J17

2. Show that (a – b)(a – b) = a 2 – 2ab + b 2
Consider a square of side a units. The square can be divided into two squares: one of
side (a – b) and another of side b; and two identical rectangles of side b and (a – b). That
is:
(a) square A has sides (a – b),
(b) rectangles B and C have sides b by (a – b),
(c) square D has side b.
Area of the big square = a 2 ………………………(i)
Area of square A = (a – b)(a – b)
= (a – b) 2
Area of rectangle B = (a – b)b
= ab - b 2
Area of rectangle C
= (a – b)b
= ab – b 2
Area of square D = b 2
Total area = (a – b) 2 + ab – b 2 + ab – b 2 + b 2
= (a – b) 2 + 2ab – b 2 ……………….(ii)
Equating (i) and (ii) gives:
(a – b) 2 + 2ab – b 2 = a 2
Therefore, (a – b) 2 = a 2 – 2ab + b 2 .
Note: (a + b) 2 ≠ a 2 + b 2 and (a – b) 2 ≠ a 2 – b 2
3. Show that (a + b)(a – b) = a 2 – b 2
Consider a square of side a units. See figure below.
The square is divided into:
(a) square A of side b,
(b) square D of side (a – b),
(c) rectangles B and C of sides b by a – b.
Area of the big square = a 2 ………………………(i)