# 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.

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## 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]