Group Event 9
G9.1-G9.3
In the figure, $B C$ is parallel to $D E$.
If $A B: B C: B F: C F: F E=5: 4: 2: 3: 5$ and the area of $\triangle B C F$ is 12 , find
G9.1 the area of $\triangle B D F$,
G9.2 the area of $\triangle F D E$,
G9.3 the area of $\triangle A B C$.
G9.1 $\triangle B C F \sim \triangle E D F$ (equiangular)
$D F: E F: D E=C E: F B: B C$ (ratio of sides, $\sim \Delta \mathrm{s}$ )
$$
D F=3 \times \frac{5}{2}=7.5, D E=4 \times \frac{5}{2}=10
$$

The area of $\triangle B D F=12 \times \frac{7.5}{3}=30$
G9.2 The area of $\triangle F D E=30 \times \frac{5}{2}=75$
G9.3 The area of $\triangle C E F=12 \times \frac{5}{2}=30$
The area of $B C E D=12+30+30+75=147$
$\triangle A B C \sim \triangle A D E$ (equiangular)
Area of $\triangle A B C:$ area of $\triangle A D E=B C^{2}: D E^{2}=4^{2}: 10^{2}=4: 25$
Let the area of $\triangle A B C$ be $y$
$$
\begin{array}{l}
y:(y+147)=4: 25 \\
4 y+588=25 y \\
21 y=588 \\
y=\text { Area of } \triangle A B C=28
\end{array}
$$

G9.4 If the volume of a sphere is increased by $72.8 \%$, then the surface area of the sphere is increased by $x \%$. Find $x$.
Let the original radius of the sphere be $r$ and the new radius be $R$
$$
\begin{array}{l}
\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi r^{3} \cdot(1+72.8 \%) \\
\left(\frac{R}{r}\right)^{3}=1.728=1.2^{3} \\
\Rightarrow R=1.2 r \\
\Rightarrow x=20
\end{array}
$$