Question

the sum of the areas of 2 squares isv468m^2. if the defference of their perimeters isv24m. find the sides of two squares

Answer 1

#AlexaRousey here!!

Let the side of first square be x nd second be y.

Area of first square = x^2
Area of second square = y^2

Perimeter of first square = 4*x = 4x
Perimeter of second square = 4*y = 4y.

According to question ;

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

& 4x - 4y = 24m

=) 4(x-y) = 24m

=) x -y = 24/4 = 6m

Squaring both sides ;

=) (x-y) ^2 = 6^2

=) x^2 + y^2 - 2xy = 36

=) 468 (by 1st eq) - 2xy = 36

=) 468 - 36 = 2xy

=) 432 = 2xy

=) 432/2 = 216 = xy

Since (x+y) ^2 = x^2 +y^2 + 2xy

=) (x+y) ^2 = 468 + 2(216)

=) (x+y) ^2 = 900

=) (x+y) = 30

Final equations are;

=) x+y = 30
=) x -y = 6

Add both eq ;

=) 2x = 36m

=) x = 36/2 = 18m

Put the value of x in eq1;

=) 18 + y = 30

=) y = 30-18 = 12m.

Hence sides are 18m and 12m.

Thanks!!