In Delta ABC,AB = AC and m angle A = 90^circ. | vedclass

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(B) Let $AB = AC = x$.In $\Delta ABC$,$m \angle A = 90^\circ$.By the Pythagorean theorem,$AB^2 + AC^2 = BC^2$.Substituting the given values: $x^2 + x^

Vedclass · Accepted Answer

$\frac{1}{2} a^2$

Vedclass · Answer

(B) Let $AB = AC = x$.In $\Delta ABC$,$m \angle A = 90^\circ$.By the Pythagorean theorem,$AB^2 + AC^2 = BC^2$.Substituting the given values: $x^2 + x^2 = (\sqrt{2} a)^2$.$2x^2 = 2a^2$.$x^2 = a^2$,so $x = a$ (since $a \in R^+$).Thus,$AB = AC = a$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes AC$.Area $= \frac{1}{2} 	imes a 	imes a = \frac{1}{2} a^2$.

In $\Delta ABC$,$AB = AC$ and $m \angle A = 90^\circ$. If $BC = \sqrt{2} a$,then the area of $\Delta ABC$ is $\ldots \ldots \ldots \ldots$ $(a \in R^+)$.

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