Using two cuts, we want to divide the \(6\) m by \(6\) m grid shown into three regions of equal area.
One way to do so is by making a horizontal cut through \(H\) and a second horizontal cut through \(K\). This method of cutting the grid works, but is not very creative.
To make things a little more interesting, we must still make two straight cuts, but each cut must start at point \(P\). Each of these two cuts will pass through a point on the outer perimeter of the grid.
Find the length of each cut. Round your answer to one decimal.
The area of the entire \(6\) m by \(6\) m square grid is \(6\times 6 = 36\mbox{ m}^2\). Since the square is divided into three regions of equal area, the area of each region must be \(\dfrac{36}{3} = 12\mbox{ m}^2\).
Consider the line through \(P\) that passes through some point on side \(QM\). Let \(A\) be the point where this line intersects \(QM\).
Since \(\angle PMQ = 90^{\circ}\), \(\triangle PMA\) is a right-angled triangle with base \(PM = 6\) m and height \(MA\).
Using the formula \(\text{area} = \dfrac{\text{base}\times\text{height}}{2}\), we have \(\mbox{area of }\triangle PMA = \dfrac{6\times MA}{2} = 3\times MA\).
We need the area of \(\triangle PMA\) to be \(12\mbox{ m}^2\). Therefore, \(3 \times MA = 12\), and so \(MA = 4\) m. Since \(H\) is the point on \(QM\) with \(MH = 4\) m, we must have \(A =H\). Therefore, one line passes through the point \(H\).
Since \(\triangle PMA\) is a right-angled triangle, using the Pythagorean Theorem we have \[\begin{aligned} PA^2 &= PM^2 + MA^2\\ &= 6^2 + 4^2\\ &= 36 + 16\\ &= 52 \end{aligned}\] Therefore, \(PA = \sqrt{52} \approx 7.2\), since \(PA >0\).
Consider the line through \(P\) that passes through some point on side \(RQ\). Let \(B\) be the point where this line intersects \(RQ\).
Since \(\angle PRQ = 90^{\circ}\), \(\triangle PRB\) is a right-angled triangle with height \(PR = 6\) m and base \(RB\).
Using the formula \(\text{area} = \dfrac{\text{base}\times\text{height}}{2}\), we have \(\mbox{area of } \triangle PRB = \dfrac{RB\times 6}{2} = 3\times RB\).
We need the area of \(\triangle PRB\) to be \(12\mbox{ m}^2\). Therefore, \(3\times RB = 12\), and so \(RB = 4\) m. Since \(V\) is the point on \(RQ\) with \(RV = 4\) m, we must have \(B =V\). Therefore, the other line passes through the point \(V\).
Therefore, one line passes through point \(H\) and the other passes through point \(V\).
Since \(\triangle PRB\) is a right-angled triangle, using the Pythagorean Theorem we have \[\begin{aligned} PB^2 &= PR^2 + RB^2\\ &= 6^2 + 4^2\\ &= 36 + 16\\ &= 52 \end{aligned}\] Therefore, \(PB = \sqrt{52}\approx 7.2\), since \(PB>0\).
Therefore, the length of each cut is approximately \(7.2\) m.
Extension:
Try dividing the grid into three regions of equal area using three cuts. (Each cut does not necessarily need to be to the outer perimeter of the grid.)