Q5. - Algebra 2

Question

What is the area of the square ABCD?

(A) 27 units ${}^{\text{2}}$

(B) $9\sqrt{2}$ units ${}^{\text{2}}$

(C) 18 units ${}^{\text{2}}$

(D) $12\sqrt{2}$ units ${}^{\text{2}}$

Step-by-Step Solution

Answer

The area of square ABCD is (C) 18 units².

Step 1 – Evaluate the length of one side of the square AB

In the graph, the points of the vertices of square are given. They are$A(-4,2),$ $B(-1,5),C(2,2)$and $D(-1,1).$

To evaluate the length of one side, we have to evaluate the distance between two points.

The distance between two points $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Step 2 – Evaluate the distance between A and B

Therefore, the length of AB is

$\begin{array}{c}=\sqrt{{\left(-1+4\right)}^2+{\left(5-2\right)}^2}\\ =\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}\\ =\sqrt{9+9}\\ =\sqrt{18}\end{array}$

Step 3 – Evaluate the area of square ABCD

We know that all the sides of the square have same length.

Area of square$=s^2$

We know that $s=\sqrt{18}$units.

Therefore, area of square ABCD is

$\begin{array}{l}={\left(\sqrt{18}\right)}^2\\ =\sqrt{18}\times \sqrt{18}\end{array}$

=18 units ²