Please submit your concern

Question:

Shyam, a fertilizer salesman, sells directly to farmers. He visits two villages A and B. Shyam starts from A, and travels 50 meters to the East, then 50 meters North-East at exactly 45° to his earlier direction, and then another 50 meters East to reach village B. If the shortest distance between villages A and B is in the form of $a \sqrt{b + \sqrt{c}}$ meters, find the value of a + b + c.

Explanation:

Total distance travelled by Shyam is;

$50 + 50 + \frac{50}{\sqrt{2}}$ km towards east, and $\frac{50}{\sqrt{2}}$ km towards north

Hence, smallest distance, say d, between village A and Village B is;

$d = \sqrt{({(100 + 25 \sqrt{2})}^{2} + {(25 \sqrt{2})}^{2})}$

Hence, $d^{2} = {(100 + 25 \sqrt{2})}^{2} + {(25 \sqrt{2})}^{2} = a^{2} (b + \sqrt{c})$

But, ${(100 + 25 \sqrt{2})}^{2} + {(25 \sqrt{2})}^{2} = 2500 (5 + 2 \sqrt{2})$

$= 2500 (5 + \sqrt{8})$

Hence, a² = 2500, b = 5 and c = 8

Hence, a + b + c = 50 + 5 + 8 = 63

Hence, option (e).

Please submit your concern

Feedback