Question

101. Find the length of the largest pole that can be
placed in a hall that is 10 m long, 10 broad and
4 m high.
(a) 14.6 m (b) 18.6 m
(c) 16.6 m (d) 20.20 m​

Answer 1

$\large\underline{\sf{Given- }}$

• Length of Hall, l = 10 m

• Breadth of Hall, b = 10 m

• Height of hall, h = 4 m

$\large\underline{\sf{To\:Find - }}$

• Length of longest pole.

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

Let us consider a cuboid having length l, breadth b and height h, then longest pole that can be placed in the cuboid is diagonal of cuboid, d given by

$\boxed{ \sf \: Longest_{(pole)}=d \: = \sqrt{ {l}^{2} + {b}^{2} + {h}^{2}}}$

$\large\underline{\sf{Solution-}}$

Given that

• Length of Hall, l = 10 m

• Breadth of Hall, b = 10 m

• Height of hall, h = 4 m

So,

• Longest pole that can be placed in room is

$\rm :\longmapsto\:Longest_{(pole)}=d = \sqrt{ {l}^{2} + {b}^{2} + {h}^{2} }$

$\rm :\longmapsto\:Longest_{(pole)}=d = \sqrt{ {(10)}^{2} + {(10)}^{2} + {(4)}^{2} }$

$\rm :\longmapsto\:Longest_{(pole)}=d = \sqrt{100 + 100 + 16}$

$\rm :\longmapsto\:Longest_{(pole)}=d = \sqrt{216}$

$\rm :\longmapsto\:Longest_{(pole)}=d = \sqrt{6 \times 6 \times 6}$

$\rm :\longmapsto\:Longest_{(pole)}=d = 6 \sqrt{6}$

$\rm :\longmapsto\:Longest_{(pole)}=d = 14.6 \: m$

$\bf \: Hence, \: Option \: (a) \: is \: correct.$

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Additional Information :-

$\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$