The sides of a right angled triangle are all **integers**.Two sides are **primes **that differ by 50.The smallest possible value of the third side is ............

- 60
- 57
- 53
- 49

Since,the third side differ by a by 50

so the third side = 50 + a.

Now, by pythagoras theorem,

${a}^{2}+{b}^{2}=(a+50{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}+{b}^{2}={a}^{2}+100a+2500\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=100a+2500\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Now,the minimum value of a to satisfy the equation and b to be an integer,is 11

i,e, ${b}^{2}=100\times 11+2500\phantom{\rule{0ex}{0ex}}\Rightarrow b=\sqrt{3600}=60\phantom{\rule{0ex}{0ex}}$

Thus,the sides of the triangle are 11,60,61

Thus,the smallest possible value of third side is 60.

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