Question

IF EACH EDGE OF A CUBE IS INCREASED BY 50 PERCENT,WHAT WILL BE THE INCREASE PERCENT IN THE SURFACE AREA OF THE CUBE?​

Answer 1

Answer:

Let x be the edge of a cube.

Surface area of the cube having edge x = 6x2 ………..(1)

As given, a new edge after increasing the existing edge by 50%, we get

The new edge = x + 50 x /100

The new edge = = 3x/2

Surface area of the cube having edge 3x/2 = 6 x (3x/2)2= (27/2)x2……..(2)

Subtract equation (1) from (2) to find the increase in the Surface Area:

Increase in the Surface Area = (27/2)x2 – 6x2

Increase in the Surface Area = = (15/2)x2

Now,

Percentage increase in the surface area = ((15/2)x2 / 6x2) x 100

= 15/12 x 100

= 125%

Therefore, the percentage increase in the surface area of a cube is 125.

Answer 2

Answer:

Let the original side be a then

Let the original side be a thenOriginal surface area S$= {6a}^{2}$

When side is increased by 50%, side becomes $\frac{3}{2}$ a then

Surface of area s$= \frac{27}{2} {a}^{2}$

Hence,

Surface area is increases by x $\frac{ \frac{27}{2} {a}^{2} - {6a}^{2} }{ {6a}^{2} } \times 100 = 125\%$

This is the required solution.

Answer 3

Answer:

Let the original side be a then

Let the original side be a thenOriginal surface area S$= {6a}^{2}$

When side is increased by 50%, side becomes $\frac{3}{2}$ a then

Surface of area s$= \frac{27}{2} {a}^{2}$

Hence,

Surface area is increases by x $\frac{ \frac{27}{2} {a}^{2} - {6a}^{2} }{ {6a}^{2} } \times 100 = 125\%$

This is the required solution.