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(B) Let the sides of the triangle be $a-d$,$a$,and $a+d$,where $d > 0$.Since it is a right-angled triangle,by the Pythagorean theorem:$(a-d)^2 + a^2 =

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$6$

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(B) Let the sides of the triangle be $a-d$,$a$,and $a+d$,where $d > 0$.Since it is a right-angled triangle,by the Pythagorean theorem:$(a-d)^2 + a^2 = (a+d)^2$$a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2$$a^2 - 4ad = 0$Since $a 
eq 0$,we have $a = 4d$.The sides are $3d$,$4d$,and $5d$.The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 24$.$\frac{1}{2} 	imes 3d 	imes 4d = 24$$6d^2 = 24$$d^2 = 4 \Rightarrow d = 2$.The sides are $3(2) = 6$,$4(2) = 8$,and $5(2) = 10$.The smallest side is $6$.

The sides of a right-angled triangle are in arithmetic progression. If the triangle has an area of $24$,then what is the length of its smallest side?

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