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Solids [Surface Area and Volume of 3-D Solids] — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSSolids [Surface Area and Volume of 3-D Solids]4 Marks
Question
The dimensions of a rectangular box are in the ratio $4: 2 : 3$. The difference between the cost of covering it with paper at $Rs. 12 \sim$per$ \sim m^2$ and with paper at the rate of $13.50 \sim$per $\sim m^2$ is $Rs. 1,248$. Find the dimensions of the box.
✓

Answer

Given dimensions of a rectangular box are in the ratio $4: 2 : 3$
Therefore, the total surface area of the box $= 2$
$[4 x \times 2 x+3 x+4 \times 3 x]$
$= 2 ( 8x^2 + 6x^2 + 12x^2 ) m^2$
Difference between cost of covering the box with paper at $Rs. 12$ per $m^2$ and with paper at $Rs. 13.50$ per $m^2 = Rs. 1,248$
$\Rightarrow 52x^2[ 13 .5 - 12 ] = 1248$
$\Rightarrow 52 \times x^2 \times 1.5=1248$
$\Rightarrow 78 \times x^2 = 1248$
$\Rightarrow x^2=\frac{1248}{78}$
$\Rightarrow x^2 = 16$
$\Rightarrow x = 4 \dots...[$ length, width and height cannot be negative $]$
Thus, the domensions of the rectungular box are$ : 4 \times 4\ m, 2 \times 4\ m, 3 \times 4\ m$
Thus, the dimensions are $16\ m, 8\ m$ and $12\ m.$

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