Question

4. Three cubes of iron whose edges are 6 cm, 8 cm
and 10 cm respectively are melted and formed into
a single cube. The edge of the new cube formed is
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 18 cm
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Answer 1

Step-by-step explanation:

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Answer 2

Question :

Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. The edge of the new cube formed is

(a)12 cm

(b)14 cm

(c)16 cm

(d)18 cm

Solution :

$\underline{\bold{Given:}}$

• Edges of three cubes of iron= 6 cm, 8 cm and 10 cm.

$\underline{\bold{To\:Find:}}$

• The edge of the new cube.

Let the edges of three cubes of iron be X= 6 cm, Y=8 cm and Z=10 cm.

Let the edge of new cube be M.

Formula used:

$\boxed{\pink{Edge\: of\: Cube=\sqrt[3]{Volume}}}$

$\boxed{\pink{Volume\: of\: Cube= {(Edge)}^{3} }}$

Atq,

$\implies Volume\: of \:new \:cube=Sum\:of\:volume \:of \:three \:cubes \:of\: iron\\\implies M^3=(X^3+Y^3+Z^3)\:cm^3\\\implies M^3=(6^3+8^3+10^3)\:cm^3\\\implies M^3=(216+512+1000)\:cm^3\\\implies M^3= 1728\:cm^3\\\implies M=\sqrt[3]{1728}\:cm\\\implies M=\sqrt[3]{12\times 12\times 12}\:cm \\\implies M=12\:cm$

$\fbox{\green {\therefore{(a)The\: edge\: of \:the\: new \:cube = 12\:cm}}}$

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