## RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

**Other Exercises**

- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

**Mark the correct alternative in each of the following:
**

**Question 1.**

**If (4, 19) is a solution of the equation y = ax + 3, then a =**

**(a) 3**

**(b) 4**

**(c) 5**

**(d) 6**

**Solution:**

∵ (4, 19) is a solution of equation

y = ax + 3

∴ x = 4, y= 19 will satisfy the equation

∴ 19 = a x 4 + 3 = 4a + 3

4a = 19-3 = 16 ⇒ a= \(\frac { 16 }{ 4 }\) = 4

∴ a = 4

**(b)**

**Question 2.
**

**If (a, 4) lies on the graph of 3x + y = 10, then the value of a is**

**(a) 3**

**(b) 1**

**(c) 2**

**(d) 4**

**Solution:**

∵ (a, 4) is the solution of the equation 3x + y = 10

∴ x = a, y = 4 will satisfy the equation

∴ Substituting the value of x and y in the equation

3 xa + 4= 10 ⇒ 3a =10- 4 = 6

⇒ a = \(\frac { 6 }{ 3 }\) = 2

∴ a = 2

**(c)**

**Question 3.
**

**The graph of the linear equation 2x – y= 4 cuts x-axis at**

**(a) (2, 0)**

**(b) (-2, 0)**

**(c) (0, -4)**

**(d) (0, 4)**

**Solution:**

∵ graph of the equation,

2x – y = 4 cuts x-axis

∴ y = 0

∴ 2x – 0 = 4 ⇒ 2x = 4

⇒ x = \(\frac { 4 }{ 2 }\) = 2

∴ The line cuts x-axis at (2, 0)

**(a)**

**Question 4.
**

**How many linear equations are satisfied by x = 2 and y = -3 ?**

**(a) Only one**

**(b) Two**

**(c) Three**

**(d) Infinitely many**

**Solution:**

∵ From a point, infinitely number of lines can pass.

∴ The solution x = 2, y = -3 is the solution of infinitely many linear equations.

**(d)**

**Question 5.
**

**The equation x – 2 = 0 on number line is represented by**

**(a) aline**

**(b) a point**

**(c) infinitely many lines**

**(d) two lines**

**Solution:**

The equation x – 2 = 0

⇒ x = 2

∴ It is representing by a point on a number line.

**(b)**

**Question 6.
**

**x = 2, y = -1 is a solution of the linear equation**

**(a) x + 2y = 0**

**(b) x + 2y = 4**

**(c) 2x + y = 0**

**(d) 2x + y = 5**

**Solution:**

x = 2, y = -1

Substituting the values of x and y in the equations one by one, we get (a) x + 2y = 0

⇒ 2 + 2(-1) = 0

⇒ 2 – 2 = 0

⇒ 0 = 0 which is true

**(a)**

**Question 7.
**

**If (2k – 1, k) is a solution of the equation 10x – 9y = 12, then k =**

**(a) 1**

**(b) 2**

**(c) 3**

**(d) 4**

**Solution:**

∵ (2k – 1, k) is a solution of the equation 10x – 9y = 12

Substituting the value of x and y in the equation

10(2k – 1) – 9k = 12

⇒ 20k – 10-9k= 12

⇒ 20k – 9k = 12 + 10

⇒ 11k = 22

⇒ k =\(\frac { 22 }{ 11 }\) = 2

^{ }∴ k = 2

**(b)**

**Question 8.
**

**The distance between the graph of the equation x = – 3 and x = 2 is**

**(a) 1**

**(b) 2**

**(c) 3**

**(d) 5**

**Solution:**

The distance between the graphs of the equation

x = -3 and x = 2 will be

2(-3) = 2+ 3 = 5

**(b)**

**Question 9.
**

**The distance between the graphs of the equations y = -1 and y = 3 is**

**(a) 2**

**(b) 4**

**(c) 3**

**(d) 1**

**Solution:**

The distance between the graphs of the equation

y = -1 and y = 3

is 3 – (-1) = 3 + 1 = 4

**(b)**

**Question 10.
**

**If the graph of the equation 4x + 3y = 12 cuts the co-ordinate axes at A and B, then hypotenuse of right triangle AOB is of length**

**(a) 4 units**

**(b) 3 units**

**(c) 5 units**

**(d) none of these**

**Solution:**

Equation is 4x + 3y = 12

If it cuts the x-axis, then y = 0

∴ 4x x 3 x 0 = 12

⇒ 4x = 12 ⇒ x = \(\frac { 12 }{ 4 }\) = 3

OA = 3 units

∴ The point of intersection of x-axis is (3, 0)

Again if it cuts the y-axis, then x = 0 , Y= 0

∴ 4x x 3 x 0 = 12

⇒ 4x = 12 ⇒ x = \(\frac { 12 }{ 3 }\) = 4

⇒ OB = 4 units

∴ The point of intersection is (0, 4)

∴ In right ΔAOB,

AB

^{2}= AO

^{2}+ OB

^{2 }= (3)

^{2}+ (4)

^{2}

= 9 + 16 = 25

= (5)

^{2 }∴ AB = 5 units

**(c)**

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS are helpful to complete your math homework.

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