Question 1102350
<br>
As the problem says, when an equilateral triangle is inscribed in another by joining the midpoints of the sides of the first triangle, the side length of the second triangle is half the length of the first.<br>
So the perimeter of the first triangle is 48; that means the second has a perimeter of 24, the third a perimeter of 12, and the fourth a perimeter of 6.  The sum fo the perimeters is 48+24+12+6 = 90