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The sides of a right triangle are such that the longest side is $4$ m more than the shortest side and the third side is $2$ m less than the longest side. Find the length of each side of the triangle. Also, find the difference between the numerical values of the area and the perimeter of the given triangle.
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Let the length of shortest side be $x$ m
$\therefore$ length of longest side = $(x + 4)$ m
and length of third side = $(x + 2)$ m
Now, $(x + 4)^2 = x^2 + (x + 2)^2$
$\Rightarrow x^2 - 4x - 12 = 0$
$\Rightarrow (x – 6)(x + 2) = 0$
$\Rightarrow x = 6$
$\therefore$ sides are $6$ m, $8$ m and $10$ m
Area = $\frac{1}{2} \times 6 \times 8 = 24$ m$^2$
Perimeter = $6+8+10 = 24$ m
Difference = $0$
$\therefore$ length of longest side = $(x + 4)$ m
and length of third side = $(x + 2)$ m
Now, $(x + 4)^2 = x^2 + (x + 2)^2$
$\Rightarrow x^2 - 4x - 12 = 0$
$\Rightarrow (x – 6)(x + 2) = 0$
$\Rightarrow x = 6$
$\therefore$ sides are $6$ m, $8$ m and $10$ m
Area = $\frac{1}{2} \times 6 \times 8 = 24$ m$^2$
Perimeter = $6+8+10 = 24$ m
Difference = $0$