Question

a well of 14m depth with a diameter 6 m is dug and the earth got by digging is evenly spread out to format a rectangular platform of base 22m×9th . find the height of the platform​

Answer 1

HeY MatE!!

HerE IS YouR AnsweR---



Given dimensions of cylindrical well--

Height of cylindrical well=14m --- (h)

Diameter of well= 6m

so, radius = $\frac{6}{2} = 3 \: m$

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when the soil which is dug, it is spread out to format a rectangular platform --

let the height of platform be h

Given dimensions of rectangular platform-

length of platform =22m--- (l)

breadth of platform= 9m--- (b)

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[To find the height ,we need to compare the volumes of both solids ]

$vol. \: of \: well = \: vol. \: of \: platform \\ \\ = > \pi {r}^{2}h = l \times b \times h \\ \\ = > \frac{22}{7} \ \times {3}^{2} \times 14 = 22 \times 9 \times h \\ \\ = > 22 \times 9 \times 2 = 22 \times 9h \\ \\ = > \frac{22 \times 9 \times 2}{22 \times 9} = h \\ \\ = > 2 = h$

therefore, the height of platform is 2m.

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✨Please Mark It As BRAINLIEST✨

Answer 2

We recall that,

Volume of cylinder $=\frac{\pi D^2h}{4}$

The volume of the rectangular platform $=l\times b\times h$

Given: The volume of tank dig is equal to that of a rectangular platform

$\frac{\pi \times6^2\times 14}{4}=22\times9\times h$

       $396=198\times h$

           $h=2m$