SOLUTION
269

Define lengths $x, y$ and $z$ as shown in the diagram, so that the larger equilateral triangle has side length $x+z$.
We have $x=\frac{10}{\sin 60}=\frac{20}{\sqrt{3}}$.
Also, $y=\frac{10}{\tan 60}=\frac{10}{\sqrt{3}}$ and hence $z=10-\frac{10}{\sqrt{3}}$.
The perimeter of the larger equilateral triangle is
$$
3(x+z)=3\left(\frac{20}{\sqrt{3}}+10-\frac{10}{\sqrt{3}}\right)=3\left(\frac{10}{\sqrt{3}}+10\right)=30+10 \sqrt{3} \text {. }
$$

Therefore, $a=30, b=10$ and $c=3$ and so $7 a+5 b+3 p=210+50+9=269$.
15. The perimeter of the square in the figure is 40 . The perimeter of the larger equilateral triangle in the figure is $a+b \sqrt{p}$, where $p$ is a prime number. What is the value of $7 a+5 b+3 p$ ?