A rectangle has a perimeter of 116 cm and it's length is 1 cm more than twice its width. Find the detentions of a rectangle given that its perimeter is 116 cm and its length is 1 cm more than twice its width.​

width = w

length = 2w+1

perimeter = 2(length +width)

110 =2(2w+1+w)

110 = 2(3w+1)

110 = 6w+2

108 =6w

18 = w

width = w = 18cm

length = 2w+1 =2(18)+1 =36+1 =37cm

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1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )