# NCERT Solutions for Class 9 Maths Chapter 8

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 and Exercise 8.2 in English Medium and Hindi Medium updated for new academic session 2020-21 based on latest NCERT Books of new session. Download Prashnavali 8.1 and Prashnavali 8.2 in Hindi Medium free PDF download for academic session 2020-21. UP Board students are now using NCERT Books for UP board course. So, they also take the help from these 9th Maths solutions. Download UP Board Solutions for class 9 Maths Chapter 8 in Hindi Medium free. Class 9 Maths Solutions 2020-2021 are in Hindi Medium and English Medium for all the students following NCERT Books 2020-21 for the academic session 2020-2021.

NCERT Solutions for all other subjects are also available in downloadable form. Videos related to each questions and Ex. 8.1 and Ex. 8.2 are also given with complete descriptions. Download Class 9 Maths App for Offline use or Download Kaksha 9 Ganit App for offline use.## NCERT Solutions for Class 9 Maths Chapter 8

Class: 9 | Maths (English and Hindi Medium) |

Chapter 8: | Quadrilaterals |

### 9th Maths Chapter 8 Solutions in English & Hindi Medium

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1 & 8.2 sols in English as well as medium for CBSE, UP Board, Uttarakhand, Bihar and Gujrat board, wherever the NCERT Books 2020-21 are prescribed as a course book. Download Offline Apps for session 2020-21, which work well even without internet.

### 9th Maths Exercise 8.1 Solutions

### 9th Maths Exercise 8.2 Solutions

### Study Material for Session 2020-2021

#### Class 9 Maths Exercise 8.1 Solution in Hindi Medium

#### Class 9 Maths Exercise 8.1 Solutions in Videos

#### Class 9 Maths Exercise 8.2 Solutions in Hindi Medium Video

#### Class 9 Maths Exercise 8.2 Solution in Videos

##### What are the properties of a parallelogram?

In a parallelogram

Opposite sides are equal

Opposite angles are equal

Diagonals bisect each other.

##### In which type of quadrilaterals, diagonals bisect each other?

The diagonals of following quadrilateral bisect each other:

Parallelogram

Rectangle

Square

Rhombus

#### Important Notes on 9th Maths Chapter 8

1. Sum of the all angles of a quadrilateral is 360.

2. A quadrilateral in which one pair of opposite sides are parallel, is called trapezium.

3. A quadrilateral in which both pairs of opposite sides are parallel, is called parallelogram.

4. A parallelogram in which one of its angle is right angle, is called a rectangle.

5. A parallelogram in which all side are equal, is called rhombus.

6. A rectangle with all sides equal, is a square.

7. Diagonals of parallelogram divides it into two congruent triangles.

8. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

9. A line drawn through the mid-point of a side of a triangle parallel to another side bisects the third side.

10. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, taken in order, is a parallelogram.

### Important Questions on 9th Maths Chapter 8

Therefore, the second angle = 5x,

Third angle = 9x and

Fourth angle = 13x

Sum of all angles of a quadrilateral is 360°.

Therefore, 3x + 5x + 9x + 13x = 360°

⇒ 30x = 360°

⇒ x = (360°)/30 =12°

Hence, the first angle = 3 × 12° =36°,

The second angle = 5 × 12° = 60°,

Third angle = 9 × 12° = 108°

The forth angle = 13 × 12° = 156°

S भुजा DA का मध्य-बिंदु हैं [∵ दिया है]

R भुजा DC का मध्य-बिंदु हैं [∵ दिया है]

अतः, SR || AC और SR = 1/2 AC

[∵ मध्य-बिंदु प्रमेय]

To Prove: ABCD is a rectangle.

Solution: In ΔABC and ΔBAD,

BC = AD [∵ Opposite sides of a parallelogram are equal]

AC = BD [∵ Given]

AB = AB [∵ Common]

Hence, ΔABC ≅ ΔBAD [∵SSS Congruency rule]

∠ABC = ∠BAD [∵ CPCT]

But, ∠ABC + ∠BAD = 180° [∵ Co-interior angles]

⇒ 2∠BAD = 180° [∵ ∠ABC = ∠BAD]

⇒ ∠BAD = (180°)/2 = 90°

A parallelogram with one of its angle is 90° is a rectangle.

Hence, ABCD is a rectangle.

To prove: ABCD is a rhombus.

Solution: In ΔAOB and ΔAOD,

BO = DO [∵ Given]

∠AOB = ∠AOD [∵ Each 90°]

AO = AO [∵ Common]

Hence, ΔAOB ≅ ΔAOD [∵SAS Congruency rule]

AB = AD [∵ CPCT]

Similarly, AB = BC and BC = CD

Now, all the four sides of quadrilateral ABCD are equal.

Hence, ABCD is a rhombus.

M भुजा AB का मध्य-बिंदु है [∵ दिया है]

तथा DM || BC [∵ दिया है]

अतः, D भुजा AC का मध्य-बिंदु होगा [∵ मध्य-बिंदु प्रमेय के विलोम से]