In Delta ABC,m angle B = 90^circ. If AC - AB | vedclass

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(B) In $\Delta ABC$,we are given $AC - AB = 27$ and $AC - BC = 6$.Therefore,$AB = AC - 27$ and $BC = AC - 6$.Since $\Delta ABC$ is a right-angled tria

Vedclass · Accepted Answer

$120$

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(B) In $\Delta ABC$,we are given $AC - AB = 27$ and $AC - BC = 6$.Therefore,$AB = AC - 27$ and $BC = AC - 6$.Since $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^\circ$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$AC^2 = (AC - 27)^2 + (AC - 6)^2$$AC^2 = (AC^2 - 54AC + 729) + (AC^2 - 12AC + 36)$$AC^2 = 2AC^2 - 66AC + 765$$0 = AC^2 - 66AC + 765$Factoring the quadratic equation:$(AC - 51)(AC - 15) = 0$So,$AC = 51$ or $AC = 15$.If $AC = 15$,then $AB = 15 - 27 = -12$,which is impossible as the length of a side cannot be negative.Thus,$AC = 51$.Now,$AB = 51 - 27 = 24$ and $BC = 51 - 6 = 45$.The perimeter of $\Delta ABC = AB + BC + AC = 24 + 45 + 51 = 120$.

In $\Delta ABC$,$m \angle B = 90^\circ$. If $AC - AB = 27$ and $AC - BC = 6$,find the perimeter of $\Delta ABC$.

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