Solution
L = 2B – 1, Diagonal = 17. D² = L²+ B²
⇒ 289 = (2B –1)²+ B²⇒ B = 8 & L = 15.
Also 8, 15, 17 is a Pythagorean triplet.
So answer can be reached directly.
L = 2B – 1, Diagonal = 17. D² = L²+ B²
⇒ 289 = (2B –1)²+ B²⇒ B = 8 & L = 15.
Also 8, 15, 17 is a Pythagorean triplet.
So answer can be reached directly.