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Problem of the Week
Problem C and Solution
Squared Up

Problem

Square \(ABCD\) is drawn on graph paper with three of the vertices located at \(A(0,4)\), \(B(3,0)\), and \(C(7,3)\), as shown.

Points A, B, and C are plotted on a grid with a line from A to B and a line from B to C.

Determine the area of square \(ABCD\).

Solution

Solution 1

In this solution we start by determining the coordinates of the fourth vertex. To do this, we observe that to get from \(A\) to \(B\) we move \(3\) units right and \(4\) units down. Then, to get from \(B\) to \(C\) we move \(4\) units right and \(3\) units up. Continuing the pattern, to get from \(C\) to \(D\) we will move \(3\) units left and \(4\) units up to reach \(D(4,7)\). As a check, to get from \(D\) to \(A\) we move \(4\) units left and \(3\) units down, as desired.

Vertex D with coordinates (4,7) is plotted on the grid, completing square ABCD.

Alternatively you could use a protractor to draw sides \(AD\) and \(CD\) on grid paper, knowing that the corners of a square have right angles. This would also give you \(D(4,7)\).

Next we draw rectangle \(EFGH\) with horizontal and vertical sides around \(ABCD\) so that each vertex of \(ABCD\) lies on one of the sides of the rectangle. This rectangle ends up being a square with side length \(7\).

Square EFGH is plotted on the grid. E is the top-left vertex (on the y-axis), H is the top-right vertex, G is the bottom right-vertex (on the x-axis), and F is the bottom-left vertex (at the origin).

Inside \(EFGH\) is square \(ABCD\) as well as four identical right-angled triangles, namely \(\triangle AED\), \(\triangle AFB\), \(\triangle BGC\), and \(\triangle CHD\). Each of these right-angled triangles has a base of \(4\) and a height of \(3\).

Finally, we calculate the area of \(ABCD\). Since the triangles are all identical, their total area is equal to four times the area of any one of them. \[\begin{aligned} \text{Area }ABCD &= \text{Area }EFGH - 4 \times \text{Area }\triangle AED\\ &= 7 \times 7 - 4 \times (4 \times 3 \div 2)\\ &= 49 - 4 \times 6\\ &= 49 - 24 = 25 \end{aligned}\] Therefore, the area of \(ABCD\) is \(25 \text{ units}^2\).

Solution 2

In this solution we will calculate the area of \(ABCD\) without determining the coordinates of \(D\). Instead we will use the Pythagorean Theorem.

Since \(ABCD\) is a square, in order to calculate its area we need to find the length of only one of its sides. Let \(O(0,0)\) be the point at the origin. Then \(\triangle AOB\) is a right-angled triangle.

Triangle AOB, with a right angle at vertex O, plotted on the grid.

The length of side \(OA\) is \(4\) and the length of side \(OB\) is \(3\). We can use the Pythagorean Theorem to calculate \(AB^2\), which is the area of square \(ABCD\). \[\begin{aligned} AB^2 &= OA^2 + OB^2\\ &= 4^2+3^2\\ &= 16+9=25 \end{aligned}\]

Therefore, the area of \(ABCD\) is \(25 \text{ units}^2\).