# Problem of the Week Problem C and Solution Penned In

## Problem

A square enclosure, labelled $$ABCD$$, is sketched out on a piece of graph paper.

Three of the vertices of the square $$ABCD$$ are located at $$A(0,3)$$, $$B(4,0)$$, and $$C(7,4)$$.

Determine the area of the square enclosure.

## Solution

Solution 1

In this solution we will determine the area of $$ABCD$$ without using the Pythagorean Theorem.

We will first determine the coordinates of the fourth vertex $$D$$. To do so, observe that to get from $$A$$ to $$B$$, you would go down $$3$$ units and right $$4$$ units. To get from $$B$$ to $$C$$, you move $$3$$ units to the right and then $$4$$ units up. Continuing the pattern, going up $$3$$ units and left $$4$$ units, you get to $$D(3,7)$$. Continuing, as a check, go left $$3$$ units and down $$4$$ units, and you arrive back at $$A$$. The coordinates of $$D$$ are $$(3,7)$$.

Draw a box with horizontal and vertical sides so that each vertex of the square $$ABCD$$ is on one of the sides of the box. This creates a large square with sides of length $$7$$ containing four congruent triangles and square $$ABCD$$. Each of the triangles has a base $$4$$ units long and height $$3$$ units long.

\begin{aligned} \text{Area }ABCD&=\text{Area of Large Square}-4\times \text{Area of One Triangle}\\ &=\text{Length}\times \text{Width}-4\times(\mbox{Base}\times \text{Height}\div 2)\\ &=7\times 7 - 4\times (4\times 3 \div 2)\\ &=49-4\times 6\\ &=49-24\\ &=25\end{aligned}

Therefore, the area of the enclosure is $$25$$ units$$^2$$.

Solution 2

In this solution we will determine the area of $$ABCD$$ using the Pythagorean Theorem.

Since $$ABCD$$ is a square, it is only necessary to find the length of one side. We can determine the area by squaring the length of the side.

Label the origin $$O$$.
$$OA$$, the distance from the origin to point $$A$$ on the $$y$$-axis, is $$3$$ units. $$OB$$, the distance from the origin to point $$B$$ on the $$x$$-axis, is $$4$$ units. Since $$A$$ lies on the $$y$$-axis and $$B$$ lies on the $$x$$-axis, $$OAB$$ forms a right-angled triangle.

Using the Pythagorean Theorem in right-angled $$\triangle OAB$$, we can find $$AB^2$$ which is $$AB \times AB$$, the area of the square.

\begin{aligned} AB^2&=OA^2+OB^2\\ &=3^2+4^2\\ &=9+16\\ &=25\end{aligned}

Therefore, the area of the enclosure is $$25$$ units$$^2$$.