Question

The sides of a triangle, taken in order, are
each 3 cm longer than the preceding side.
If the perimeter of the triangle is 81 cm,
find the length of its sides​

Answer 1

Solution :

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.4,5.8){ A }\put(11.2,5.8){ B }\put(8.4,10){ C }\put(6,6){\line(2,3){2.5}}\put(11,6){\line(-2,3){2.5}}\put(6,6){\line(1,0){5}}\put(5,7.9){ x+3\:cm }\put(11,7.9){ x+6\:cm }\put(8,5){ x\:cm }\end{picture}$

ABC is a triangle in which

AB = x

Preceding side of AB let we consider AC is 3 cm longer

AC = x + 3

So, BC is preceding side of AC will 3 cm longer

BC = x + 3 + 3 = x + 6

Perimeter of triangle = 81 cm

x + x + 3 + x + 6 = 81

3x + 9 = 81

3x = 81 - 9

$x \: =\: \dfrac{81-9}{3}$

$x \: =\: \dfrac{72}{3}$

$x \: =\: 24 \: cm$

AC = 24 + 3 = 27 cm

BC = 24 + 6 = 30 cm

Answer 2

Answer:

let the sides be x, x+3 , x+6

81 = x + x+3 + x+6

81 = 3x + 9

72 = 3x

72/3 = x

x = 24

so the sides of triangle are 24cm, 27cm & 30cm

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