Question

The dimensions of a cuboid are 44cm,21cm,12cm. It is melted and a cone of hight 24 cm is made . Find the radius of it's base

Answer 1

volume of a cuboid=l*b*h

=44*21*12

=11088

volume of cuboid =volume of melted cone

or,11088=1/3πr²h

or,11088=1/3*22/7*r²*24

or,11088=176/7*r²

or,11088*7/176=r²

or,441=r²

r=21

radius of a base=21

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Answer 2

Answer: 21 cm.

Given:

$\sf Dimensions\:of\:cuboid:44\:cm,21\:cm,12\:cm.$

$\sf Height\:of\:the\:cone(h)=24\:cm$

To find:

$\sf Radius\:of\:the\:base\:of\:the\:cone(r).$

Formula used:

$\boxed{\sf Volume\:of\:cuboid=Length(l)\times Breadth(b)\times Height(h)}$

$\boxed{\sf Volume\:of\:cone=\frac{1}{3}\pi(radius(r))^2Height(h) }$

Explanation:

$\sf According\:to\:the\:question,cuboid\:is\:melted\:to\:form\:cone.Therefore,$

$\boxed{\sf Volume\:of\:cone=Volume\:of\:cuboid}$

$\Rightarrow\sf \dfrac{1}{3}\pi r^2h=l\times b\times h$

$\Rightarrow\sf \dfrac{1}{3}\times\dfrac{22}{7} (r)^2 \times24=44\times 21\times 12$

$\Rightarrow\sf \dfrac{22}{7} (r)^2 \times8=924\times 12$

$\Rightarrow\sf \dfrac{22}{7} (r)^2 \times8=11,088$

$\Rightarrow\sf \dfrac{22}{7} (r)^2=\dfrac{11,088}{8}$

$\Rightarrow\sf \dfrac{22}{7} (r)^2=1386$

$\Rightarrow\sf r^2=1386\times\dfrac{7}{22}$

$\Rightarrow\sf r^2=63\times 7$

$\Rightarrow\sf r^2=441$

$\Rightarrow\sf r=\sqrt{441}=21$

$\Rightarrow\boxed{\sf r=21\:cm}$

Therefore, the radius is 21 cm.