## Question

### Solution

Correct option is

7920

Since (AB)2 + (AC)2 = 50 = (BC)2 the lines AB and AC are at right angle.

It is given that all side are integers and since the triangle is right angled, they from a Pythagorean tripled, so one leg must be even. As two sides are primes, both cannot be the legs of the right angled triangle.

So, let AC = p, a prime, BC = p + 50 and AB = a (even)

Then     a2 = (p + 50)2 – p2 = 100 (p + 25) Since a is least possible integer and p is a prime

We get a = 60 when p = 11.

So smallest value of the third side is 60.

S = 60 + 11 + 61 = 132

S = 60 × 132 = 7920. #### SIMILAR QUESTIONS

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