Question

The side of a square exceeds the side of the another square by 4 cm and the sum of the area of the two squares is 400 square cm find the dimensions of the square

Answer 1

Answer:

12 cm and 16 cm

Step-by-step explanation:

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

Thus sides or dimensions of square are 12cm and 16 cm respectively

Answer 2

Answer:

Let the sides of Square A be “x” cm,

and the sides of Square B be “x + 4” cm.

Hence, sum of area of two squares A abd B are :

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

x^2 + (x^2 + 8x + 16) = 400

2x^2 + 8x + 16 = 400

x^2 + 4x + 8 = 200

x^2 + 4x - 192 = 0

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

therefore, x = 12, or -16.

Substitute x = 12 into first equation, we have

144 + (16)^2 = 400

144 + 256 = 400 QED.

dimension of Square A is 12cm

dimension of Square B is 16cm