Question

The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths $39,40,41$. The length of the main diagonal of the cuboid which joins a pair of opposite corners

• A. $49$
• B. $49\sqrt{2}$
• C. $60$
• D. $60\sqrt{2}$

Answer 1

Answer: (A)

$49$

$l^2+b^2={39}^2,b^2+h^2={40}^2,h^2+l^2={41}^2$
Adding
$2(l^2+b^2+h^2)={39}^2+{40}^2+{41}^2$
$\sqrt{l^2+b^2+c^2}=\sqrt{\frac{{39}^2+{40}^2+{41}^2}{2}}=\sqrt{\frac{(40-1{)}^2+{40}^2+(40+1{)}^2}{2}}$
$=\sqrt{\frac{3(40{)}^2+2}{2}}=\sqrt{\frac{4802}{2}}=\sqrt{2401}$
$=49$