Question

A 20 m deep well with diameter 7 m is dug, and the earth from digging is evenly spread out to form a platform 22 m by 14 m.

Find the height of the platform.

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Answer

:

  • Radius of the well: $$3.5 \text{ m}$$
  • Depth of the well: $$20 \text{ m}$$
  • Dimensions of the rectangular platform: $$22 \text{ m} \times 14 \text{ m}$$

Volume of earth dug out:

$$ \begin{aligned} \text{Volume} &= \pi r^{2} h \ &= \pi \times (3.5)^{2} \times 20 \ &= \pi \times 12.25 \times 20 \ &= \pi \times 245 \ &\approx 22 \times 35 = 770 \text{ m}^3 \end{aligned} $$

Area of the platform:

$$ 22 \times 14 \text{ m}^2 \quad (\text{Area of a rectangle} = l \times b) $$

Let the height of the platform be H meters.

Volume of the platform is equal to the volume of the earth dug out:

$$ \begin{aligned} 22 \times 14 \times H &= 770 \text{ m}^3 \ H &= \frac{770}{22 \times 14} \ H &= \frac{770}{308} \ H &= 2.5 \text{ m} \end{aligned} $$

Therefore, the height of the platform is 2.5 meters.


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