Question

Let the sides of a triangle ABC are all integers with A as the origin. If (2, –1) and (3, 6) are points on the line AB and AC respectively (lines AB andAC may be extended to contain these points), and length of any two sides are primes that differ by 50. If a is least possible lengths of the third side and S is the least possible perimeter of the triangle then aS is equal to

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.

                                                                     

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Q7

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Q8

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Q10

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