If each side of a cube is increased by 10%. what is the % increase in surface area of the cube?
Answers
Answer:
21%
Step-by-step explanation:
Let the original length of side be 10a.
Thus, surface area = 6side²
= 6(10a)² = 600a²
When it is increased by 10%:
New length = original + 10% of original
= 10a + 10% of 10a
= 10a + (10/100 x 10a)
= 10a + a = 11a
Surface area = 6(11a)² = 726a²
Increase in SA=new-orig. = 726a²-600a²
= 126a²
Therefore,
Increase % = increase/original x 100%
= 126a²/600a² x 100%
= 21%
% increase in surface area of the cube is 21%.
Answer:
21%
Step-by-step explanation:
Let the original length of side be x.
Thus, surface area = 6side²
= 6(x)²
= 6x².
When it is increased by 10%:
New length = original + 10% of original
= x+10% of x
= x+(10/100*x)
= x+x/10
=11x/10.
Surface area = 6(11x/10)²
=6*121x²/100
=363x²/50.
Increase in SA=new-original
=363x²/50-6x²
=63x²/50.
Therefore,
Increase % = increase/original x 100%
= 63x²/50/6x²x100%
= 21%.
% increase in surface area of the cube is 21%.