Question

the sum of the area of two squares is 640 m^2 square if difference in their perimeters is 64 find the side of two squares

Answer 1

Let the side of the first square be x.

Then the Area of the first square will be x^2 and the perimeter will be 4 * s = 4x.

Let the side of the second square be y.

Then the Area of the second square will be y^2 and the perimeter will be 4 * s = 4y.

Given that Sum of the area of two squares is 640m^2.

x^2 + y^2 = 640   ------ (1)

Given that the differences in their perimeters is 64.

4x - 4y = 64

x - y = 16

x = 16 + y   ----- (2)

Substitute (2) in (1), we get

(16 + y)^2 + y^2 = 640

We know that (a + b)^2 = a^2 + b^2 + 2ab.

Here a = 16, b = y.

16^2 + y^2 + 2 * 16 * y + y^2 = 640

256 + y^2 + 32y + y^2 = 640

256 + 32y + 2y^2 = 640

 2y^2 + 32y = 384

y^2 + 16y = 192

y^2 + 16y - 192 = 0

y^2 + 24y - 8y - 192 = 0

y(y + 24) - 8(y + 24) = 0

(y +24) = 0 (or) (y - 8) = 0

y + 24 = 0  (or) y - 8 = 0

y = -24     y = 8.

The value of y cannot be -ve.Therefore the value is y = 8.

Substitute y = 8 in (2), we get

x = 16 + 8

x = 24.

Therefore the Side of the first square x = 24m.

                  The side of the second square y = 8m.


Hope this helps!

Answer 2

Hello users .....
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Solution:-
let, side of first square = x unit
And
 side of second square = y unit

We know that:
Area of square = a² square unit
And
perimeter of square = 4a unit

Where, a is the side of square.

Here,
Area of first square = x² square unit
And
Area of second square = y² square unit 

And
perimeter of first square = 4x unit
And
perimeter of second square = 4y unit 

Now,
According to condition
(1.) sum of the area of two squares is 640 m²

=> x² + y² = 640 m²  ..........(1.)

And
(2.) difference in their perimeters is 64.

=> 4x - 4y = 64 
=> x - y = 16  ..........(2.) 
=> x = 16 + y

Now,
putting the value of y in equation (1.)
we get
=> x² + y² = 640

=> (16 + y)² + y² = 640

=> 16² + y² + 32 y + y² = 640

=> 2 y² + 32 y = 640 - 256  

=> 2 y² + 32 y = 384

=> 2 (y² + 16 y ) = 2 × 192

=> y² + 16 y - 192 = 0

=> y² + 24 y - 8 y - 192 = 0 

=> y(y + 24) - 8 ( y + 24) = 0 

=> (y + 24 )(y - 8) = 0 

=> y + 24 = 0  or y - 8 = 0 

=> y = -24 or y =  8  

Here, side of square can't be -ve 

=> y  = 8 unit

 now,
putting the value of y in equation (2.) 
We get
x - y = 16 

=> x - 8  = 16 

=> x = 16 + 8 = 24 unit

Hence;
Side of first square = x = 24 unit 
And
Side of second square = y = 8 unit Answer 

Hope it helps :)