GIVEN: The perimeter of triangle STU is 125.
The segments TW=21 cm
VW=30 cm
VT=24 cm
TO FIND: The length of the segment SU.
SOLUTION:
As corresponding sides must be same. so we have,
[tex]\frac{TV}{TS}=\frac{TW}{TU}\\\\ \frac{24}{TS}=\frac{21}{TU}\\\frac{TU}{TS}=\frac{7}{8}=k(say)\\[/tex]
Then, TU=7k and TS=8k
[tex]\frac{TV}{TS}=\frac{VW}{SU}\\\\ \frac{24}{TS}=\frac{30}{SU}\\\frac{24}{8k}=\frac{30}{SU}\\SU=10k[/tex]
As the perimeter of triangle STU is 125.
so,
8·k+7·k+10·k=125
⇒25.k=125
⇒k=5
Therefore, SU=10×5=50
