Square exceeds the side of another square by 4 cm and sum of the areas of two squares is 400 square cm find the dimension of the square
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Let the squares be A & B and their sides be x & y respectively.
Now , y exceeds x by 4 cm.
Thus y = (x+4) cm
Now sum of areas of A & B is 400 sq.cm
Here Area of square =( side)^2
; x^2 + y^2 = 400 sq.cm
: x^ 2 + (x+4)^2 = 400
x ^ 2 + x^2 + 8x + 16 = 400 .... using (a+b) ^2
2x^2 + 8x= 400-16
2x [ x +4] = 384
: x^2 +4x = 192...384÷2
x^2+ 4x-192=0
: x^2 + 16x - 12x-192=0
x (x+16)-12 ( x+16)=0
: (x+16)(x-12)=0
(x+16)= 0 ; x = - 16 but. a side can't be negative
(x-12)=0 ; x = 12
y = x+4 i.e 12+4= 16
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