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A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.​

Answer 1

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Given, the external dimensions of a closed wooden box are 78 cm, 19 cm.

Let the external height be  h cm respectively. 

As the thickness is  1.5 cm , the internal length, breadth and height of the closed box are 78−2(1.5) = 75 cm, 19−2(1.5) = 16 cm

and h−2(1.5) = (h−3) cm respectively.

Given, capacity of the box = 15cudm = 15000cucm 

As the box is cuboidal, its volume = length × breadth  × height 

⇒[75×16×(h−3)] = 15000

⇒(h−3) = 12.5 cm 

⇒h = 15.5 cm

Answer 2

ǫᴜᴇsᴛɪᴏɴ →

A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.

ᴀɴsᴡᴇʀ →

Given,

the external dimensions of a closed wooden box are 78 cm, 19 cm.

Let the external height be h cm respectively.

As the thickness is 1.5 cm , the internal length, breadth and height of the closed box are 78−2(1.5) = 75 cm, 19 − 2(1.5) = 16 cm

and h−2(1.5) = (h−3) cm respectively.

Given,

$\sf{capacity \: of \: the \: box = 15cudm = 15000cucm }$

$\sf{As \: the \: box \: is \: cuboidal, }$

$\sf{its \: volume = length × breadth × height }$

$\longmapsto\sf{(75×16×(h−3))=15000}$

$\longmapsto\sf{(h−3)=12.5 cm }$

$\boxed{h=15.5 cm}$