# Question 2

The following picture shows a (not to scale) triangle which is divided into four regions. The areas of three of the regions are shown. Find the area of the shaded region.

## Solution by Chums

We decorate the diagram as shown:

Let the desired area be \(U = x+y\).

Now, \(\sin A = \sin(\pi-D) = \sin D\). By the sine rule, \(\frac{1}{2}ab\sin A=10\) and \(\frac{1}{2}cf\sin D=\frac{1}{2}cf\sin A=8\).

As a result, \(8/cf = 10/cb\), so \(5f=4b\).

Similarly, \(\frac{1}{2}cb\sin D=10\) and \(\frac{1}{2}gb\sin D=5\) so \(2g=c\).

Also, \(\frac{1}{2}fg\sin A=y\) and \(\frac{1}{2}cb\sin A=10\), and that gives us \(\frac{y}{fg} = \frac{10}{cb}\), and so \(2ygb = ycb = 10fg\), and so \(yb = 5f = 4b\), and so \(y=4\). It remains to work out \(x\).

Now, \(x=\frac{1}{2}\sin F\), and the whole triangle has \[x+27 = \frac{1}{2}(m+d)(n+e)\sin F = x + \frac{1}{2}(me+dn+de)\sin F.\]

Hence \(\frac{1}{2}(me+dn+de)\sin F = 27\).

Also, by considering two other triangles we get \(\frac{1}{2}m(n+e)\sin F = x+9\) and \(\frac{1}{2}n(m+d)\sin F=x+12\). Subtracting our formula for \(x\) gives us \(\frac{1}{2}me\sin F=9\) and \(\frac{1}{2}nd\sin F=12\).

But as \(\frac{1}{2}me\sin F + \frac{1}{2}nd\sin F + \frac{1}{2}de\sin F=27\), we get \(\frac{1}{2}de\sin F = 6\).

This gives us that \(\frac{6}{de} = \frac{12}{nd}\), and so \(e = n/2\). This finally gives us that \(x = \frac{1}{2}mn\sin F = 2\frac{1}{2}me\sin F = 2\cdot 9=18\).

Hence \(U = x+y = 18+4 = 22\).