Question

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

Answer 1

Let the original side of cube be 'a'
Therefore, Original Surface area = 6a²
Now,
Increase in side = 50%

Therefore,

Side = a(50% of a)
= a(50/100 x a)
= a + a/2
= 2a+a/2
= 3a/2

New Surface Area = 6a² = 6 x 3a/2 x 3a/2 = 27a²/2

It's increase Surface Area= New Area-Original area
= 27a²/2-6a²
= 27a²-12a²/2 = 15a²/2

Therefore,
Percentage of Increase in Surface Area= (15a²/2/6a² x 100)%
= (1.25 x 100)%
= 125 %

Answer 2

Let the original side of the cube be x units.

So, the Total Surface Area previously = 6x^2

Now, when side is increased by 50%

New side = x + 50% of x.

$\implies{ \bf{x + \frac{50}{100} \times x}}$

$\implies \sf \: x + \frac{x}{2}$

$\implies \ \frac{2x + x}{2}$

$\implies \: \sf \: \frac{3x}{2}$

Total Surface area of new cube =
$\sf \: { 6 \times (\frac{3x}{2} )}^{2}$

$\implies \sf 6 \times \frac{9 {x}^{2} }{4}$

$\implies \: \sf \: \frac{{54x}^{2}}{4}$

$\implies \: \sf \: \frac{{27x}^{2}}{2}$

Increase in total surface area = New area - older area

$\implies \: \sf \: \frac{27 {x}^{2} }{2} - {6 {x}^{2} }$

$\implies$ 27 x^2 -12x^2 /2

$\implies$ 15x^2/2

Percentage increase =

$\sf \: \frac{increase\:in\:area \: }{older\: area } \times 100$

$\implies \: \sf \: \frac{ \frac{15{x}^{2} }{2} }{{6x}^{2}}} \times 100$

$\implies$ 5/4 ×100

= 125%