Solution

The Bobsie twins share an L-shaped room. The area of the entire room is \(11.2\text{ m}^2\). The twins are not getting along, so their parents decide to partition the room with tape so that each child has exactly the same area.

The layout of their room is represented by \(ABCDEF\) in the diagram. The partitioning tape, indicated by a dashed line, will travel from \(C\) to a point \(G\) on \(AF\).

\(ABCDEF\) is a six-sided polygon with all right angles, forming a shape like a capital \(L\). To trace its perimeter, start at the top left vertex, \(A\), move down to the bottom left vertex, \(F\), then move right \(4\text{ m}\) to the bottom right vertex, \(E\), then up an unknown amount to \(D\) then to the left the same unknown amount to \(C\), then up \(2\text{ m}\) to reach \(B\) on the top side, then left to return to \(A\).

\(G\) is on side \(AF\) below and to the left of vertex \(C\). A slanted dashed line joins \(C\) and \(G\), dividing the polygon into two regions.

Where should \(G\) be located on \(AF\) in order to split the room into two smaller rooms of equal area?

Solution

Let \(x\) represent the length of \(CD\), in metres. Since \(DE=CD\), then \(DE=x.\)

Extend \(CD\) to intersect \(AF\) at \(H\). This creates two rectangles \(ABCH\) and \(DEFH\) with \(AB \parallel DH \parallel EF\). Also, \(AB=EF-CD=4-x.\)

We can now find the value of \(x\) using areas. \[\begin{aligned} \text{Area }ABCDEF&= \text{Area }ABCH+\text{Area }DEFH\\ 11.2&= (AB\times BC)+(DE\times EF)\\ 11.2&= (4-x)(2)+x(4)\\ 11.2&= 8-2x+4x\\ 3.2&= 2x\\ 1.6&= x \end{aligned}\] Since \(x=1.6\text{ m}\), \(CD=DE=1.6\text{ m}\), and \(AB=4-x=2.4\text{ m}\). Also, \(AH=BC=2\text{ m}\), and \(AF=DE+BC=1.6+2=3.6\text{ m}.\)

Now, the area of \(ABCH\) is \(2.4\times 2=4.8\text{ m}^2\) and the area of area \(DEFH\) is \(4\times 1.6=6.4\text{ m}^2\). Since \(6.4 > 4.8\), then \(G\) must lie on \(HF.\)

Let \(y\) represent the length of \(GH\), in metres. A diagram with updated information is below.

\(ABCG\) is a trapezoid with opposite parallel sides \(BC=2\) and \(AG=2+y\). \(AB\) is perpendicular to both \(BC\) and \(AG\), and \(AB=2.4\text{ m}\). We also know that the area of trapezoid \(ABCG\) is half the area of \(ABCDEF\), so the area of trapezoid \(ABCG\) is \(5.6\text{ m}^2.\)

Therefore, \[\begin{aligned} \text{Area of Trapezoid }ABCG&=\frac{AB\times (BC+AG)}{2}\\ 5.6&=\frac{2.4\times (2+2+y)}{2}\\ 5.6&=1.2\times (4+y)\\ 5.6&=4.8+1.2y\\ 0.8&=1.2y \end{aligned}\] Thus, \(y = \dfrac{0.8}{1.2} = \dfrac{8}{12} = \dfrac{2}{3}\). Since \(AG=2+y\), we have \(AG=2+\dfrac{2}{3}=\dfrac{8}{3}\text{ m}.\)

Also, since \(GF=AF-AG\), we have \(GF=3.6-\dfrac{8}{3}=\dfrac{18}{5}-\dfrac{8}{3}=\dfrac{54-40}{15}=\dfrac{14}{15}\text{ m}.\)

Therefore, \(G\) should be positioned \(\dfrac{14}{15}\text{ m}\) from \(F\), and \(\dfrac{8}{3}\text{ m}\) from \(A.\)

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