Question

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

Answer 1

Let the side of the small square be x

Then the side of the bigger square will be x+4

The sum of the areas of both square is equal to 400 cm square

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

x^2 + x^2 + 4^2 + 2 × x × 4 = 400

2x^2 + 16 + 8x = 400

Now divide the whole equation by 2

Then you will get x^2 + 8 + 4x = 200

x^2 + 4x = 192

Now by completing square method :

x^2 + 2 × 2x = 192

x^2 +2 × 2x + 2^2 = 192 + 2^2

(x+2)^2 = 192 + 4

(x+2)^2 = 196

(x+2) = under root of 196

(x+2) = + 14

+ve -ve

x = +14 -2 x = -14 -2

x = 12 x = -16 {skip}

So, the length of the small square's side is 12 cm while the length of the bigger square's side is 12 + 4 = 16 cm

I hope this will help.....

Mark as brainliest !!!!!!!!

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