Question

An inverted cone of vertical height 12cm and Base radius 9cm contains water to a depth of 4 cm. Find the area of the interior surface of the cone not in contact with the water.​

Answer 1

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$ we have to find interior surface of cone not in contact with water ??

$\huge\underline\blue{\sf Given:}$ Vertical H =12

Base radius = 9cm

water is to depth = 4cm

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$\huge\underline\purple{\mathcal Answer:-}$

we know that ,

$slant \: height(l) = \sqrt{(r^{2} } + {h}^{2} ) \\ \\ \\ l = \sqrt{( {12}^{2} } + {9}^{2} ) \\ \\ l = \sqrt{225} \\ \\ l = 15cm$

now, we also know that ,

$surface \: area \: of \: cone = \\pi \times r \times l \\ \\ \\ surface \: area \: = \pi \times 9 \times 15 \\ \\ surface \: area = 135\pi$

Now, Since Water will also form a cone, so we can easily calculate the Radius of water cone , using proportions : -------

$\frac{12}{9} = \frac{4}{r} \\ \\ r = 3cm$

Now, slant height of water in the cone is :-----

$slant \: height(l) = \frac{15 \times 4}{12} \\ \\ l = 5cm$

so, interior surface of the cone in contact with the water is :-----

$area \: with \: water \: = \pi \times 5 \times 3 = 15\pi$

so, finally the area of the cone not in contact with the water = =

$\large\red{\boxed{\sf</strong><strong> </strong><strong>(135\pi - 15\pi) = 120\pi</strong><strong>}}$

$\huge\blue{</strong><strong>Nice\</strong><strong>:</strong><strong>Question\</strong><strong>:</strong><strong>Thanks</strong><strong>}$