A gold biscuit of dimensions 5.9cm×3.2cm×0.3cm was used to form a hollow box of thickness 0.2cm if the external length and width of the box are 2cm and 1.5cm find the height of the box
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External Height of the Box = 4 cm if 5.9cm×3.2cm×0.3cm was used to form a hollow box of thickness 0.2cm with external length and width 2cm and 1.5cm
Step-by-step explanation:
A gold biscuit of dimensions 5.9cm×3.2cm×0.3cm
=> Volume of Gold Biscuit = 5.9 * 3.2 * 0.3 cm³
= 5.664 cm³
hollow box of thickness 0.2cm
external length and width of the box are 2cm and 1.5cm
Let Say Height = H cm
External Volume = 2 * 1.5 * H = 3H cm³
Internal Volume = ( 2 - 2 * 0.2)(1.5 - 2 *0.2)(H - 2 * 0.2)
= 1.6 * 1.1 * (H - 0.4)
= 1.76H - 0.704 cm³
3H - (1.76H - 0.704) = 5.664
=> 1.24H = 4.96
=> H = 4 cm
External Height of the Box = 4 cm
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