Question

The dimensions of cuboid are 44cm , 21cm , 12cm. It is melted and a cone of height 24cm is made.Find the radius of its base.

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Answer 1

Note :-

While melting a substance to another substance the volume of both the substance remains same.

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$formulae \: related \: to \: cone :- \\ \\ slant \: height(l) = \sqrt{ {h}^{2} } + \sqrt{ {r}^{?} } \\ curved \: surface \: area \: = \pi \: rl \\ total \: surface \: area \: = \pi \: r (r + l) \\ volume = \frac{1}{3} \times \pi \: {r}^{2} h \\ \\ formulae \: related \: to \: cuboid:- \\ \\ lateral \: surface \: area \: = 2h(l + b) \\ total \: surface \: area = 2(lb + bh + hl) \\ volume = l \times b \times h$

Answer 2

The dimensions of the cuboid are 44 cm,

21 cm and 12 cm.

Let the radius of the cone ber cm.

Height of the cone, h = 24 cm

It is given that cuboid is melted to form a

cone.

.: Volume of metal in cone = Volume of metal in cuboid

⇒$\dfrac{1}{3}$πr²h=44×21×12(volume of cuboid=$\dfrac{length}{times breadth}$×height)

⇒$\dfrac{1}{3}$×$\dfrac{22}{7}$×r²×24=44×21×12

⇒

$r = \sqrt{ \frac{44 \times 21 \times 12 \times 21}{22 \times 24} } = \sqrt{21 \times 21} = 21cm$

thus,the radius of the base of cone is 21 cm

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