Perimeter and legs

Determine the perimeter of a right triangle if the length of one leg is 75% length of the second leg and its content area is 24 cm2.

Result

x =  24

Solution:

S=ab/2=24 224=0.75a2 a=224/0.75=8 cm b=0.75a=6 cm c2=a2+b2 c=a2+b2=10 cm x=a+b+c=8+6+10=24S = ab/2 = 24 \ \\ 2\cdot 24 = 0.75 a^2 \ \\ a = \sqrt{ 2\cdot 24 / 0.75 } = 8 \ cm \ \\ b = 0.75 a= 6 \ cm \ \\ c^2 = a^2 + b^2 \ \\ c = \sqrt{ a^2 + b^2 } = 10 \ cm \ \\ x = a+b+c = 8 + 6 + 10 = 24



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