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The internal and external radii of a hollow hemisphere are $5\sqrt{2}$ cm and $10$ cm respectively. A cone of height $5\sqrt{7}$ cm and radius $5\sqrt{2}$ cm is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of $\pi$. (Use $\sqrt{2} = 1.4$)
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Sol. Let internal and external radii be $r_1 = 5\sqrt{2}$ cm and $r_2 = 10$ cm respectively.
$\therefore$ Slant height $(l)$ of the cone $= \sqrt{(5\sqrt{2})^2 + (5\sqrt{7})^2} = 15$ cm (I) (1 Mark)
Now, the total surface area of the object $= 2\pi r_2^2 + \pi r_1 l + \pi(r_2^2 - r_1^2)$ (II) (1 Mark)
$= \pi (2\times 10^2 + 5\sqrt{2} \times 15 + 10^2 - (5\sqrt{2})^2)$ (III) (1 Mark)
$= 355 \pi$ cm$^2$
$\therefore$ Slant height $(l)$ of the cone $= \sqrt{(5\sqrt{2})^2 + (5\sqrt{7})^2} = 15$ cm (I) (1 Mark)
Now, the total surface area of the object $= 2\pi r_2^2 + \pi r_1 l + \pi(r_2^2 - r_1^2)$ (II) (1 Mark)
$= \pi (2\times 10^2 + 5\sqrt{2} \times 15 + 10^2 - (5\sqrt{2})^2)$ (III) (1 Mark)
$= 355 \pi$ cm$^2$