Let AD be x cm and AB be y cm
Perimeter of the rectangle ABCD = 2x + 2y = 20 cm
x + y = 10
x = 10 -
y ------------ (1)
Total areas of the 4 squares = 2x^2 + 2y^2 = 80 cm^2
x^2
+ y^2 =
40 -------------
(2)
Substitute x = 10 - y into equation (2) ie
(10
- y)^2 + y^2 = 40
100 - 20y + y^2 + y^2 = 40
2y^2 - 20 y + 60 = 0
y^2 - 10 y + 30 = 0
no real solution for y
Hence, the question is WRONG !
Alternatively, we can use the guess and check method
x y Perimeter of the rectangle ABCD
= 20 cm Areas of 4 squares = 80 cm^2
1 9 1+9+1+9 =
20
2(1)^2 + 2(9)^2 = 163
2 8 2+8+2+8 =
20
2(2)^2 + 2(8)^2 = 136
3 7 3+7+3+7 =
20
2(3)^2 + 2(7)^2 = 116
4 6 4+6+4+6 =
20
2(4)^2 + 2(6)^2 = 104
5 5 5+5+5+5 =
20
2(5)^2 + 2(5)^2 = 100
6 4 6+4+6+4 =
20
2(6)^2 + 2(4)^2 = 104
So, it is impossible for the areas of the 4squares to be 80
cm^2. Hence, the question is Wrong !
