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# A cubical box has each edge$10cm$ and another cuboidal box is $12.5cm$ long, $10cm$wide and $8cm$ 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?

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## Given,Edge of a cubical box = $10cm$Length of the cuboidal box, l = $12.5cm$Breadth of the cuboidal box, b = $10cm$Height of cuboidal box, h = $8cm$(i) We know that -The lateral surface area of a cube =$4×\left({edge}^{2}\right)$So, the Lateral surface area of a cubical box =$4×\left({edge}^{2}\right)$ =$4×\left({10}^{2}\right)$ $=400{cm}^{2}$And, The lateral surface area of a cuboidSo, the Lateral surface area of the cuboidal box $=2\left(l+b\right)×h\phantom{\rule{0ex}{0ex}}=2\left(12.5+10\right)×8{cm}^{2}\phantom{\rule{0ex}{0ex}}=2×22.5×8{cm}^{2}\phantom{\rule{0ex}{0ex}}=360{cm}^{2}$ So, the lateral surface area of the cubical box is greater than the cuboidal box by $\left(400c{m}^{2}–360c{m}^{2}\right)=40c{m}^{2}\phantom{\rule{0ex}{0ex}}$.(ii) Now,The total surface area of a cubical box $=6×{\left(edge\right)}^{2}\phantom{\rule{0ex}{0ex}}=6×{\left(10\right)}^{2}=6×100\phantom{\rule{0ex}{0ex}}=600c{m}^{2}$The total surface area of the cuboidal box is $=2\left(lb+bh+lh\right)\phantom{\rule{0ex}{0ex}}=2×\left(12.5×10+10×8+8×12.5\right)\phantom{\rule{0ex}{0ex}}=2×\left(125+80+100\right)\phantom{\rule{0ex}{0ex}}=2×305\phantom{\rule{0ex}{0ex}}=610{cm}^{2}$Therefore, the total surface area of the cuboidal box is greater than the cubical box by $\left(610c{m}^{2}–600c{m}^{2}\right)=10c{m}^{2}$.  Suggest Corrections  39      Similar questions  Explore more