One side of a triangle is 1 in. longer than the shortest side and is 1 in. shorter than the longest side. The perimeter is 17 in. What are the dimensions of the triangle?

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Answer 1 · 2018-01-03T04:29:52

Measure of the three sides of the triangle $(4 (\frac{2}{3}), 5 (\frac{2}{3}), 6 (\frac{2}{3}))$ $color(blue)((4(2/3), 5(2/3), 6(2/3))$

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The sides are a, b, c.

Given a = b - 1 as also c = b + 1;

Further perimeter $p = a + b + c = 17$ $p = a + b + c = 17$

$:. p = 17 = a + b + c = b - 1 + b + b + 1 = 3b$

$b = 17 / 3 = 5 (2/3)$

$:. a = b - 1 = 5 (2/3) - 4 (2/3)$

$c = b + 1 5 (2/3) + 1 6 (2/3)$

Measure of the three sides of the triangle is $color(blue) ((4(2/3), 5(2/3), 6(2/3))$

Answer 2 · 2018-01-03T04:38:02

$4(2)/3$ in; $5(2)/3$ in.; $6(2)/3$ in.

Suppose the shortest side of the triangle = $x$ in.
Then the other side is $x+1$ in.

Therefore, the longest side = $x+1+1$ = $x+2$ in.

It is given that the perimeter = 17 in.

and, as perimeter is the sum of all the sides, it can be written in an equation form as:
$x+(x+1)+(x+2)$ = 17

or, $3x+3 = 17$
or, $3x = 17-3$ =14

or, $x$ = $14/3$ in. = $4(2)/3$ in.(shortest length)

and, $x+1$ = $14/3 +1$ = $17/3$ = $5(2)/3$ in. (mid-length)

and, $x+2$ = $14/3+2$ = $20/3$ = $6(2)/3$ in. (longest side)