9. Three equilateral triangles are cut from the corners of a large equilateral triangle to form an irregular hexagon, as shown in the diagram.
The perimeter of the large equilateral triangle is $60 \mathrm{~cm}$. The perimeter of the irregular hexagon is $40 \mathrm{~cm}$. What is the sum of the perimeters of the triangles that were cut from the large triangle?
A $60 \mathrm{~cm}$
B $66 \mathrm{~cm}$
C $72 \mathrm{~cm}$
D $75 \mathrm{~cm}$
E $81 \mathrm{~cm}$
9. A Let the lengths of the sides of the equilateral triangles that are cut off be $x \mathrm{~cm}, y \mathrm{~cm}$ and $z \mathrm{~cm}$, as shown in the diagram.
The length of a side of the large equilateral triangle is $\frac{1}{3} \times 60 \mathrm{~cm}=20 \mathrm{~cm}$. The perimeter of the irregular hexagon is $40 \mathrm{~cm}$. Therefore we have
$40=x+(20-x-y)+y+(20-y-z)+z+(20-z-x)$. Hence $40=60-(x+y+z)$ and therefore $x+y+z=20$. Therefore the sum of the perimeters of the triangles cut off is $(3 x+3 y+3 z) \mathrm{cm}=60 \mathrm{~cm}$.