# If the sides of a triangle measure 8,9,and 10,find the longest side of a similar triangle whose perimeter is 81?

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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.