Question

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

Answer 1

$\Large{\underline{\underline{\mathfrak{\green{\bf{Question}}}}}}$

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

$\Large{\underline{\underline{\mathfrak{\green{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• Sum of the areas of two squares is 468 m²
• the difference of their perimeters is 24 m

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• side of both squares

$\Large{\underline{\underline{\mathfrak{\green{\bf{Explanation}}}}}}$

Let,

• Side of first square = x meter.
• Side of second square = y meter.

★ Area of first square = x² meter²

★ Area of second square = y² meter²

A/C to question,

( Sum of the areas of two squares is 468 m2 )

$\small\sf{\:(area\:of\:first\:square)+(area\:of\:second\:square)\:=\:468}$

$\mapsto\sf{\:(x^2+y^2)\:=\:468.......(1)}$

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★ perimeter of first square = 4x meter

★ perimeter of second square = 4y meter

Again, A/C to question,

( the difference of their perimeters is 24 m)

$\small\sf{\:( perimeter\:of\:first\:square )\:+\:( perimeter\:of\:second\:square)\:=\:24}$

$\mapsto\sf{\:(4x-4y)\:=\:24}$

$\mapsto\sf{\:4(x-y)\:=\:24}$

$\mapsto\sf{\:(x-y)\:=\:\dfrac{\cancel{24}}{\cancel{4}}}$

$\mapsto\sf{\:(x-y)\:=\:6......(2)}$

______________________

We Know,

★ (a-b)² = a² + b² - 2ab

So,

$\mapsto\sf{\:(x-y)^2\:=\:x^2+y^2-2xy}$

$\:\:\:\:\:\small\sf{\:( keep\:value\:by\:(1)\:and\:(2) )}$

$\mapsto\sf{\:(6)^2\:=\:468-2xy}$

$\mapsto\sf{\:-2xy\:=\:36-468}$

$\mapsto\sf{\:2xy\:=\:432......(3)}$

______________________

We know,

★$\sf{\:(x+y)\:=\:\sqrt{(x - y)^2+4xy}}$

$\:\:\:\:\:\small\sf{\:( keep\:value\:by\:(2)\:and\:(3) )}$

$\mapsto\sf{\:(x+y)\:=\:\sqrt{(6)^2+2\times (432)}}$

$\mapsto\sf{\:(x+y)\:=\:\sqrt{(36+864)}}$

$\mapsto\sf{\:(x+y)\:=\:\sqrt{(900)}}$

$\mapsto\sf{\:(x+y)\:=\:30........(4)}$

_____________________

addition of equ (2) and (4)

$\mapsto\sf{\:2x\:=\:36}$

$\mapsto\sf{\:x\:=\:\dfrac{\cancel{36}}{\cancel{2}}}$

$\mapsto\sf{\:x\:=\:18}$

______________________

Keep value of x in equ(1) ,

$\mapsto\sf{\:18-y\:=\:6}$

$\mapsto\sf{\:y\:=\:-6+18}$

$\mapsto\sf{\:y\:=\:12}$

______________________

Thus:-

• Side of first square = 18 meter
• Side of second square = 12 meter

______________________

$\Large{\underline{\underline{\mathfrak{\green{\bf{Answer Verification}}}}}}$

( Sum of the areas of two squares is 468 m² )

$\small\sf{\:(area\:of\:first\:square)+(area\:of\:second\:square)\:=\:468}$

$\mapsto\sf{\:(18)^2\:+\:(12)^2\:=\:468}$

$\mapsto\sf{\:324+144\:=\:468}$

$\mapsto\sf{\:468\:=\:468}$

L.H.S.=R.H.S.

______________________

Again,

( the difference of their perimeters is 24 m)

$\small\sf{\:( perimeter\:of\:first\:square )\:+\:( perimeter\:of\:second\:square)\:=\:24}$

$\mapsto\sf{\:4\times 18\:-\:4\times 12\:=\:24}$

$\mapsto\sf{\:72\:-\:48\:=\:24}$

$\mapsto\sf{\:24\:=\:24}$

L.H.S. = R.H.S

_____________________

so, we can say that our solution is absolutely correct

______________________

Answer 2

Given :

Sum of areas of 2 squares = 468m²

The difference of their perimeter = 24cm

To Find:

The sides of the two square

Solution:

Let the sides of two squares be x m and y m respectively .

Case 1 .

→ Sum of the areas of two squares is 468 m² .

A/Q,

∵ x² + y² = 468 . ...........(1) .

[ ∵ area of square = side² . ]

Case 2 .

→ The difference of their perimeters is 24 m .

A/Q,

∵ 4x - 4y = 24 .

[ ∵ Perimeter of square = 4 × side . ]

⇒ 4( x - y ) = 24 .

⇒ x - y = 24/4 .

⇒ x - y = 6 .

∴ y = x - 6 ..........(2) .

From equation (1) and (2) , we get

∵ x² + ( x - 6 )² = 468 .

⇒ x² + x² - 12x + 36 = 468 .

⇒ 2x² - 12x + 36 - 468 = 0 .

⇒ 2x² - 12x - 432 = 0 .

⇒ 2( x² - 6x - 216 ) = 0 .

⇒ x² - 6x - 216 = 0 .

⇒ x² - 18x + 12x - 216 = 0 .

⇒ x( x - 18 ) + 12( x - 18 ) = 0 .

⇒ ( x + 12 ) ( x - 18 ) = 0 .

⇒ x + 12 = 0 and x - 18 = 0 .

⇒ x = - 12m [ rejected ] . and x = 18m .

∴ x = 18 m

Put the value of 'x' in equation (2), we get

∵ y = x - 6 .

⇒ y = 18 - 6 .

∴ y = 12 m .