Question

The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a .. A) square (B) rectangle (C) rhombus( D )trapaezium

Answer 1

AB= √(9-9)²+(6-0)²
=6
BC= √(-9-9)²+(6-6)²
=18
CD = √(-9+9)²+(0-6)²
=6
DA = √(-9-9)²+(0-0)²
=18
so,it's opposite sides are equal,hence it is rectangle.

Answer 2

Given:

Points:

A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0)

To find:

the type of figure formed

Solution:

We first calculate the distance between the points:

AB=$\sqrt{(9-9)^2+(6-0)^2}$ = 6

BC=$\sqrt{(9-(-9))^2+(6-6)^2}$ =18

CD=$\sqrt{(-9-(-9))^2+(6-0)^2}$ =6

AD=$\sqrt{(9-(-9))^2+(0-0)^2}$ =18

Since, AB=CD and BC=AD, the points represent vertices in a rectangle.

Hence, the points are the vertices of a rectangle. (option B)