In Delta ABC,m angle B = 90^{circ},AC - BC = | vedclass

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(C) Let $AB = x$. Given $BC - AB = 7$,so $BC = x + 7$. Given $AC - BC = 2$,so $AC = BC + 2 = (x + 7) + 2 = x + 9$. Since $\Delta ABC$ is a right-angle

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$40$

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(C) Let $AB = x$. Given $BC - AB = 7$,so $BC = x + 7$. Given $AC - BC = 2$,so $AC = BC + 2 = (x + 7) + 2 = x + 9$. Since $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$,by the Pythagorean theorem: $AB^2 + BC^2 = AC^2$ $x^2 + (x + 7)^2 = (x + 9)^2$ $x^2 + x^2 + 14x + 49 = x^2 + 18x + 81$ $2x^2 + 14x + 49 = x^2 + 18x + 81$ $x^2 - 4x - 32 = 0$ $(x - 8)(x + 4) = 0$ Since $x$ represents a length,$x = 8$. Thus,$AB = 8$,$BC = 8 + 7 = 15$,and $AC = 8 + 9 = 17$. The perimeter of $\Delta ABC = AB + BC + AC = 8 + 15 + 17 = 40$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$AC - BC = 2$ and $BC - AB = 7$. Find the perimeter of $\Delta ABC$.

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