Question

The sum of the areas of two squares is 640cm ^2and the difference of their perimeter iss 64cm then find the sides,of the two squares

Answer 1

Answer:

Given:

The sum of the areas of two squares is 640 cm² and the difference of their perimeter iss 64 cm.

Find:

Find the sides of the two squares.

Calculations:

Let 'x and y' be the two sides of the square's.

$\tt\rightarrow x^2+y^2=640$

$\tt\rightarrow 4x-4y=64$

$\tt\rightarrow x-y=16$

$\tt\rightarrow (x-y)^2=x^2+y^2-2xy$

$\tt\rightarrow 256=640-2xy$

$\tt\rightarrow xy=192$

$\tt\rightarrow y= \frac{192}{x}$

$\tt\rightarrow x-y=16$

$\tt\rightarrow x- \frac{192}{x} =16$

$\tt\rightarrow x^2-16x-192=0$

$\tt\rightarrow (x=24)or (-8)$

$\tt\rightarrow (x = 24) \: and \: ( y = 8)$

Therefore, the value of x and y = 24, 8 and also 24 and 8 are the two sides of square.

Answer 2

Step-by-step explanation:

Given :

• Sum of areas of two squares is 640 cm²
• Difference of their perimeter is 64 cm.

To find :

• Sides of both squares.

Solution :

Let a be the side of first square and A be the side of another square

[ say first square > second one ]

$\sf{\underline{According\:to\:question-}}$

• a² + A² = 640............(1)
• 4a - 4A = 64..............(2)

From (2),

: $\implies$ a - A = 16 ......(3)

Now, We know that,

$\large{\boxed{\sf{\red{(a\:-\:b)^2\:=\:a^2\:+\:b^2\:-\:2ab}}}}$

: $\implies$ ( a - A )² = a² + A² - 2aA

$\small{\sf{\blue{Putting\:the\:values\:-}}}$

: $\implies$ ( 16 )² = 640 - 2aA

: $\implies$ 2aA = 640 - 256

: $\implies$ A = $\sf{\dfrac{384}{2a}}$

: $\implies$ A = $\sf{\dfrac{192}{a}}$

Now, put the value of a in eq (3).

: $\implies$ ( $\sf{a\:-\dfrac{192}{a} ) \:=\:16}$

: $\implies$ a² - 16a - 192 = 0

By splitting middle term, we get,

: $\implies$ a = 24 and a = -8

Rejecting the negative value, we get,

$\large{\sf{\pink{Side\:of\:first\:square\:=\:24\:cm}}}$

After putting the value of a in (3), we get,

$\large{\purple{\sf{Side\:of\:second\:square\:=\:8\:cm}}}$