Question

given a triangle with a perimeter of 56in. we know the first side is 3 inches less than twice the length of the second side, and the third side is 10 inches more than half the length of the second side. if the second side length is h, how long is the longest side? write an equation to solve for the variable.

Answer 1

Answer:

Side A = 25 in

Side B = 14 in

Side C = 17 in

Step-by-step explanation:

Perimeter of a triangle = 56 in

Side B = x

Side A = 2x - 3

Side C = 1/2x + 10

Perimeter of a triangle = side A + Side B + side C

56 = (2x -3) + x + (1/2x + 10)

56 = 2x - 3 + x + 1/2x + 10

56 + 3 - 10 = 2x + x + 1/2x

49 = 3 1/2x

Note: 3 1/2x = 3.5x

49 = 3.5x

x = 49/3.5

x = 14 in

Side B = x = 14 in

Side A = 2x - 3

= 2(14) - 3

= 28 - 3

= 25 in

Side C = 1/2x + 10

= 1/2(14) + 10

= 7 + 10

= 17 in