Question

lateral surface area of a cuboid is486m*m. its breadth is12m and length and height are in the ratio 5:3find the length and height of cuboid

Answer 1

Dear Student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Since}\mathrm{the}\mathrm{length}\mathrm{and}\mathrm{height}\mathrm{are}\mathrm{in}\mathrm{the}\mathrm{ratio}5:3. \\ \mathrm{So},\mathrm{suppose}\mathrm{the}\mathrm{length}\mathrm{and}\mathrm{height}\mathrm{of}\mathrm{cuboid}\mathrm{be}5\mathrm{x}\mathrm{and}3\mathrm{x}\mathrm{respectively}. \\ \mathrm{breadth}=12\mathrm{m} \\ \mathrm{Lateral}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cuboid}=2\mathrm{h}\left(\mathrm{l}+\mathrm{b}\right)=2\times 3\mathrm{x}\left(5\mathrm{x}+12\right)=486{\mathrm{m}}^2 \\ \Rightarrow 2\times 3\mathrm{x}\left(5\mathrm{x}+12\right)=486 \\ \Rightarrow 30{\mathrm{x}}^2+72\mathrm{x}=486 \\ \Rightarrow 30{\mathrm{x}}^2+72\mathrm{x}-486=0 \\ \Rightarrow 6\left(5{\mathrm{x}}^2+12\mathrm{x}-81\right)=0 \\ \Rightarrow 5{\mathrm{x}}^2+12\mathrm{x}-81=0 \\ \Rightarrow \mathrm{x}=\frac{-12\pm \sqrt{{12}^2-4\times 12\times \left(-81\right)}}{2\times 5}=\frac{-12\pm 42}{10} \\ \Rightarrow \mathrm{x}=\frac{-12+42}{10}=3\mathrm{and}\mathrm{x}=\frac{-12-42}{10}=-5.4 \\ \mathrm{Since}\mathrm{length}\mathrm{can}\text{'}\mathrm{t}\mathrm{be}\mathrm{negative},\mathrm{so}\mathrm{x}=3 \\ \mathrm{So}\mathrm{length}\mathrm{of}\mathrm{cuboid}=5\times 3=15\mathrm{m} \\ \mathrm{And}\mathrm{height}\mathrm{of}\mathrm{cuboid}=3\times 3=9\mathrm{m} \end{gathered}$$
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