In a right-angled triangle ABC, the measures | vedclass

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(B) $1$. Let the angles be $a-d$, $a$, and $a+d$. Since the sum of angles in a triangle is $180^\circ$, $(a-d) + a + (a+d) = 180^\circ$, which gives $

Vedclass · Accepted Answer

$8\sqrt{3} 	ext{ sq. units}$

Vedclass · Answer

(B) $1$. Let the angles be $a-d$, $a$, and $a+d$. Since the sum of angles in a triangle is $180^\circ$, $(a-d) + a + (a+d) = 180^\circ$, which gives $3a = 180^\circ$, so $a = 60^\circ$.$2$. Since it is a right-angled triangle, one angle must be $90^\circ$. Thus, $a+d = 90^\circ \implies 60^\circ + d = 90^\circ \implies d = 30^\circ$.$3$. The angles are $60^\circ - 30^\circ = 30^\circ$, $60^\circ$, and $90^\circ$.$4$. In a $30^\circ-60^\circ-90^\circ$ triangle, the sides are in the ratio $1 : \sqrt{3} : 2$. The smallest side is opposite the smallest angle $(30^\circ)$.$5$. Given the smallest side is $4$, the sides are $4$, $4\sqrt{3}$, and $8$.$6$. The area of the right-angled triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 4\sqrt{3} = 8\sqrt{3} 	ext{ sq. units}$.

In a right-angled $\triangle ABC$, the measures of the angles are in an Arithmetic Progression ($A$.$P$.). If its smallest side is $4 \text{ units}$, then the area of $\triangle ABC$ is:

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