math trig problem with similar triangles?

Question

if trangle abc is similar to triangle def , ab=3cm , bc=5cm ,
ca=7cm , and the permeter of triangle def is 24 cm , how would you find the lengths of the sides of triangle def

i think that the first triangle has a permeter of 15 cm and you ue this with proportions of the second triangle to find the permeter of triangle def ... but i am having trouble doing the work to find this could someone please show me the work to solve this....

actually i need the sides of triangle def not permeter

Answer 1

You've got the right idea.

First find the proportion. 24/15 = 1.6

Then multiply each side of ABC by 1.6 to get DEF.

de = ab*1.6 = 3*1.6 = 4.8cm
ef = bc*1.6 = 5*1.6 = 8cm
fd = ca*1.6 = 7*1.6 = 11.2cm

Answer 2

let the sides of triangle DEF are x, y , z

the sides of triangle ABC are 3, 5, and 7

x + y + z = 24

since both are similar triangles

x/3 = y/5 = z/7 = k, where k = some constant

x = 3k

y = 5k

z = 7k

adding

(x+y+z) = k(3+5+7) = 15k

24 = 15k

k = 24/15 = 8/3

x = 3k = 3(8/3) = 8 cm

y = 5k = 5(8/3) = 40/3 cm

z = 7k = 7(8/3) = 56/3 cm

the sides of triangle DEF are 8, 40/3 and 56/3 cm

Answer 3

If the triangles are similar, then the perimeters are proportional, as well as the individual sides. In this case, you would multiply the perimeter of triangle ABC by 1.6 to get the perimeter of triangle DEF (since 24 / 15 = 1.6). Therefore, you need to multiply the given lengths of sides AB, BC, and CA by 1.6 to get the lengths of the sides of triangle DEF.

Answer 4

24/15 = 1.6
So de = 1.6*3 = 4.8
cf= = 1.6*5 = 8
fd = 1.6*7 = 11.2
There sum is 24 and the sides of the two triangles are proportional.

Answer 5

in a similar triangle the angles are equal and the sides are porprotional to each other

15/24=3/de=5/ef=7/fd

de=4.8
ef=8
ef=11.2

Answer 6

a million)certain, 4sin 21 cos 21 = 2 sin 40 2 2) you do not "practice" it, by using very undeniable reality which could not an "identity" (not actual for ALL x). it truly is an "Equation" and also you therapy it: sin2x+cosx = 0 => 2sinxcosx + cosx = 0 => cosx(2sinx+a million)=0 =>cosx = 0, sinx=-a million/2 => x=ok*PI + PI/2, x=2k*PI - PI/4,x=2k*PI+5PI/4 => (for 0 a million-sin^2 d = sin^2 d - sin d => 2 sin^2 d - sin d - a million= 0 => sin d = (a million+- sqrt(a million+8))/4 => sin d= a million, sin d= -a million/2 => d = 2k*PI + PI/2, 2k*PI - PI/4, 2k*PI + 5PI/4

Answer 7

use pythagoriam