Question

sum of the areas of two squares is 850sq.m. if the difference of their perimeter is 40m, find the sides of the two squares.​

Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Sum of the areas of two squares is 850m².

• The difference of their perimeter is 40m.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Sides of 2 squares

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

1st square:-

• Let the side of 1st square be A.

• Then, perimeter of 1st square = 4A

2nd square:-

• Let the side of 2nd square be A'.

• Then, perimeter of 2nd square = 4A'

Now,

• Sum of areas of both the squares = 850m²

Substituting the values,

• A² + A'² = 850 $\longrightarrow \boxed{1}$

Now,

• Difference of the Perimeters of the squares = 40 m.

Substituting the values,

• 4A - 4A' = 40

• 4 (A - A') = 40

• $\sf A \: - \: A' \: = \: \dfrac{40}{4}$

• $\sf A \: - \: A' \: = \: \dfrac{\cancel{40}}{\cancel{4}}$

• A - A' = 10

• A = 10 + A'

Now,

Substituting A = 10 + A' in $\boxed{1}$

• (10 + A')² + A'² = 850

• 10² + 2 * 10 * A' + A'² + A'² = 850

• 100 + 20A' + 2A'² = 850

Dividing the whole equation by 2,

• $\sf \dfrac{100\: +\: 20A'\: + \:2A'^{2} \: = \:850}{2}$

• 50 + 10A' + A'² = 425

• A'² + 10A' = 425 - 50

• A'² + 10A' = 375

• A'² + 10A' - 375 = 0

Taking PSF method,

• P = - 375 * A'² = - 375A'²

• S = + 10A'

• F = + 25A' , - 15A'

Now,

• A'² + 25A' - 15A' - 375 = 0

• A' (A' + 25) - 15 (A' + 25) = 0

• (A' - 15) (A' + 25) = 0

• A' - 15 = 0 , A' + 25 = 0

• A' = 15 , A' = - 25

Since, side of Square cannot be negative.

• A' = 15

Substituting A' = 15 in A - A' = 10,

• A - 15 = 10

• A = 10 + 15

• A = 25

Therefore,

• $\underline{\boxed{\therefore{\blue{\sf Sides\: of \:the\: squares\: = \: 25 \: , \: 15.}}}}$

Answer 2

Question:-

• Sum of the areas of two squares is 850sq.m. if the difference of their perimeter is 40m, find the sides of the two squares.

To Find:-

• Find the sides of the square.

Solution:-

Given ,

• Sum of the areas of two squares is 850sq.m.

• Difference of their perimeter is 40m

According to the Question:-

$\sf\longrightarrow \: { x }^{ 2 } + { y }^{ 2 } = 850$ . . . . ( i )

$\sf\longrightarrow \: 4x - 4y = 40$

$\sf\longrightarrow \: 4( x - y ) = 40$

$\sf\longrightarrow \: x - y = \cancel\dfrac { 40 } { 4 }$

$\sf\longrightarrow \: x - y = 10$

$\sf\longrightarrow \: x = 10 + y$

Substitute x in equation ( i ):-

$\sf\longrightarrow \: { ( 10 + y ) }^{ 2 } + { y }^{ 2 } = 850$

$\sf\longrightarrow \: 100 + { y }^{ 2 } + 20y + { y }^{ 2 } = 850$

$\sf\longrightarrow \: 2{ y }^{ 2 } + 20y - 750 = 0$

$\sf\longrightarrow \: 2( { y }^{ 2 } + 10y - 375) = 0$

$\sf\longrightarrow \: { y }^{ 2 } + 25y - 15y - 375 = 0$

$\sf\longrightarrow \: y( y - 25 ) - 15( y - 25 ) = 0$

$\sf\longrightarrow \: ( y - 25 ) ( y - 15 ) = 0$

$\sf\longrightarrow \: y = 25 \: ; \: 15$

Hence ,

• Sides are 25 and 15