Question

From a solid cylinder whose height is $2.4 cm$ and diameter $1.4 cm$ , a conical cavity of the same height and base is hollowed out as shown in the below figure. Find the total surface area of the remaining solid.

Question Image

Accepted Answer

Let $h$ and $r$ be the height and radius of the cone and cylinder. Let $l$ be the slant height of the cone.

Given that, $h = 2.4 cm$ and $d = 1.4 cm; r = 0.7 cm$

Here,

Total surface area of the remaining solid $=$ C.S.A of the cylinder $+$ C.S.A of the cone $+$ Area of the bottom

$= 2 π r h + π r l + π r^{2}$ sq.units

Now, $l = \sqrt{r^{2} + h^{2}} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5 cm$

$l = 2.5 cm$

Area of the remaining solid $= 2 π r h + π r l + π r^{2}$ sq.units

$= π r (2 h + l + r)$

$= \frac{22}{7} \times 0.7 \times [(2 \times 2.4) + 2.5 + 0.7]$

$= 17.6$

Therefore, total surface area of the remaining solid is $17.6 m^{2}$

Tamil Nadu Board Solutions for Chapter: Mensuration, Exercise 7: Solved Example

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Mensuration, Exercise 7: Solved Example

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