The sides of a triangle are distinct positive | vedclass

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(B) Let the sides of the triangle be $a-d, a, a+d$ where $a$ and $d$ are positive integers and $d > 0$.The smallest side is $a-d = 10$,which implies $

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$9$

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(B) Let the sides of the triangle be $a-d, a, a+d$ where $a$ and $d$ are positive integers and $d > 0$.The smallest side is $a-d = 10$,which implies $a = 10+d$.The sides are $10, 10+d, 10+2d$.For these to form a triangle,the sum of the two smaller sides must be greater than the largest side:$10 + (10+d) > 10+2d$$20+d > 10+2d$$10 > d$Since $d$ is a positive integer,$d$ can take values from $1, 2, 3, 4, 5, 6, 7, 8, 9$.Thus,there are $9$ possible values for $d$,and consequently $9$ such triangles.

The sides of a triangle are distinct positive integers in an arithmetic progression. If the smallest side is $10$,the number of such triangles is

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