Question

Sum of the areas of two squares is 468m2.
If the difference of their perimeters is 24 m, then
find the sides of the two squares.​

Answer 1

Answer:

18 and 12

Step-by-step explanation:

Given :-

Sum of areas of two squares is 468 m^2 ---1

Difference of their perimeters is 24 m ----2

To Find :-

Sides of two squares.

Solution :-

Let side of 1st square be = x

Let side of 2nd square be = y

According to the question :-

$x^{2} +y^{2} = 468$ -----1

and

( 4x ) - ( 4y ) = 24 -----2

Dividing the equation 2 by 4.

x - y = 6

y = x - 6

Substituting value of y in 1.

$x^{2} + (x-6)^{2} = 468$

Opening bracket.

$x^{2} +x^{2} -12x + 36 = 468$

Equating with zero.

$2x^{2} + 12x -468 +36 = 0$

$2x^{2} - 12x - 432=0$

Dividing the whole equation by 2.

$x^{2} -6x-216=0$

Factorization :-

$x^{2} -18x + 12x-216 =0$

$x(x-18)+12(x-18)=0$

$(x-18)(x+12)=0$

Finding value of x.

x - 18 = 0

x = 18

and

x  + 12 = 0

x = -12

Note :- Side cannot be negative so we will take x = 12.

Finding the sides of the square :-

Side of 1st square = 18 m

Side of 2nd square = 12 m

Answer 2

Let side of one square be = M

Let side of another square be = N

${M}^{2}\: +\: {N}^{2}$ = 468 ______ (eq 1)

We know that ..

Perimeter of square = 4 × side

=> 4M - 4N = 24 

=> M - N = 6

=> M = 6 + N _______ (eq 2)

=> (6 + N)² + N² = 468

=> 36 + 2N² + 12N = 468

=> 2N² + 12N - 432 = 0

=> N² + 6N - 216 = 0

=> N² + 18N - 12N - 216 = 0

=> N(N + 18) - 12(N + 18) = 0

=> (N - 12) (N + 18) = 0

=> N = 12, - 18 (-18 neglected because side can never be negative)

Put value of of N in (eq 2)

=> M = 6 + 12

=> M = 18

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Side of one square = 18 m (M)

Side of another square = 12 m (N)

___________ [ ANSWER ]

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✡ VERIFICATION :

According to question we have equation : M = 6 + N

From above calculations we have M = 18 and N = 12

Put them in above equation

→ 18 = 6 + 12

→ 18 = 18

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