In Delta ABC,m angle B = 90^{circ} and AC = 1 | vedclass

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(A) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$ and $AB = BC$.Let $AB = BC = x$.According to the Pythagorean the

Vedclass · Accepted Answer

$98$

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(A) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$ and $AB = BC$.Let $AB = BC = x$.According to the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.Substituting the values,we get $x^2 + x^2 = (14\sqrt{2})^2$.$2x^2 = 196 	imes 2$.$2x^2 = 392$.$x^2 = 196$.$x = 14$.Thus,the base $BC = 14$ and the height $AB = 14$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 14 	imes 14 = 98$ square units.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $AC = 14\sqrt{2}$. If $AB = BC$,find the area of $\Delta ABC$.

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