# 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 $$U = x+y = 18+4 = 22$$.