Question

An edge of a cube is increased by 10%. Find the percentage by which the surface area of the cube has increased.

Answer 1

Let ‘a’ be the edge of the cube.
Surface area of cube = 6a2 sq units
Given length of edge is increased by 10%
Hence new measure of edge = a + 10% of a = (11a/10)
Surface area of cube (after increase) = 6(11a/10)2
                                                     = (121/100) x 6a2 sq units
Change in surface area = [(121/100) x 6a2 ] – 6a2
                                  = (21/100) x 6a2
% increase =  [(21/100) x 6a2] x 100/ [6a2 ] 
                 = 21%

Answer 2

Let the edge of the cube be 'l'

Now,
Total surface area of the cube = 6*l*l

Now,
when the edge of the cube is increased let the length of the edge be 'l 1'.

Now,
l 1 = l + 10% of l
= l + 10/100 l
= 11/10 l

Now.

the new Total surface area= 6*l 1* l 1.
= 6* 11/10 l * 11/10 l
= 6 * l * l * 121/100
= 6*l*l * 121 %

But we know that the 6*l*l = the original surface area

The new surface area = 121% of original surface area.

Now the increase in the surface area
= 121% - 100%
= 21%

Hope this helps.....