - 2 - 7

⇒ - 2y = - 9

⇒ y =

Hence we get x = 7 and

Note: We can also solve these system of equations by making the coefficients of x

to be the same in both the equations.

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

Y = 4X+2

Y = X+8

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

Ans :- x = 2 ; y =10

Explanation :-

⇒ y = 4x + 2 — eq 1

⇒ y = x + 8—- eq 2

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

4x + 2 = x + 8 ⇒ 4x – x = 8 - 2

⇒ 3x = 8 - 2 ⇒ 3x = 6

⇒ x =2

Step 2 :- substitute value of x and find y

⇒ y = x + 8 ⇒ y = 2 + 8

∴ y= 10

∴ x = 2 and y = 10 is the solution of the given pair of equations.

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

Y = 4X+2

Y = X+8

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

Ans :- x = 2 ; y =10

Explanation :-

⇒ y = 4x + 2 — eq 1

⇒ y = x + 8—- eq 2

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

4x + 2 = x + 8 ⇒ 4x – x = 8 - 2

⇒ 3x = 8 - 2 ⇒ 3x = 6

⇒ x =2

Step 2 :- substitute value of x and find y

⇒ y = x + 8 ⇒ y = 2 + 8

∴ y= 10

∴ x = 2 and y = 10 is the solution of the given pair of equations.

### If lines l and m are parallel, find the value of x.

HINT: Use the property of parallel lines angle rules.

Complete step by step solution:

Here we have 2 parallel lines m and land a transversal intersecting these parallel

lines.

Here, forms co-interior angles and they add up to

(co-interior angles)

Hence the value of

### If lines l and m are parallel, find the value of x.

HINT: Use the property of parallel lines angle rules.

Complete step by step solution:

Here we have 2 parallel lines m and land a transversal intersecting these parallel

lines.

Here, forms co-interior angles and they add up to

(co-interior angles)

Hence the value of

Prove that: PR = PQ + QR

Give a two-column proof

SOL – In the figure, we can see that PR coincides with

PQ + QR.

Acc. to Euclid’s Axiom, things which coincide with one another are equal to one another.

Hence Proved

NOTE – Alternative method

We can also prove using Segment addition postulate which states that if a line segment has two endpoints, A and C, a third point B lies on the line segment AC if and only if this equation AB + BC = AC

Since in figure given above, Q lies on the line segment PR Þ PR = PQ + QR.

Prove that: PR = PQ + QR

Give a two-column proof

SOL – In the figure, we can see that PR coincides with

PQ + QR.

Acc. to Euclid’s Axiom, things which coincide with one another are equal to one another.

Hence Proved

NOTE – Alternative method

We can also prove using Segment addition postulate which states that if a line segment has two endpoints, A and C, a third point B lies on the line segment AC if and only if this equation AB + BC = AC

Since in figure given above, Q lies on the line segment PR Þ PR = PQ + QR.

### Find x and hence find 𝑚∠𝐶𝑂𝐷

- Step by step explanation:
- Given:

𝑚∠COD = (x + 20)°

𝑚∠COP = 2x°.

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

∠POD is right angle hence ∠POD = 90^{o}.

and

∠POD = ∠COD + ∠COP

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

∠POD = ∠COD + ∠COP

90 = (x + 20) + 2x

90 = 2x + x + 20

90 = 3x + 20

3x = 90 - 20

3x = 70

x =

- Step 3:
- ∠COD = x + 20

+ 20

+

+

- Final Answer:

### Find x and hence find 𝑚∠𝐶𝑂𝐷

- Step by step explanation:
- Given:

𝑚∠COD = (x + 20)°

𝑚∠COP = 2x°.

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

∠POD is right angle hence ∠POD = 90^{o}.

and

∠POD = ∠COD + ∠COP

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

∠POD = ∠COD + ∠COP

90 = (x + 20) + 2x

90 = 2x + x + 20

90 = 3x + 20

3x = 90 - 20

3x = 70

x =

- Step 3:
- ∠COD = x + 20

+ 20

+

+

- Final Answer:

### Solve the system of equations by elimination :

X - Y = 4

2X + Y = 5

HINT: Perform any arithmetic operation and then find.

Complete step by step solution:

Let x - y = 4…(i)

and 2x + y = 5….(ii)

On multiplying (i) with 2, we get 2( x - y = 4)

⇒ 2x - 2y = 8…(iii)

Now, we have the coefficients of in (ii) and (iii) to be the same.

On subtracting (ii) from (iii),

we get LHS to be 2x - 2y - (2x + y) = - 2y - y = - 3y

and RHS to be 8 - 5 = 3

On equating LHS and RHS, we have - 3y = 3

⇒ y = - 1

On substituting the value of in (i), we get - (- 1) = 4

⇒x + 1 = 4

⇒x = 4 - 1

⇒x = 3

Hence we get x = 3 and y = - 1

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 - Y = 4

2X + Y = 5

HINT: Perform any arithmetic operation and then find.

Complete step by step solution:

Let x - y = 4…(i)

and 2x + y = 5….(ii)

On multiplying (i) with 2, we get 2( x - y = 4)

⇒ 2x - 2y = 8…(iii)

Now, we have the coefficients of in (ii) and (iii) to be the same.

On subtracting (ii) from (iii),

we get LHS to be 2x - 2y - (2x + y) = - 2y - y = - 3y

and RHS to be 8 - 5 = 3

On equating LHS and RHS, we have - 3y = 3

⇒ y = - 1

On substituting the value of in (i), we get - (- 1) = 4

⇒x + 1 = 4

⇒x = 4 - 1

⇒x = 3

Hence we get x = 3 and y = - 1

Note: We can also solve these system of equations by making the coefficients of y

to be the same in both the equations.

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

X = -Y + 3

3X - 2Y = -1

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

Ans :-

Explanation :-

— eq 1

—- eq 2

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

Y = 2

Step 2 :- substitute value of y and find x

x = 1 and y = 2 is the solution of the given pair of equations

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

X = -Y + 3

3X - 2Y = -1

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

Ans :-

Explanation :-

— eq 1

—- eq 2

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

Y = 2

Step 2 :- substitute value of y and find x