Question

the side of a square exceeds the side of another square by 4cm and the sum of areas of two squares is 400cm square.Find the dimensions of the square.​

Answer 1

Answer:

let the side of 2nd square be=x

then side of 1st square =x+4

sum of area =400

x^2+(x+4)^2=400

x^2+4x-192=0

x^2+16x-12x-192=0

x(x+16)-12(x+16)=0

x=12 and -16. ( but side can't be negative)

hence

side of 2nd square=12cm

and that of 1st= 16cm

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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