Question

Suppose the length of a cube is increased by $10\%$ and its breadth is decreased by $10\%$. Will the volume of the new cuboid be the same as that of the cube? What about the total surface areas? If they change, what would be the percentage change in both the cases?

Answer 1

Let $a$ be the length of the side of cube.

Then volume of cube $=a^3$

Suppose length of cube increased by $10\%$ i.e. New Length $L=a+\frac{10a}{100}$

And breadth is decreased by $10\%$. i.e. New breadth $B=a-\frac{10a}{100}$

So, the given with side $a$ becomes a cuboid with length $L$, breadth $B$ and height $H=a$

We know that,

Volume of a cuboid $=Length\times Breadth\times Height$

$=L\times B\times H$

$=(a+\frac{10a}{100})(a-\frac{10a}{100})a$

$=(a^2-\frac{100a^2}{(100{)}^2})a$

$=(a^2-\frac{a^2}{100})a$

$=\frac{99a^3}{100}<a^3$

Now, difference between volumes of cube and cuboid $=a^3-\frac{99a^3}{100}=\frac{a^3}{100}$

Thus, the percentage change in volume $=\frac{\frac{a^3}{100}}{a^3}\times 100=1\%$

Now, Total surface area of a cube $=6a^2$

Total surface area of cuboid $=2(LB+BH+LH)$

$=2((a+\frac{10a}{100})(a-\frac{10a}{100})+(a-\frac{10a}{100})a+(a+\frac{10a}{100})a)$

$=2\left((a^2-\frac{a^2}{100}+a^2+a^2)\right)$

$=2\left(\frac{299a^2}{100}\right)$

$=5.98a^2<6a^2$

Thus, the surface area decreases.

And percentage decrease in total surface area $=\frac{{0.02}^2}{6a^2}\times 100=0.33\%$