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) trapezium

Answer 1

(b) rectangle
$A\left(9,0\right),B\left(9,6\right),C\left(-9,6\right)\text{and}D\left(-9,0\right)$ are the given vertices.
Then,
$$\begin{gathered} A{B}^2={\left(9-9\right)}^2+{\left(6-0\right)}^2 \\ ={\left(0\right)}^2+{\left(6\right)}^2 \\ =0+36 \\ =36\text{units} \\ B{C}^2={\left(-9-9\right)}^2+{\left(6-6\right)}^2 \\ ={\left(-18\right)}^2+{\left(0\right)}^2 \\ =324+0 \\ =324\text{units} \\ C{D}^2={\left(-9+9\right)}^2+{\left(0-6\right)}^2 \\ ={\left(0\right)}^2+{\left(-6\right)}^2 \\ =0+36 \\ =36\text{units} \\ D{A}^2={\left(-9-9\right)}^2+{\left(0-0\right)}^2 \\ ={\left(-18\right)}^2+{\left(0\right)}^2 \\ =324+0 \\ =324\text{units} \end{gathered}$$
Therefore, we have:
$A{B}^2=C{D}^2\text{and}B{C}^2=D{A}^2$
Now, the diagonals are:
$$\begin{gathered} A{C}^2={\left(-9-9\right)}^2+{\left(6-0\right)}^2 \\ ={\left(-18\right)}^2+{\left(6\right)}^2 \\ =324+36 \\ =360\text{units} \\ B{D}^2={\left(-9-9\right)}^2+{\left(0-6\right)}^2 \\ ={\left(-18\right)}^2+{\left(-6\right)}^2 \\ =324+36 \\ =360\text{units} \end{gathered}$$
Therefore,
$A{C}^2=B{D}^2$
Hence, ABCD is a rectangle.