In Delta ABC,m angle B = 90^{circ} and AB : A | vedclass

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(B) Given $AB : AC = 24 : 25$.Let $AB = 24k$ and $AC = 25k$,where $k > 0$.In right-angled $\Delta ABC$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$

Vedclass · Accepted Answer

$112$

Vedclass · Answer

(B) Given $AB : AC = 24 : 25$.Let $AB = 24k$ and $AC = 25k$,where $k > 0$.In right-angled $\Delta ABC$,by the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$(25k)^2 = (24k)^2 + (14)^2$$625k^2 = 576k^2 + 196$$625k^2 - 576k^2 = 196$$49k^2 = 196$$k^2 = 4$Since $k > 0$,$k = 2$.Now,$AB = 24 	imes 2 = 48$ and $AC = 25 	imes 2 = 50$.The perimeter of $\Delta ABC = AB + BC + AC = 48 + 14 + 50 = 112$.Thus,the perimeter of $\Delta ABC$ is $112$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $AB : AC = 24 : 25$. If $BC = 14$,find the perimeter of $\Delta ABC$.

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