Question

The sum of the areas of two squares is 468 m^2 . if the difference between their perimeters is 24 m . find the side of the two squares.

Answer 1

Answer:

here it is

Step-by-step explanation:

please note that in using 'P' for perimeter

let the side of square 1 be x metres

P of square 1 =4*side=4x

Difference of perimeters is 24

P of square 1-P of square 2=24

4x-P of square 2=24

4x-24=P of square 2

Now,

P of square 2=4x-24

4×(side of square 2)=24

side of square 2=4x-24/4=4(x-6)/4=x-6

Hence,

side of square 1 is x

and side of square 2 is 6

Also,

sum of area of squares is 468m²

A of square 1+A of square 2=468m²

(side of square 1)² +(side of square 2)²= 468

x²+(x-6)²= 468

x²+x-2×x×6+6²=468

x²+x²-12x+36-468=0

2x²-12x-432=0

Dividing both sides by 2

2x²-12x-432/2=0/2

x²-6x-216=0. (it is a quadratic equation)

D=b²-4ac

D=( -6)² -4(1)(-216)

=36+4×216

=36+864

=900

Using the Quadratic Formula

x=-b+-√b²-4ac/2a

x=-(-6)+-√900/2(1)

x=6+-√900/2

x=6+-√9×100/2

x=6+-√3²×10²/2

x=6+-3×10/2

x=6+-30/2

Solving

x=6+30/2. x=6-30/2

=36/2. = -24/2

=18. = -12

So x=18. &x= -12

Since x is a side of a square, it cannot be negative

So x=18

Side of square 1=x=18

Side of square 2=x-6=12

Hope it helps.