Question

Sum of the area of two squares is 656 cm^2. If the difference of their sides is 4cm find the sides of the two squares.

Answer 1

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HERE IS YOUR ANSWER
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Given:
sum of areas of 2 squares = $656 {cm}^{2}$

Let the side of 1st square be x cm.
then,
length of side of other square = (x + 4) cm.

Area of 1st square = ${x}^{2} cm$

Area of 2nd square = ${(x + 4)}^{2} cm \: = \: ( {x}^{2} + 16 \: + 8x)cm.$

»» sum of areas of 2 squares = ${2x}^{2} + 16 + 8x$

=> ${2x}^{2} + 16 + 8x = 656$

$=> {2x}^{2} + 8x - 640 = 0 \\ => \: {x}^{2} + 4x - 320 = 0\\ => {x}^{2} + 20x - 16x - 320 = 0\\ = > x(x + 20) - 16(x + 20) = 0 \\ = > (x + 20)(x - 16) =0$

either,
x = 16 or, x = -20

since length can't be negative, so, side of 1st square = 16 cm

length of 2nd square = 20 cm

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Answer 2

Let A1 & A2 be two squares .

Given : side of square A1= x cm
side of square A2=( x +4 ) cm

Area of square A1 = side × side = x× x= x²
Area of square A2 = side × side = (x+4)× (x+4) =(x+4)²=  x²+8x +16
[(a+b)² = a² +2ab +b²]

Area of square A1 + Area of square A2 = 656 ( GIVEN)
x² + x²+8x +16 = 656
2x² +8x +16 = 656
2x² +8x = 656 -16
2x² +8x = 640
2x² +8x - 640= 0
2(x² +4x -320)= 0
x² +4x -320 = 0
x² + 20x -16x -320= 0
x( x +20) - 16(x +20)= 0

(x- 16)(x+20)= 0
(x- 16) = 0  or  (x+20)= 0
x = 16   or   x  = -20

The length of the side of a square cannot be negative. Therefore x = 16 .

side of square A1= x cm = 16 cm
side of square A2=( x +4 ) cm = 16 +4= 20 cm

Hence, side of square A1= 16 cm &
side of square A2= 20 cm.

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