If the sides of a triangle measure 8,9,and 10,find the longest side of a similar triangle whose perimeter is 81?
First,we'll calculate the perimetero fthe original triangle, adding up the three lengths of the sides:
P1 = 8+9+10
P1 = 27
Now, we'll divide the perimeter of the similar triangle, namely 81, by the perimeter of the original triangle:
P2 = 81
P2/P1 = 81/27
P2/P1 = 3 units
So, the rapport is of 3 units. We'll multiply the length of each side of the original triangle by 3:
8*3 = 24 units
9*3 = 27 units
10*3 = 30 units
The longest length of the side of the similar triangle is of 30 units.
Similar triangles bear the same ratio of the correspoding sides.
If a , b and c are the sides of one triangle and d, e and f are he correspoding sides of the similar triangle , then :
a/d = b/e= c/f each = k say....(1)
Then a = dk, b = ek and c = fk.
Therefore (a+b+c) = (d+e+f)k.
Or (a+b+c)/(d+e+f) = k.
Given a=8, b=9 and c = 10. So a+b+c= 8+9+10 = 27. Also d+e+f = 81.
Therefore (a+b+c)/(d+e+f) = 27/81 = 1/3.
Therefore from (1) :
d = a/k = 8/(1/3) = 24
e = b/k = 9/(1/3) = 27.
f = c/k = 10/(1/3) = 30.
So the sides of the similar triangle whose perimeter is 81 are: 24,27 and 30.