- ✓$49$
- B$49 \sqrt{2}$
- C$60$
- D$60 \sqrt{2}$
Let the length, breadth and height of cuboid is $l, b$ and $h$ respectively.
$Given, l^2+h^2=39^2$
$\Rightarrow b^2+h^2=40^2$
$\Rightarrow \quad l^2+b^2=41^2$
$\Rightarrow \quad 2\left(l^2+b^2+h^2\right)=39^2+40^2+41^2$
$\Rightarrow \quad l^2+b^2+h^2=2401$
$\therefore$ Length of longest diagonal
$=\sqrt{l^2+b^2+h^2}$
$=\sqrt{2401}=49$
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$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}, n \geq 1$
$b_1=1 \text { and } b_n=a_{n-1}+a_{n+1}, n \geq 2.$
Then which of the following options is/are correct?
$(1)$ $a_1+a_2+a_3+\ldots . .+a_n=a_{n+2}-1$ for all $n \geq 1$
$(2)$ $\sum_{n=1}^{\infty} \frac{ a _{ n }}{10^{ n }}=\frac{10}{89}$
$(3)$ $\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\frac{8}{89}$
$(4)$ $b=\alpha^n+\beta^n$ for all $n>1$