Let the length(l) breadth (b) and height ( h) be the external dimension of an open box and thickness be x.
Given that.
External length of an open box (l) = 36 cm
External breadth of an open (b) = 25 cm
And external height of an open box (h) = 16.5 cm
∴ External volume of an open box
=l× b× h =36×25×16.5=14850 cm3 Since, the thickness of the iron (x) = 1.5 cm
So, internal length of an open box
(l1)=l–2x =36×2×1.5=36–3=33 cm Therefore, internal breadth of an open box (b2) = b – 2x
=25−2×1.5=25−3=22cm And internal height of an open box ( h2) = ( h – x )
= 16.5 - 1.5 = 15 cm
So, internal volume of an open box = ( l – 2x ) ( b – 2x ) ( h –x )
=33×22×15=10890 cm3 Therefore, required iron to construct an open box
= External volume of an open box – internal volume of an open box
=1480−10890=3960 cm3 Hence, required iron to construct an open box is
3960 cm3 Given that,
1 cm3 of iron weights
=7.5g=7.51000 kg=0.0075 kg ∴3960 cm3 of iron weights=3960×0.0075=29.7k