Question

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Answer 1

Perimeter of the ΔJKL is 48 cm

Solution:

Given $\triangle \mathrm{RST} \sim \Delta \mathrm{JKL}$.

RS = 8 cm and JK = 12 cm

Perimeter of $\triangle R S T$ = 32 cm

Let us find the perimeter of $\Delta J K L$.

Property of similarity triangles:

If two triangles are similar, then the ratio of their perimeters is equal to the ratio of their corresponding sides.

$\frac{\text{Perimeter of}\ \triangle RST }{{\text{Perimeter of}\ \triangle JKL}} =\frac{RS}{JK}$

$\Rightarrow\frac{32 }{{\text{Perimeter of}\ \triangle JKL}} =\frac{8}{12}$

Do cross multiplication.

$\Rightarrow 32\times 12 ={8}\times {\text{Perimeter of}\ \triangle JKL}$

$\Rightarrow 384 ={8}\times {\text{Perimeter of}\ \triangle JKL}$

Divide both sides of the equation by 8.

$\Rightarrow 48 = {\text{Perimeter of}\ \triangle JKL}$

Hence perimeter of the ΔJKL is 48 cm.