Rectangle \(ACEG\) has \(B\) on \(AC\) and \(F\) on \(EG\) such that \(BF\) is parallel to \(CE\). Also, \(D\) is on \(CE\) and \(H\) is on \(AG\) such that \(HD\) is parallel to \(AC\), and \(BF\) intersects \(HD\) at \(J\). The area of rectangle \(ABJH\) is \(6\text{ cm}^2\) and the area of rectangle \(JDEF\) is \(15\text{ cm}^2\).
If the dimensions of rectangles \(ABJH\) and \(JDEF\), in centimetres, are integers, then determine the largest possible area of rectangle \(ACEG\). Note that the diagram is just an illustration and is not intended to be to scale.
Let \(AB = x\), \(AH = y\), \(JD = a\) and \(JF = b\).
Then, $$\begin{align} HJ &= GF = AB = x\\ BJ &= CD = AH= y\\ BC &= FE = JD = a\\ HG &= DE = JF = b \end{align}$$
Thus, we have $$\begin{align} \text{area}(ACEG) &= \text{area}(ABJH) + \text{area}(BCDJ) + \text{area}(JDEF) + \text{area}(HJFG)\\ &= 6 + ya + 15 + xb\\ &= 21 + ya + xb \end{align}$$
Since the area of rectangle \(ABJH\) is \(6\text{ cm}^2\) and the side lengths of \(ABJH\) are integers, then the side lengths must be \(1\) and \(6\) or \(2\) and \(3\). That is, \(x = 1\) cm and \(y = 6\) cm, \(x = 6\) cm and \(y = 1\) cm, \(x = 2\) cm and \(y = 3\) cm, or \(x = 3\) cm and \(y = 2\) cm.
Since the area of rectangle \(JDEF\) is \(15\text{ cm}^2\) and the side lengths of \(JDEF\) are integers, then the side lengths must be \(1\) and \(15\) or \(3\) and \(5\). That is, \(a = 1\) cm and \(b = 15\) cm, \(a = 15\) cm and \(b = 1\) cm, \(a = 3\) cm and \(b = 5\) cm, or \(a = 5\) cm and \(b = 3\) cm.
To maximize the area, we need to pick the values of \(x\), \(y\), \(a\), and \(b\) which make \(ya + xb\) as large as possible. We will now break into cases based on the possible side lengths of \(ABJH\) and \(JDEF\) and calculate the area of \(ACEG\) in each case. We do not need to try all \(16\) possible pairings, because trying \(x = 1\) cm and \(y=6\) cm with the four possibilities of \(a\) and \(b\) will give the same \(4\) areas, in some order, as trying \(x=6\) cm and \(y=1\) cm with the four possibilities of \(a\) and \(b\). Similarly, trying \(x = 2\) cm and \(y=3\) cm with the four possibilities of \(a\) and \(b\) will give the same \(4\) areas, in some order, as trying \(x=3\) cm and \(y=2\) cm with the four possibilities of \(a\) and \(b\). (As an extension, we will leave it to you to think about why this is the case.)
Case 1: \(x = 1\) cm, \(y = 6\) cm, \(a=1\) cm, \(b=15\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 6(1) + 1(15) = 42\text{ cm}^2\).
Case 2: \(x = 1\) cm, \(y = 6\) cm, \(a=15\) cm, \(b=1\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 6(15) + 1(1) = 112\text{ cm}^2\).
Case 3: \(x = 1\) cm, \(y = 6\) cm, \(a=3\) cm, \(b=5\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 6(3) + 1(5) = 44\text{ cm}^2\).
Case 4: \(x = 1\) cm, \(y = 6\) cm, \(a=5\) cm, \(b=3\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 6(5) + 1(3) = 54\text{ cm}^2\).
Case 5: \(x = 2\) cm, \(y = 3\) cm, \(a=1\), \(b=15\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 3(1) + 2(15) = 54\text{ cm}^2\).
Case 6: \(x = 2\) cm, \(y = 3\) cm, \(a=15\), \(b=1\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 3(15) + 2(1) = 68\text{ cm}^2\).
Case 7: \(x = 2\) cm, \(y = 3\) cm, \(a=3\), \(b=5\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 3(3) + 2(5) = 40\text{ cm}^2\).
Case 8: \(x = 2\) cm, \(y = 3\) cm, \(a=5\), \(b=3\) cm
Then \(\text{area}(ACEG) = 21 + ya + xb = 21 + 3(5) + 2(3) = 42\text{ cm}^2\).
We see that the maximum area is \(112\text{ cm}^2\), and occurs when \(x = 1\) cm, \(y = 6\) cm and \(a=15\) cm, \(b=1\) cm. It will also occur when \(x = 6\) cm, \(y = 1\) cm and \(a=1\) cm, \(b=15\) cm.
The following diagrams show the calculated values placed on the original diagram. The diagram given in the problem was definitely not drawn to scale! Both solutions produce rectangles with dimensions \(7\text{ cm}\) by \(16\text{ cm}\), and area \(112\text{ cm}^2\).