A cubical box has each edge $ 10 \mathrm{~cm} $ and another cuboidal box is $ 12.5 \mathrm{~cm} $ long, $ 10 \mathrm{~cm} $ wide and $ 8 \mathrm{~cm} $ high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?

Given:

A cubical box has each edge \( 10 \mathrm{~cm} \) and another cuboidal box is \( 12.5 \mathrm{~cm} \) long, \( 10 \mathrm{~cm} \) wide and \( 8 \mathrm{~cm} \) high.

To do:

We have to find:

(i) Which box has the greater lateral surface area and by how much and
(ii) Which box has the smaller total surface area and by how much.

Solution:

We have,

The edge of the cubical box $1$ $=10\ cm$
 The length of the cuboidal box $2$ $l=12.5\ cm$

The breadth of the cuboidal box $2$ $b=10\ cm$

The height of the cuboidal box $2$ $h=8\ cm$

(i) The lateral surface area of cubical box $1=4(edge^2)$

$=4(10^2)$

$=4(100)$

$=400\ cm^2$

The lateral surface area of cuboidal box $=2(lh+bh)$

$=2(12.5\times8+10\times8)$

$=2(100+80)$

$=2(180)$

$=360\ cm^2$

Therefore,

The lateral surface area of the cubical box is more than the cuboidal box.

This implies,

$=(400-360)$

$=40\ cm^2$

Therefore, by $40\ cm^2$ cubical box has greater lateral surface area then cuboidal box.

(ii) We know that,

The total lateral surface area of the cubical box $=6(edge^2)$

This implies,

$=6(10^2)$

$=6(100)$

$=600\ cm^2$

The total lateral surface area of the cuboidal box $=2(lh+bh+lb)$

This implies,

$=2(12.5\times8+10\times8+12.5\times10)$

$=2(305)$

$=610\ cm^2$

Therefore, 

The total lateral surface area of the cubical box is smaller than the cuboidal box

This implies,

$610-600=10\ cm^2$

Therefore, by $10\ cm^2$ cubical box has smaller lateral surface area then cuboidal box.

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Updated on: 10-Oct-2022

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