Question

An edge of cube is incteased by10%.find the %age by which the surface atra increased.

Answer 1

Initial side of the cube = a initial surface are of the cube=$6a^2$ ......1
New side of the cube=a+10%of a 
                                  = 11a/10new surface area= $6*(11a/10)^2$ ........2
% increase = new surface area-old surface area/ old surface area $*$ 100

2-1/1 $*$ 100
= 21% ( ans)

Answer 2

Initial length of side of the cube = a
Initial Surface area = 6 a²

If the length is increased by 10%;
10% of a = (10/100)*a = a/10 = 0.1a
length of side of the cube = a+0.1a = 1.1a
Final surface area=  6(1.1a)² = 6*1.21a² = 7.26 a²

Percentage increase is

$\frac{(Final\ surface\ area\ -\ initial\ surface\ area)}{initial\ surface\ area} *100 \\ \\ \frac{7.26a^2 -\ 6a^2}{6a^2} *100= \frac{1.26}{6}*100=21\ percent$