Question

the side of square exceeds the side of another square by 4cm and the sum of the area of the two squares is 400 cm2. find the dimensions of the 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
solve
12cm and 16cm sides.

Hope that it helps. . .

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