Question

the sum of the area of two squares is 468 m^2 and the perimeters are 24 m. tell the sides of the squares?

Answer 1

Given,

Sum of the area of two squares is 468 m^2

Difference between the perimeter of these sqaures is 24 m.

To find ,

length of the sides of these sqaures.

Main solution :-

Let the side of 1st sqaures be equal to x metre.

Let the side of the second sqaure be equal to y metre.

So,

Area of sqaure 1 = x^2

Area of sqaure 2 = y^2

Perimeter of sqaure 1 = 4x

Perimeter of sqaure 2 = 4y

According to question,

x^2 + y^2 = 468 m^2 (i)

4x - 4y = 24m

To find, x and y

4x - 4y = 24

4(x - y) = 24

x -y = 24/4

x-y = 6 (ii)

x = 6 +y (iii)

Putting this in (i)

x^2 + y^2 = 468

(6+y)^2 +y^2 = 468

36 + y^2 + 12 y +y^2 = 468

2y^2 + 12y -432 = 0

2(y^2 +6y -216 ) = 0

y^2 +6y -216 = 0

y^2+18y-12y-216=0

y(y+18)-12(y+18)=0

(y+18)(y-12)=0

Now, y = -18 or y = +12

Since, length can't be negative. We will take only the positive value of y .

Now,

From equation (iii)

x = 6 + y

= 6 + 12

= 18

Answer :- Length of side of one sqaure is 18 m and length of side of other sqaure is 12 m

Answer 2

$\huge \bold \color {red} {Hello Mate}$

$\boxed {GIVEN :}$

sum of area of two squares =>
${468m}^{2}$
difference between perimeter of the squares =>
$24m$

$\boxed{WeHaveToFind}$

length of the side of the squares

$\boxed {Solution}$

let the side of a square = x metres

let the side of the second square = y metre q

$\boxed {Therefore}$

area of square 1 = x^2
area of square 2=y^2
perimeter of square 1 = 4x
perimeter of square 2 = 4y

$\boxed {A. T. Q}$

${x}^{2} + {y}^{2} = {468m}^{2} - - - (1)$
4x-4y=24m

To find x and y
4(x-y) =24

X-y = 24/4

X-y=6---------------------(2)

Put this in 1
X=6+y--------------------(3)

${x}^{2} + {y}^{2} = {468m}^{2}$
${(6 + y)}^{2} + {y}^{2} = 468$
36+y^2+12y+y^2=468
2y^2+12y-468=0
2(y^2+6y-216)=0

Y^2+6y-216=0
Y^2+18y-12y-216=0
(y+18)(y-12)=0

Now, y =-18 or y =12y

since length cannot be negative we take only the positive value