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Let ${r_1}$ and ${r_2}$ be the radius of cylinder and conical heap respectively.

Now we know that volume of cylinder $ = \pi {r_1}^2h$

And volume of conical heap $ = \dfrac{1}{3}\pi {r_2}^2h$

$ \Rightarrow $ Volume of bucket full of sand will be $\pi {r_1}^2h$

$ \Rightarrow $ Volume of cone $ = \dfrac{1}{3}\pi {r_2}^2h$

At first we will find volume of sand using the below mentioned formula-

$ \Rightarrow $ $\pi {r_1}^2h$

$ = \dfrac{{22}}{7} \times 18 \times 18 \times 32$

= 32585.1429------(i)

Also we know that cylindrical buckets are emptied into conical heaps so both will have the same amount of sand.

Now we will find the volume of the cylindrical bucket by using the formula of volume of conical heap.

Therefore , Volume of conical heap $ = \dfrac{1}{3}\pi {r_2}^2h$------(ii)

Comparing equation (i) and (ii) we get,

Or $\dfrac{1}{3} \times \dfrac{{22}}{7} \times {r_2}^2 \times h = 32585.1429$

Now we calculate the value of $r_2^2$

$ \Rightarrow $$r_2^2$ = 32585.1429$ \times $ 7$ \times $$\dfrac{1}{{22}} \times \dfrac{1}{8}$

Or $r_2^2 = \sqrt {1296} $

$ \Rightarrow $radius of conical heap = 36

$ \Rightarrow $(Slant height) L = $\sqrt {r_2^2 + {h^2}} $

$ \Rightarrow $$\sqrt {1296 + 576} $

$ \Rightarrow $$\sqrt {1872} $

$ \Rightarrow $L = 43.26

Since we have to find the value of slant height of conical heap to one place of decimal, the answer will be 43.3 cm.

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