Maths-

General

Easy

Question

# Find the value of 𝑚 & 𝑛 to make a true statement.

(𝑚𝑥 + 𝑛𝑦)^{2} = 4𝑥^{2} + 12𝑥𝑦 + 9𝑦^{2}

Hint:

### The methods used to find the product of binomials are called special products.

Multiplying a number by itself is often called squaring.

For example (*x* + 3)(*x* + 3) = (*x* + 3)2

## The correct answer is: (2,2) and (-2,-2).

### (mx + ny)^{2} can be written as (mx + ny)(mx + ny)

(mx + ny)(mx + ny) = mx(mx + ny) + ny(mx + ny)

= mx(mx) + mx(ny) + ny(mx) + ny(ny)

= m^{2}x^{2} + mnxy + mnxy + n^{2}y^{2}

= m^{2}x^{2} + 2mnxy + n^{2}y^{2}

Now, m^{2}x^{2} + 2mnxy + n^{2}y^{2} = 4𝑥^{2} + 12𝑥𝑦 + 9𝑦^{2}

Comparing both sides, we get

m^{2} = 4, n = 9, 2mn = 12

So, m = +2 or -2 , n = +3 or -3

Considering 2mn = 12, there are two combinations possible

- m = +2 and n = +2
- m = -2 and n = -2

Final Answer:

Hence, the values of (m, n) are (2,2) and (-2,-2).

### Related Questions to study

Maths-

### 37. In an academic contest correct answers earn 12 points and incorrect answers lose 5

points. In the final round, school A starts with 165 points and gives the same number

of correct and incorrect answers. School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A. The game ends with the two schools tied.

i)Which equation models the scoring in the final round and the outcome of the contest

Answer:

○ Take the variable value as x or any alphabet.

correct answer = 12 points

incorrect answers = -5 points

School A starts with 165 points and gives the same number of correct and incorrect answers.

School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A

○ Step 1:

○ Let the number of correct answers given by school A be x.

So, the number of incorrect answers is also x.

At school A starts with 165 points. After giving x correct and a incorrect answers points will be

165 + 12x - 5x

School B starts with 65 points. Schools are given the same number of correct answers as school A and no incorrect answers. So, their points will be

65 + 12x

○ Step 2:

○ As both schools tied

∴ 165 + 12x - 5x = 65 + 12x

Option B. 165 + 12x - 5x = 65 + 12x

- Hint:

○ Take the variable value as x or any alphabet.

- Step by step explanation:

correct answer = 12 points

incorrect answers = -5 points

School A starts with 165 points and gives the same number of correct and incorrect answers.

School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A

○ Step 1:

○ Let the number of correct answers given by school A be x.

So, the number of incorrect answers is also x.

At school A starts with 165 points. After giving x correct and a incorrect answers points will be

165 + 12x - 5x

School B starts with 65 points. Schools are given the same number of correct answers as school A and no incorrect answers. So, their points will be

65 + 12x

○ Step 2:

○ As both schools tied

∴ 165 + 12x - 5x = 65 + 12x

- Final Answer:

Option B. 165 + 12x - 5x = 65 + 12x

### 37. In an academic contest correct answers earn 12 points and incorrect answers lose 5

points. In the final round, school A starts with 165 points and gives the same number

of correct and incorrect answers. School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A. The game ends with the two schools tied.

i)Which equation models the scoring in the final round and the outcome of the contest

Maths-General

Answer:

○ Take the variable value as x or any alphabet.

correct answer = 12 points

incorrect answers = -5 points

School A starts with 165 points and gives the same number of correct and incorrect answers.

School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A

○ Step 1:

○ Let the number of correct answers given by school A be x.

So, the number of incorrect answers is also x.

At school A starts with 165 points. After giving x correct and a incorrect answers points will be

165 + 12x - 5x

School B starts with 65 points. Schools are given the same number of correct answers as school A and no incorrect answers. So, their points will be

65 + 12x

○ Step 2:

○ As both schools tied

∴ 165 + 12x - 5x = 65 + 12x

Option B. 165 + 12x - 5x = 65 + 12x

- Hint:

○ Take the variable value as x or any alphabet.

- Step by step explanation:

correct answer = 12 points

incorrect answers = -5 points

School A starts with 165 points and gives the same number of correct and incorrect answers.

School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A

○ Step 1:

○ Let the number of correct answers given by school A be x.

So, the number of incorrect answers is also x.

At school A starts with 165 points. After giving x correct and a incorrect answers points will be

165 + 12x - 5x

School B starts with 65 points. Schools are given the same number of correct answers as school A and no incorrect answers. So, their points will be

65 + 12x

○ Step 2:

○ As both schools tied

∴ 165 + 12x - 5x = 65 + 12x

- Final Answer:

Option B. 165 + 12x - 5x = 65 + 12x

Maths-

### Find the value of x. Identify the theorem used to find the answer.

Answer:

AC = 4x – 4

AD is perpendicular bisector at BC.

A is point on AD

A is equidistant from B and C.

So,

AB = AC

2x = 4x – 4

4 = 4x – 2x

4 = 2x

x = 2

Perpendicular bisector theorem is used.

- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.

- Step by step explanation:
- Given:

AC = 4x – 4

AD is perpendicular bisector at BC.

- Step 1:
- In

A is point on AD

A is equidistant from B and C.

So,

AB = AC

2x = 4x – 4

4 = 4x – 2x

4 = 2x

x = 2

- Final Answer:

Perpendicular bisector theorem is used.

### Find the value of x. Identify the theorem used to find the answer.

Maths-General

Answer:

AC = 4x – 4

AD is perpendicular bisector at BC.

A is point on AD

A is equidistant from B and C.

So,

AB = AC

2x = 4x – 4

4 = 4x – 2x

4 = 2x

x = 2

Perpendicular bisector theorem is used.

- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.

- Step by step explanation:
- Given:

AC = 4x – 4

AD is perpendicular bisector at BC.

- Step 1:
- In

A is point on AD

A is equidistant from B and C.

So,

AB = AC

2x = 4x – 4

4 = 4x – 2x

4 = 2x

x = 2

- Final Answer:

Perpendicular bisector theorem is used.

Maths-

### Where is the circumcentre located in any right triangle? Write a coordinate proof of this result.

Answer:

O = (0, 0)

A = (2a, 0)

B = (0, 2b).

The midpoint of BC is given by,

So, the perpendicular bisector will intersect BC at M (a, b).

Equation of line BC is

And point M (a, b) satisfy above equation

b = b

Hence, point M (a, b) lies on BC.

- Hints:

- Distance between two points having coordinates (x
_{1}, y_{1}) and (x_{2}, y_{2}) is given by formula:

- Distance =

- Step by step explanation:
- Step 1:
- Let triangle ABO,

O = (0, 0)

A = (2a, 0)

B = (0, 2b).

- Step 1:
- Let triangle ABO, where:

The midpoint of BC is given by,

So, the perpendicular bisector will intersect BC at M (a, b).

Equation of line BC is

And point M (a, b) satisfy above equation

b = b

Hence, point M (a, b) lies on BC.

- Final Answer:

### Where is the circumcentre located in any right triangle? Write a coordinate proof of this result.

Maths-General

Answer:

O = (0, 0)

A = (2a, 0)

B = (0, 2b).

The midpoint of BC is given by,

So, the perpendicular bisector will intersect BC at M (a, b).

Equation of line BC is

And point M (a, b) satisfy above equation

b = b

Hence, point M (a, b) lies on BC.

- Hints:

- Distance between two points having coordinates (x
_{1}, y_{1}) and (x_{2}, y_{2}) is given by formula:

- Distance =

- Step by step explanation:
- Step 1:
- Let triangle ABO,

O = (0, 0)

A = (2a, 0)

B = (0, 2b).

- Step 1:
- Let triangle ABO, where:

The midpoint of BC is given by,

So, the perpendicular bisector will intersect BC at M (a, b).

Equation of line BC is

And point M (a, b) satisfy above equation

b = b

Hence, point M (a, b) lies on BC.

- Final Answer:

Maths-

### Ayush is choosing between two health clubs. Health club 1: Membership R s 22 and

Monthly fee R s 24.50. Health club 2: Membership R s 47.00 , monthly fee R s 18.25

. After how many months will the total cost for each health club be the same ?

Answer:

Health club 1: Membership R s 22 and Monthly fee R s 24.50

Health club 2: Membership R s 47.00 , monthly fee R s 18.25.

○ Step 1:

○ Let the number of months after which the total cost is equal be x.

So, for Health club 1:

After x months, total cost will be =

Membership Rs 22 + fee for x months

22 + 24.50x

Health club 2:

After x months, total cost will be =

Membership R s 47 + fee for x months

47 + 18.25x

○ Step 2:

○ Equalize both costs to get number of months

22 + 24.50x = 47 + 18.25x

24.50x - 18.25x= 47 - 22

6.25x = 25

x =

x = 4

- Hint:

- Step by step explanation:

Health club 1: Membership R s 22 and Monthly fee R s 24.50

Health club 2: Membership R s 47.00 , monthly fee R s 18.25.

○ Step 1:

○ Let the number of months after which the total cost is equal be x.

So, for Health club 1:

After x months, total cost will be =

Membership Rs 22 + fee for x months

22 + 24.50x

Health club 2:

After x months, total cost will be =

Membership R s 47 + fee for x months

47 + 18.25x

○ Step 2:

○ Equalize both costs to get number of months

22 + 24.50x = 47 + 18.25x

24.50x - 18.25x= 47 - 22

6.25x = 25

x =

x = 4

- Final Answer:

### Ayush is choosing between two health clubs. Health club 1: Membership R s 22 and

Monthly fee R s 24.50. Health club 2: Membership R s 47.00 , monthly fee R s 18.25

. After how many months will the total cost for each health club be the same ?

Maths-General

Answer:

Health club 1: Membership R s 22 and Monthly fee R s 24.50

Health club 2: Membership R s 47.00 , monthly fee R s 18.25.

○ Step 1:

○ Let the number of months after which the total cost is equal be x.

So, for Health club 1:

After x months, total cost will be =

Membership Rs 22 + fee for x months

22 + 24.50x

Health club 2:

After x months, total cost will be =

Membership R s 47 + fee for x months

47 + 18.25x

○ Step 2:

○ Equalize both costs to get number of months

22 + 24.50x = 47 + 18.25x

24.50x - 18.25x= 47 - 22

6.25x = 25

x =

x = 4

- Hint:

- Step by step explanation:

Health club 1: Membership R s 22 and Monthly fee R s 24.50

Health club 2: Membership R s 47.00 , monthly fee R s 18.25.

○ Step 1:

○ Let the number of months after which the total cost is equal be x.

So, for Health club 1:

After x months, total cost will be =

Membership Rs 22 + fee for x months

22 + 24.50x

Health club 2:

After x months, total cost will be =

Membership R s 47 + fee for x months

47 + 18.25x

○ Step 2:

○ Equalize both costs to get number of months

22 + 24.50x = 47 + 18.25x

24.50x - 18.25x= 47 - 22

6.25x = 25

x =

x = 4

- Final Answer:

Maths-

### Find the gradient and y- Intercept of the line

Hint:

Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.

Step by step solution:

The given equation of the line is

x + 2y = 14

We need to convert this equation in the slope-intercept form of the line, which is

y = mx + c

Rewriting the given equation, we have

2y = 14 - x

Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get

Gradient =

y-intercept = 7

Note:

We can find the slope and y-intercept directly from the general form of the equation too; slope = and y-intercept = , where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.

Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.

Step by step solution:

The given equation of the line is

x + 2y = 14

We need to convert this equation in the slope-intercept form of the line, which is

y = mx + c

Rewriting the given equation, we have

2y = 14 - x

Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get

Gradient =

y-intercept = 7

Note:

We can find the slope and y-intercept directly from the general form of the equation too; slope = and y-intercept = , where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.

### Find the gradient and y- Intercept of the line

Maths-General

Hint:

Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.

Step by step solution:

The given equation of the line is

x + 2y = 14

We need to convert this equation in the slope-intercept form of the line, which is

y = mx + c

Rewriting the given equation, we have

2y = 14 - x

Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get

Gradient =

y-intercept = 7

Note:

We can find the slope and y-intercept directly from the general form of the equation too; slope = and y-intercept = , where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.

Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.

Step by step solution:

The given equation of the line is

x + 2y = 14

We need to convert this equation in the slope-intercept form of the line, which is

y = mx + c

Rewriting the given equation, we have

2y = 14 - x

Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get

Gradient =

y-intercept = 7

Note:

We can find the slope and y-intercept directly from the general form of the equation too; slope = and y-intercept = , where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.

Maths-

### We can express any constant in the variable form without changing its value as

Explanation:

We have given a constant

We have to find how we can express any constant 𝑘 in the variable form without changing its value

We know that x

So After multiplying it with any constant, it will not change its value.

So,kx

Hence, Option C is correct.

- We have given a constant
- We have to find how we can express any constant 𝑘 in the variable form without changing its value.

We have given a constant

*k*We have to find how we can express any constant 𝑘 in the variable form without changing its value

We know that x

^{0}= 1So After multiplying it with any constant, it will not change its value.

So,kx

^{0}is the answerHence, Option C is correct.

### We can express any constant in the variable form without changing its value as

Maths-General

Explanation:

We have given a constant

We have to find how we can express any constant 𝑘 in the variable form without changing its value

We know that x

So After multiplying it with any constant, it will not change its value.

So,kx

Hence, Option C is correct.

- We have given a constant
- We have to find how we can express any constant 𝑘 in the variable form without changing its value.

We have given a constant

*k*We have to find how we can express any constant 𝑘 in the variable form without changing its value

We know that x

^{0}= 1So After multiplying it with any constant, it will not change its value.

So,kx

^{0}is the answerHence, Option C is correct.

Maths-

### x^{0} = ?

### x^{0} = ?

Maths-General

Maths-

### State and prove the Perpendicular Bisector Theorem.

Answer:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

So, according to SAS rule

Hence,

CA = CB

So, any point on perpendicular bisector is at equal distance from end points of line segment.

Hence proved.

- To prove:
- Perpendicular Bisector Theorem:

- Proof:
- Statement:
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.

- Step 1:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

So, according to SAS rule

Hence,

CA = CB

So, any point on perpendicular bisector is at equal distance from end points of line segment.

Hence proved.

### State and prove the Perpendicular Bisector Theorem.

Maths-General

Answer:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

So, according to SAS rule

Hence,

CA = CB

So, any point on perpendicular bisector is at equal distance from end points of line segment.

Hence proved.

- To prove:
- Perpendicular Bisector Theorem:

- Proof:
- Statement:

- Step 1:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

So, according to SAS rule

Hence,

CA = CB

So, any point on perpendicular bisector is at equal distance from end points of line segment.

Hence proved.

Maths-

### What is a monomial? Explain with an example.

Explanation:

A monomial is an expression with only one term.

The examples are: 3x, 6y

- We have to define monomial by giving an example.

A monomial is an expression with only one term.

The examples are: 3x, 6y

### What is a monomial? Explain with an example.

Maths-General

Explanation:

A monomial is an expression with only one term.

The examples are: 3x, 6y

- We have to define monomial by giving an example.

A monomial is an expression with only one term.

The examples are: 3x, 6y

Maths-

### Find the equation of a line that passes through and

Hint:

We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:

Let the given points be denoted by

(a, b) = (-3, 1)

(c, d) = (2, -14)

The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have

Cross multiplying, we get

5(y + 14) = -15(x - 2)

Expanding the factors, we have

5y + 70 = -15x + 30

Taking all the terms in the left hand side, we have

15x + 5y + 70 - 30 = 0

Finally, the equation of the line is

15x + 5y + 40 = 0

Dividing the equation throughout by 5, we get

3x + y + 8 = 0

This is the required equation.

Note:

We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.

We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:

Let the given points be denoted by

(a, b) = (-3, 1)

(c, d) = (2, -14)

The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have

Cross multiplying, we get

5(y + 14) = -15(x - 2)

Expanding the factors, we have

5y + 70 = -15x + 30

Taking all the terms in the left hand side, we have

15x + 5y + 70 - 30 = 0

Finally, the equation of the line is

15x + 5y + 40 = 0

Dividing the equation throughout by 5, we get

3x + y + 8 = 0

This is the required equation.

Note:

We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.

### Find the equation of a line that passes through and

Maths-General

Hint:

We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:

Let the given points be denoted by

(a, b) = (-3, 1)

(c, d) = (2, -14)

The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have

Cross multiplying, we get

5(y + 14) = -15(x - 2)

Expanding the factors, we have

5y + 70 = -15x + 30

Taking all the terms in the left hand side, we have

15x + 5y + 70 - 30 = 0

Finally, the equation of the line is

15x + 5y + 40 = 0

Dividing the equation throughout by 5, we get

3x + y + 8 = 0

This is the required equation.

Note:

We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.

We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:

Let the given points be denoted by

(a, b) = (-3, 1)

(c, d) = (2, -14)

The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have

Cross multiplying, we get

5(y + 14) = -15(x - 2)

Expanding the factors, we have

5y + 70 = -15x + 30

Taking all the terms in the left hand side, we have

15x + 5y + 70 - 30 = 0

Finally, the equation of the line is

15x + 5y + 40 = 0

Dividing the equation throughout by 5, we get

3x + y + 8 = 0

This is the required equation.

Note:

We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.

Maths-

### The degree of 25x^{2}y^{23} is

Explanation:

We have given an expression

We know that degree is highest power of variable present in polynomial.

So,

The degree of is

Degree of is

Degree of is

So, Total degree is

Hence, Option B is correct.

- We have been given a polynomial expression in the question for which we have to find its degree.

We have given an expression

We know that degree is highest power of variable present in polynomial.

So,

The degree of is

Degree of is

Degree of is

So, Total degree is

Hence, Option B is correct.

### The degree of 25x^{2}y^{23} is

Maths-General

Explanation:

We have given an expression

We know that degree is highest power of variable present in polynomial.

So,

The degree of is

Degree of is

Degree of is

So, Total degree is

Hence, Option B is correct.

- We have been given a polynomial expression in the question for which we have to find its degree.

We have given an expression

We know that degree is highest power of variable present in polynomial.

So,

The degree of is

Degree of is

Degree of is

So, Total degree is

Hence, Option B is correct.

Maths-

### Ritu earns R s 680 in commission and is paid R s 10.25 per hour. Karina earns R s

410 in commissions and is paid R s 12.50 per hour. What will you find if you solve

for x in the equation 10.25x + 680 = 12.5x + 480

Answer:

10.25x + 680 = 12.5x + 480

○ Step 1:

○ Solve the equation.

○ Group the like terms

10.25x + 680 = 12.5x + 480

680 - 480 = 12.5x - 10.25x

200 = 2.25x

○ Step 2:

○ Divide both side with 2.25

88.88 = x

- Hint:

- Step by step explanation:

10.25x + 680 = 12.5x + 480

○ Step 1:

○ Solve the equation.

○ Group the like terms

10.25x + 680 = 12.5x + 480

680 - 480 = 12.5x - 10.25x

200 = 2.25x

○ Step 2:

○ Divide both side with 2.25

88.88 = x

- Final Answer:

### Ritu earns R s 680 in commission and is paid R s 10.25 per hour. Karina earns R s

410 in commissions and is paid R s 12.50 per hour. What will you find if you solve

for x in the equation 10.25x + 680 = 12.5x + 480

Maths-General

Answer:

10.25x + 680 = 12.5x + 480

○ Step 1:

○ Solve the equation.

○ Group the like terms

10.25x + 680 = 12.5x + 480

680 - 480 = 12.5x - 10.25x

200 = 2.25x

○ Step 2:

○ Divide both side with 2.25

88.88 = x

- Hint:

- Step by step explanation:

10.25x + 680 = 12.5x + 480

○ Step 1:

○ Solve the equation.

○ Group the like terms

10.25x + 680 = 12.5x + 480

680 - 480 = 12.5x - 10.25x

200 = 2.25x

○ Step 2:

○ Divide both side with 2.25

88.88 = x

- Final Answer:

Maths-

### If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then the triangle is ____ isosceles.

Answer:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

ADC = BDC = 90

So, according to SAS rule

ACD BCD

Hence,

CA = CB

So, triangle is always isosceles.

Hence proved.

- Hints:

- Perpendicular bisector theorem

- Step by step explanation:
- Step 1:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

ADC = BDC = 90

^{o}So, according to SAS rule

ACD BCD

Hence,

CA = CB

So, triangle is always isosceles.

Hence proved.

- Final Answer:

### If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then the triangle is ____ isosceles.

Maths-General

Answer:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

ADC = BDC = 90

So, according to SAS rule

ACD BCD

Hence,

CA = CB

So, triangle is always isosceles.

Hence proved.

- Hints:

- Perpendicular bisector theorem

- Step by step explanation:
- Step 1:

Let arbitrary point C on perpendicular bisector.

In which,

CD is perpendicular bisector on AB.

Hence,

AD = DB

CD = CD (common)

ADC = BDC = 90

^{o}So, according to SAS rule

ACD BCD

Hence,

CA = CB

So, triangle is always isosceles.

Hence proved.

- Final Answer:

Maths-

### Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B. On which of the following segments must P be located?

Answer:

So,

As P is equidistant from A and B it lies on perpendicular bisector on AB.

- Hints:

- Perpendicular bisector theorem

- Step by step explanation:
- Given:

- Step 1:
- In △ ABC,

So,

As P is equidistant from A and B it lies on perpendicular bisector on AB.

- Final Answer:

### Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B. On which of the following segments must P be located?

Maths-General

Answer:

So,

As P is equidistant from A and B it lies on perpendicular bisector on AB.

- Hints:

- Perpendicular bisector theorem

- Step by step explanation:
- Given:

- Step 1:
- In △ ABC,

So,

As P is equidistant from A and B it lies on perpendicular bisector on AB.

- Final Answer:

Maths-

### A constant has a degree__________.

Explanation:

The degree of a constant is Zero.

By definition of degree we know that it is highest power of variable present in polynomial.

But for constant there is no polynomial.

So, The degree of constant is Zero

Hence, Option B is correct.

- We have been given a statement in the question for which we have to fill the blank by choosing the appropriate answer from the given four options.

The degree of a constant is Zero.

By definition of degree we know that it is highest power of variable present in polynomial.

But for constant there is no polynomial.

So, The degree of constant is Zero

Hence, Option B is correct.

### A constant has a degree__________.

Maths-General

Explanation:

The degree of a constant is Zero.

By definition of degree we know that it is highest power of variable present in polynomial.

But for constant there is no polynomial.

So, The degree of constant is Zero

Hence, Option B is correct.

- We have been given a statement in the question for which we have to fill the blank by choosing the appropriate answer from the given four options.

The degree of a constant is Zero.

By definition of degree we know that it is highest power of variable present in polynomial.

But for constant there is no polynomial.

So, The degree of constant is Zero

Hence, Option B is correct.