Question

solve the question in above figure please​

Answer 1

In ∆ MKL

<MKL = 30°

tan30 = ML/KL = 1/√3

=> ML = 6

In ∆ MKN

< MKN = 90°

< N = 45°

=> ML = √6²+(6√3)²

=> ML = 6√(1+√3)

MN = 6√(2+2√3)

=> perimeter of fig. KLMN =

6+6√3+6√(1+√3) +6√(2+2√3)

= 6 [ 1+√3 + √(1+√3)+√(2+2√3)]