# 10th CBSE math solution for exercise 7.1 part 3

This page 10th CBSE math solution for exercise 7.1 part 3 is going to provide you solution for every problems that you find in the exercise no 7.1

## 10th CBSE math solution for exercise 7.1 part 3

(6) Name the type of quadrilateral formed. If any,by the following points,and give reasons for your answer:

(i) (-1,-2) (1,0) (-1,2) (-3,0)

Solution:

Let the given points as A(-1,-2)  B(1,0)  C(-1,2) and D (-3,0)

Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²

Length of the side AB

Here x₁ = -1, y₁ = -2, x₂ = 1  and  y₂ = 0

= √(1-(-1))² + (0-(-2))²

= √(1+1)² + (2)²

= √4 + 4

= √8

Length of the side BC

Here x₁ = 1, y₁ = 0, x₂ = -1  and  y₂ = 2

= √(-1-1)² + (2-0)²

= √(-2)² + (2)²

= √4 + 4

= √8

Length of the side CD

Here x₁ = -1, y₁ = 2, x₂ = -3  and  y₂ = 0

= √(-3-(-1))² + (0-2)²

= √(-3+1)² + (-2)²

= √4 + 4

= √8

Length of the side DA

Here x₁ = -3, y₁ = 0, x₂ = -1  and  y₂ = -2

= √(-1-(-3))² + (-2-0)²

= √(-1+3)² + (-2)²

= √4 + 4

= √8

AB = BC = CD = DA. Since length of all sides are equal. So it forms an square.

In the page 10th CBSE math solution for exercise 7.1 part 3 we are going to see the solution of next problem

(ii) (-3,5) (3,1) (0,3) (-1,-4)

Solution:

Let the given points as A(-3,5)  B(3,1)  C(0,3) and D (-1,-4)

Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²

Length of the side AB

Here x₁ = -3, y₁ = 5, x₂ = 3  and  y₂ = 1

= √(3-(-3))² + (1-5)²

= √(3+3)² + (-4)²

= √36 + 16

= √52

Length of the side BC

Here x₁ = 3, y₁ = 1, x₂ = 0  and  y₂ = 3

= √(0-3)² + (3-1)²

= √(-3)² + (2)²

= √9 + 4

= √13

Length of the side CD

Here x₁ = 0, y₁ = 3, x₂ = -1  and  y₂ = -4

= √(-1-0)² + (-4-3)²

= √(-1)² + (-7)²

= √1 + 49

= √50

Length of the side DA

Here x₁ = -1, y₁ = -4, x₂ = -3  and  y₂ = 5

= √(-3-(-1))² + (5-(-4))²

= √(-3+1)² + (5 +4)²

= √4 + 81

= √85

The length of all sides of quadrilateral are of different. Therefore,it can be observed that it is a only a quadrilateral.