## RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1

Other Exercises

- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.

Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.

Solution:

Sum of four angles of a quadrilateral = 360°

Three angles are = 110°, 50° and 40°

∴ Fourth angle = 360° – (110° + 50° + 40°)

= 360° – 200° = 160°

Question 2.

In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measures of each angle of the quadrilateral.

Solution:

Sum of angles of a quadrilateral ABCD = 360°

Ratio in angles = 1 : 2 : 4 : 5

Let first angle = x

Second angle = 2x

Third angle = 4x

and fourth angle = 5x

∴ x + 2x + 4x + 5x = 360°

⇒ 12x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 12 }\) = 30°

∴ First angle = 30°

Second angle = 30° x 2 = 60°

Third angle = 30° x 4 = 120°

Fourth angle = 30° x 5 = 150°

Question 3.

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. [NCERT]

Solution:

Sum of four angles of a quadrilateral = 360°

Ratio in the angles = 3 : 5 : 9 : 13

Let first angle = 3x

Then second angle = 5x

Third angle = 9x

and fourth angle = 13x

∴ 3x + 5x + 9x+ 13x = 360°

⇒ 30x = 360°

⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\) = 12°

∴ First angle = 3x = 3 x 12° = 36°

Second angle = 5x = 5 x 12° = 60°

Third angle = 9x = 9 x 12° = 108°

Fourth angle = 13 x 12° = 156°

Question 4.

In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.

Prove that ∠COD = \(\frac { 1 }{ 2 }\) (∠A + ∠B).

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1 are helpful to complete your math homework.

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