Important!help me, f.2 maths(ch.4)20pts!!~

Given 2 squares of areas A1² and A2² (A1²>A2²)

express, in terms of A1 and A2,
a) the difference of the squares,

the difference of squares
= A1²-A2²
= (A1-A2)(A1+A2)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
b) the square of the difference,

the square of differences
= (|A1-A2|)²
= A1² - 2A1A2 + A2²

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
c) the sum of the square of their perimeters.

Perimeter of square A1
= 4A1
Perimeter of square A2
= 4A2
Sum of square of perimeters
= (4A1)² + (4A2)²
= 16A1² + 16A2²
= 16(A1² + A2²)

2006-11-21 20:39:31 補充：
小小補充：part (c) 是要求兩個正方形的周界的二次方的和，即是(A 正方形周界)² + (B 正方形周界)²

a)
(A1)^2-(A2)^2
=(A1+A2)(A1-A2)
b)
By a),
[(A1+A2)(A1-A2)]^2
=(A1+A2)^2(A1-A2)^2
c)
A1*4
OR
A2*4

2006-11-22 17:45:53 補充：
4)Sum of square of perimeters= (4A1)² (4A2)²= 16A1² 16A2²= 16(A1² A2²)

Given:
1) Two squares of area: (A1)^2 & (A2)^2
2) (A1)^2 > (A2)^2 [assumed as the given question missed out their names]

a) the difference of the squares = (A1)^2 - (A2)^2

b) the square of the difference = (A1 - A2)^2

c) the sum of the sqaure of their perimeter= 4(A1) + 4(A2) = 4(A1 + A2)