Question

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

Answer 1

Hint: To find the percentage increase in the surface area of the cube we have the formula $\left( \dfrac{S'-S}{S'} \right)100$, where S’ is the new Surface area of the cube and S is the old surface area of the cube.

Complete Step-by-Step Solution:
Let the length of an edge of a cube be ‘a’.
Now we have the Formula of Surface area of the cube as $S=6{{a}^{2}}$
We are given that the edge of each cube increases by 50%, that is the length of an edge of each cube increases by 50%.
If a’ is the new length of the edges of the cube after the increment then

& a'=a+\dfrac{50}{100}a \\\ & \\\ & \Rightarrow a'=\left( a+\dfrac{1}{2}a \right) \\\ & \\\ & \Rightarrow a'=a\left( 1+\dfrac{1}{2} \right) \\\ & \\\ & \Rightarrow a'=a\left( \dfrac{3}{2} \right) \\\ \end{aligned}$$ Let S’ be the new surface area obtained by putting edges as a’ Then $$S'=6{{(a')}^{2}}$$ $$\Rightarrow S'=6{{\left( a\left( \dfrac{3}{2} \right) \right)}^{2}}$$ $$\Rightarrow S'=6{{a}^{2}}\dfrac{9}{4}$$ Now the percentage increase in the area becomes $$\left( \dfrac{S'-S}{S'} \right)100=\left( \dfrac{6{{a}^{2}}\dfrac{9}{4}-6{{a}^{2}}}{6{{a}^{2}}} \right)100$$ $$\Rightarrow \left( \dfrac{S'-S}{S'} \right)100=\dfrac{6{{a}^{2}}}{6{{a}^{2}}}\left( \dfrac{\dfrac{9}{4}-1}{1} \right)100$$ $$\Rightarrow \left( \dfrac{S'-S}{S'} \right)100=\left( \dfrac{9-4}{4} \right)100$$ $$\Rightarrow \left( \dfrac{S'-S}{S'} \right)100=(1.25)(100)=125%$$ Hence, the percentage increase in the surface area of the cube becomes 125%, which is the desired answer. Note: The possibility of error in the question is that you calculate the percentage increase in the surface area of the cube, without changing the length of the edges. It is important to first calculate the new length of an edge and then go for calculating the new Surface area.