Question

A solid cone has height 12cm and base radius 6 cm has 4 cm of remove find the tsa of remaining solid

Answer 1

Height of cone = 12 cm

Base radius = 6 cm

If 4 cm is removed than, then height will be

= 12 - 4

= 8 cm

So it will no more be a cone and become a FRUSTUM.

As In the above figure,

So,

CD . AB = BE . AC

6 × 4 = BE × 12

24 = BE × 12

24/12 = BE

2cm = BE = smaller radius of frustum

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Slant height of frsutum

$= \sqrt{ {h}^{2} + (r1 - r2)}$

$= \sqrt{ {8}^{2} + ({6 - 2})^{2} }$

$= \sqrt{ {8}^{2} + {4}^{2} }$

$= \sqrt{64 + 16}$

$= 80$

$= 4 \sqrt{5} cm$

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Total Surface Area of Frustum =

Curved Surface area of frustum

+

area of above circular part

+

area of below circular part

= [pie . l . ( R + r )] + [pie . r^2] + [pie . r^2]

= (4 \sqrt{5} \times 8)\pi + 36\pi + 4\pi[/tex]

$= (32 \sqrt{5} )\pi + 36\pi + 4\pi$

$= (32 \sqrt{5} + 36 + 4)\pi$

$= (32 \sqrt{5} + 40)\pi$

$= (32 \times 2.23 + 40) \times \frac{22}{7}$

$= (71.36 + 40) \times \frac{22}{7}$

$= 111.36 \times \frac{22}{7}$

$= \frac{111.36 \times 22}{7}$

$= \frac{2449.92}{7}$

$= 349.9$

$= 350 \: \: {cm}^{2}$

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The total surface area of frustum is 350 sq.cm