F.3 MATHS

Let the original surface area be A, radius be R.
The radius of the new one be r. As the new surface area is 0.64A, so
A : 0.64A = R^2 : r^2
==> 1 : 0.64 = R^2 : r^2
==> 1 : 0.8 = R : r
==> r = 0.8R
The percentage change in radius is :
(r - R)/R * 100%
= -20%

Therefore, the radius of the sphere is decreased by 20%.

let x be the original radius,
then, the original surface area:
(x^2)*pie*4

So the new surface area:
[(x^2)*pie*4]*0.64
=(x^2)*pie*2.56

and the new radius:
{[(x^2)*pie*2.56]/4*pie}開方
={[(x^2)*16]/25}開方
=4x/5

so the percentage change is:
[x-(4x/5)]/x < (抽common factor)
={x[1-(4/5)]}/x <(simplify two x)
=0.2
so the percentage decrease in radius is 20%

喺張紙到寫一次出黎會容易睇D :D

percentage change in area = [ (new - old)/old ] x 100%
= [ ((4πr^2)(1-0.36) - 4πr^2 )/4πr^2 ] x 100%
= [ ((4πr^2)(0.64) - 4πr^2 )/4πr^2 ] x 100%
= [ (2.56πr^2 - 4πr^2 )/4πr^2 ] x 100%
= [ -1.44πr^2/4πr^2 ] x 100%
= -36%
new area/old area = (new radius / old radius)^2
let c be the new radius

4 π r^2 (1-36%) = (c/r) ^2
c/r = 0.8
c=0.8r

therefore there are 20% decrease in radius.