Maths questions???(9)

Assume the the side of right-angled triangle is x and y respectively, tangent of right-angled triangle is z.
(a) percentage decrease in perimeter
=[ (x+y+z) - (1-20%)x+(1-20%)y+(1-20%)z / (x+y+z) ]*100%
=[ (x+y+z) - 80%(x+y+z) / (x+y+z) ]*100%
=20%

(b) Area decrease in perimeter
={[ xy/2 - [(1-20%)x (1-20%)y /2 ] / xy/2 }*100%
={[ xy/2 - 0.64xy/2 ] / xy/2 }*100%
={0.36xy/2 / xy/2 }*100%
=36%

let the original lenghts of the sides be x
the new lenghts of the sides be 0.8x
the new perimeter:
0.8x+0.8x+√0.8^2+0.8^2
=2.73x units (corr. to 2d.p.)
the original perimeter :
x+x+√x^2+x^2
=3.41x units (corr to 2d.p.)
the percentage change:
(2.73x-3.41x)/3.41x *100%
=-20%
thus , the percentage decrease in 20%

B) the new area:
0.8x*0.8x/2
=0.32x^2 square units
the original area:
x*x/2
=0.5x^2 square units
the percentage change:
(0.32x^2-0.5x^2)/0.5x^2 X100%
=-36%
thus, the percentage decrease in36%