Question

The side of a square exceeds the side of square by 4 cm and the sum of the areas of two squares is 400 square

Answer 1

Let side of smaller square be x cm

The side of other square = (x+4) cm

The sum of area of squares x^2+(x+4)^2 = 400

After solving we get

12cm and 16cm sides

Hope it helps you.....

Answer 2

Answer:

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