Answer:
Since the ratio of the altitudes of the two triangles ABC and DEF is 1:3, we can say that the altitude of triangle DEF is three times that of triangle ABC.
Let's assume the altitude of triangle ABC is h, then the altitude of triangle DEF is 3h.
Now, we know that the area of a triangle is given by the formula:
Area = (1/2) * base * height
Let's apply this formula to both triangles:
Area of triangle ABC = (1/2) * base * altitude
= (1/2) * 7cm * h
= 3.5h cm^2
Area of triangle DEF = (1/2) * base * altitude
= (1/2) * 7cm * 3h
= 10.5h cm^2
Since the two triangles are similar, their areas are proportional to the squares of their corresponding sides. That is:
Area of triangle ABC / Area of triangle DEF = (AB / DE)^2
Substituting the values, we get:
3.5h / 10.5h = (AB / DE)^2
1/3 = (AB / DE)^2
Taking the square root of both sides, we get:
AB / DE = 1 / sqrt(3)
AB = (1 / sqrt(3)) * DE
Now, we know that the sides of triangle ABC are 3cm, 5cm, and 7cm.
Let's assume that the sides of triangle DEF are a, b, and c.
Since the sides of the two triangles are proportional, we can write:
a / 3 = b / 5 = c / 7 = 1 / sqrt(3)
Solving for a, b, and c, we get:
a = 3 / sqrt(3) = sqrt(3) cm
b = 5 / sqrt(3) = (5 * sqrt(3)) / 3 cm
c = 7 / sqrt(3) = (7 * sqrt(3)) / sqrt(3) = 7 cm
Therefore, the measures of the sides of triangle DEF are sqrt(3) cm, (5 * sqrt(3)) / 3 cm, and 7 cm.