Perimeter of the ΔJKL is 48 cm
Solution:
Given [tex]\triangle \mathrm{RST} \sim \Delta \mathrm{JKL}[/tex].
RS = 8 cm and JK = 12 cm
Perimeter of [tex]\triangle R S T[/tex] = 32 cm
Let us find the perimeter of [tex]\Delta J K L[/tex].
Property of similarity triangles:
If two triangles are similar, then the ratio of their perimeters is equal to the ratio of their corresponding sides.
[tex]$\frac{\text{Perimeter of}\ \triangle RST }{{\text{Perimeter of}\ \triangle JKL}} =\frac{RS}{JK}[/tex]
[tex]$\Rightarrow\frac{32 }{{\text{Perimeter of}\ \triangle JKL}} =\frac{8}{12}[/tex]
Do cross multiplication.
[tex]$\Rightarrow 32\times 12 ={8}\times {\text{Perimeter of}\ \triangle JKL}[/tex]
[tex]$\Rightarrow 384 ={8}\times {\text{Perimeter of}\ \triangle JKL}[/tex]
Divide both sides of the equation by 8.
[tex]$\Rightarrow 48 = {\text{Perimeter of}\ \triangle JKL}[/tex]
Hence perimeter of the ΔJKL is 48 cm.
