In Delta ABC,m angle B = 90^{circ}. If AB = 6 | vedclass

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(D) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$.Here,$AC$ is the hypotenuse and $AB$ is one of the legs.Using th

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$24$

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(D) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$.Here,$AC$ is the hypotenuse and $AB$ is one of the legs.Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.Substituting the given values: $10^2 = 6^2 + BC^2$.$100 = 36 + BC^2$.$BC^2 = 100 - 36 = 64$.$BC = \sqrt{64} = 8$.The area of a right-angled triangle is given by the formula: $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,base $BC = 8$ and height $AB = 6$.$	ext{Area} = \frac{1}{2} 	imes 8 	imes 6 = 4 	imes 6 = 24$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 6$ and $AC = 10$,find the area of $\Delta ABC$.

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